Mathematical relationships between volume, pressure, and temperature of a given mass of gas are referred to as Gas laws. In this article. we shall study Charle’s Law.

Charle’s Law:

The relationship between the volume of a gas and temperature was observed by Jacques Charles in 1787.

Statement:

At constant pressure the volume of a given mass of a gas increases or decreases by 1/273 of its volume at 0^oC for every degree rise or fall in temperature.

Explanation:

Let V_o be the volume of a gas at 0 °C, Let this gas be heated through t °C, Let V_t be the volume of the gas at t °C. then,

Thus V ∝ T

Alternate Statement of Charle’s Law:

Thus at constant pressure, the volume of the certain mass of enclosed gas is directly proportional to the absolute temperature of the gas.

In general

This relation is called the mathematical statement of Charle’s law.

Graphical Representation:

A graph is drawn by taking the absolute temperature on the x-axis and volume on the y-axis. The graph is as follows. This graph is also known as a V-T  diagram.

Each line of the graph represents different constant pressure. The lines are called isobar lines.

Charle’s Law and Kinetic Theory of Gases:

According to the kinetic theory of gases, gas consists of a large number of minute particles which are always in constant random motion. During this process, they collide with each other and with the walls of the container containing it. When molecules collide with the walls of the container, there is a change in the momentum of the colliding molecules. This is the cause of the pressure of the gas. According to the kinetic theory of gases, the kinetic energy of molecules is directly proportional to the absolute temperature of the gas.

Thus when gas is heated, the kinetic energy of molecules increases. Due to which the velocity of molecules increases, which results in more collision of the molecules with the walls of the container. But pressure is kept constant, hence the volume of the gas increases proportionally. Hence at constant pressure, the volume of a given mass of a gas is directly proportional to temperature.

Significance of Charle’s Law:

• Charle’s law is significant because it explains how gases behaviour at constant pressure and the relation between the absolute temperature and the volume of the gas. According to Charle’s law, at constant pressure, the volume and absolute temperature of a gas are directly proportional to each other.
• At constant pressure, the density of a gas is inversely proportional to its pressure.
• Using this concept hot air is used to fill the ballons used for meteorological purposes.

Numerical Problems:

Example – 01:

At 300 K a certain mass of a gas occupies  1 x 10^-4 dm³ by volume. Calculate the volume of the gas at 450 K at the same pressure.

Given: Initial temperature = T₁ = 300 K, Initial volume = V₁ = 1 x 10^-4 dm³ , Final temperature = T₂ = 450 K

To Find: Final volume = V₂ = ?

Solution:

By Charle’s Law

∴  V₂ =(T₂/T₁) x V₁

∴  V₂ =(450/300) x 1 x 10^-4 dm³ = (1.5) x 1 x 10^-4 dm³ = 1.5 x 10^-4 dm³

Ans: The volume of the gas at 450 K is 1.5 x 10^-4 dm³

Example – 02:

The volume of a given mass of a gas at 0 °c is 2 dm³. Calculate the new volume of the gas at the constant pressure where (i) the temperature is increased by 10 °C (ii) the temperature is decreased by 10 °C.

Solution:

Part – I: the temperature is increased by 10 °C

Given: Initial temperature = T₁ = 0 °C= 0 + 273.15 = 273.15 K, Initial volume = V₁ = 2 dm³ , Final temperature = T₂ = 0 °C + 10 °C = 10 °C = 10 + 273.15 = 283.15 K

To Find: Final volume = V₂ = ?

By Charle’s Law

∴  V₂ =(T₂/T₁) x V₁

∴  V₂ =(283.15/273.15) x 2 dm³ = 2.07 dm³

Part – II: the temperature is decreased by 10 °C

Given: Initial temperature = T₁ = 0 °C= 0 + 273.15 = 273.15 K, Initial volume = V₁ = 2 dm³ , Final temperature = T₂ = 0 °C – 10 °C = – 10 °C = – 10 + 273.15 = 263.15 K

To Find: Final volume = V₂ = ?

By Charle’s Law

∴  V₂ =(T₂/T₁) x V₁

∴  V₂ =(263.15/273.15) x 2 dm³ = 1.93 dm³

Ans: When the temperature is increased by 10 °C, the new volume of gas is 2.07 dm³. When the temperature is decreased by 10 °C, the new volume of gas is 1.93 dm³

Example – 03:

A certain mass of a gas occupies a volume of 0.2 dm³ at 273 K. Calculate the volume of the gas if the absolute temperature is doubled at the same pressure.

Given: Initial temperature = T₁ = 273 K, Initial volume = V₁ = 0.2 dm³ , Final temperature = T₂ = 2 x 273 = 546 K

To Find: Final volume = V₂ = ?

Solution:

By Charle’s Law

∴  V₂ =(T₂/T₁) x V₁

∴  V₂ =(546/273) x 0.2 dm³ = 0.4 dm³

Ans: When the temperature is doubled, the new volume of gas is 0.4 dm³

Example – 04:

When a ship is sailing in pacific ocean where the temperature is 23.4 °C, a balloon filled with 2.0 L of air. What will be the volume of the balloon when the ship reaches the Indian ocean, where the temperature is 26.1 °C.

Given: Initial temperature = T₁ = 23.4 °C = 23.4 + 273 = 296.4 K, Initial volume = V₁ =2.0 L, Final temperature = T₂ = 26.1 °C = 26.1 + 273 = 299.1 K

To Find: Final volume = V₂ = ?

Solution:

By Charle’s Law

∴  V₂ =(T₂/T₁) x V₁

∴  V₂ =(299.1/296.4) x 2.0 L =  2.018 L

Ans: The volume of the balloon in the Indian ocean is 2.018 L.

Example – 05:

A sample of a gas occupies 10 dm³ at 127 °C and 1 bar pressure. The gas is cooled to – 73 °C at the same pressure. What will be the volume of the gas?

Given: Initial temperature = T₁ = 127 °C = 127 + 273 = 400 K, Initial volume = V₁ =10 dm³, Final temperature = T₂ = – 73 °C = – 73 + 273 = 200 K

To Find: Final volume = V₂ = ?

Solution:

By Charle’s Law

∴  V₂ =(T₂/T₁) x V₁

∴  V₂ =(200/400) x 10 dm³ =  5 dm³

Ans: The new volume of the gas is 5 dm³

Example – 06:

A sample of gas is found to occupy a volume of 900 cm³ at 27 °C. Calculate the temperature at which it will occupy a volume of 300 cm³, provided the pressure is kept constant.

Given: Initial temperature = T₁ = 27 °C = 27 + 273 = 300 K, Initial volume = V₁ = 900 cm³ , Final volume = V₂ =300 cm³

To Find: Final temperature = T₂ = ?

Solution:

By Charle’s Law

∴  T₂ =(V₂/V₁) x T₁

∴  T₂ =(300 cm³/900 cm³) x 300 K  = 100 K

∴  T₂ = 100 – 273 = -173 °C

Ans: At temperature  -173 °C the volume is 300 cm³

Example – 07:

A gas occupies 100.0 mL at 50  °C and 1 atm pressure. The gas is cooled at constant pressure so that its volume is reduced to 50 mL. What is the final temperature?

Given: Initial temperature = T₁ = 50 °C = 50 + 273 = 323 K, Initial volume = V₁ = 100.0 mL , Final volume = V₂ = 50.0 mL

To Find: Final temperature = T₂ = ?

Solution:

By Charle’s Law

∴  T₂ =(V₂/V₁) x T₁

∴  T₂ =(50.0 mL/100.0 mL) x 323 K  = 161.5 K

∴  T₂ = 150 – 273 = -111.5 °C

Ans: At temperature  – 115.5 °C the volume is 50 mL.

Example – 08:

A gas occupies 100.0 mL at 50  °C and 1 atm pressure. The gas is cooled at constant pressure so that its volume is reduced to 50 mL. What is the final temperature?

Given: Initial temperature = T₁ = 50 °C = 50 + 273 = 323 K, Initial volume = V₁ = 100.0 mL , Final volume = V₂ = 50.0 mL

To Find: Final temperature = T₂ = ?

Solution:

By Charle’s Law

∴  T₂ =(V₂/V₁) x T₁

∴  T₂ =(50.0 mL/100.0 mL) x 323 K  = 161.5 K

∴  T₂ = 150 – 273 = -111.5 °C

Ans: At temperature  – 115.5 °C the volume is 50 mL.

Example – 09:

A vessel of capacity 400 cm³ contains hydrogen gas at 1 atm pressure and 7°C. In order to expel 28.57 cm³ of the gas the same pressure to what temperature the vessel should be heated?

Given: Initial temperature = T₁ = 7 °C = 7 + 273 = 280 K, Initial volume = V₁ = 400 cm³, Final volume = V₂ = 400 cm³ + 28.57 cm³ = 428.57cm³.

To Find: Final temperature = T₂ = ?

Solution:

By Charle’s Law

∴  T₂ =(V₂/V₁) x T₁

∴  T₂ =(428.57 cm³/400.0 cm³) x 280 K  = 300 K

∴  T₂ = 300 – 273 = 27 °C

Ans: At temperature  27 °C, 28.57 cm³ of the gas expels.

Example – 10:

1 L of air weighs 1.293 g at 0 °C and 1 atm pressure. At what temperature 1 L of air at 1 atm pressure will weigh 1 g.

Given: Initial temperature = T₁ = 0 °C = 0 + 273 = 273 K, Initial density = ρ₁ = 1.293 g/L, Final density = ρ₂ = 1 g/L.

To Find: Final temperature = T₂ = ?

Solution:

By Charle’s Law

∴  T₂ =(V₂/V₁) x T₁

∴  T₂ =((m/ρ₂)/(m/ρ₁)) x T₁

∴  T₂ =(ρ₁/ρ₂) x T₁

∴  T₂ =(1.293/1) x 273 = 353 K

∴  T₂ = 353 – 273 = 27 °C

Ans: At temperature  80 °C, 1 L of air weighs 1 g

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Mathematical relationships between volume, pressure, and temperature of a given mass of gas are referred to as Gas laws. In this article. we shall study Boyle’s Law.

Boyle’s Law:

In 1662, Robert Boyle, on the basis of his experiments put forward the law. It is the relation between the volume and the pressure of enclosed gas at constant temperature.

Statement:

At constant temperature, the volume of a certain mass of enclosed gas varies inversely with its pressure.

Explanation:

Let P be the pressure and V be the volume of a certain mass of enclosed gas, then at the constant temperature

P   ∝  1/V

Thus PV = Constant.

Thus, for a given amount of the gas, the product of pressure and volume is constant at constant temperature.

If V₁ and V₂ are the volumes of a gas at pressures P₁ and P₂ respectively at a constant temperature.

P₁V₁ = P₂V₂

This relation is called the mathematical statement of Boyle’s Law.

Graphical Representation:

Graph of Pressure (P) Versus Volume (V):

A graph is drawn by taking volume on the x-axis and pressure on the y-axis. The graph is as follows. This graph is also known as PV diagram.

Each curve is rectangular hyperbola and corresponds to a different constant temperature and is known as an isotherm (constant temperature plot). Higher curves correspond to higher temperatures. It should be noted that volume of the gas doubles if pressure is halved.

Graph of Pressure (P) Versus Reciprocal of Volume (1 / V):

The straight lines obtained in the graph confirms that P ∝ 1/V.

Graph of Product of Pressure and Volume (PV) Versus Pressure (P):

The graph parallel to x-axis confirms that at a particular temperature of the gas, the product of its volume and corresponding pressure is always constant.

Graphs in terms of Logarithmic Variations:

By Boyle’s law, we have PV = k = constant

Taking log of both sides

Log P + Log V = log K

∴  LogP = – LogV + log K

∴  LogP = log (1/V) + log k

Relation Between Density and Pressure:

Thus the density of the certain mass of an enclosed gas is directly proportional to its pressure.

Boyle’s Law and Kinetic Theory of Gases:

According to the kinetic theory of gases, gas consists of a large number of minute particles which are always in constant random motion. During this process, they collide with each other and with the walls of the container containing it. When molecules collide with the walls of the container, there is a change in the momentum of the colliding molecules. This is the cause of the pressure of the gas. According to the kinetic theory of gases, the kinetic energy of molecules is directly proportional to the absolute temperature of the gas. In Boyle’s law temperature is constant. Hence the kinetic energy of the molecules is a constant.

For enclosed gas, the number of molecules of a gas is constant. When the volume of a gas is reduced, the molecules are forced more closer together. Thus the density of gas increased and they collide more frequently. Hence at less volume, there are more collisions and hence more pressure. When the volume of a gas is increased, the molecules are away from each other and they collide less frequently. Hence at larger volumes, there are fewer collisions and hence less pressure. So at a constant temperature, the volume and pressure of a gas are inversely proportional.

Significance of Boyle’s Law:

• Boyle’s law is significant because it explains how gases behaviour at constant temperature and the relation between the pressure and the volume of the gas. According to Boyle’s law, at a constant temperature, the pressure and volume of a gas are inversely proportional to each other.
• At constant temperature, the density of a gas is directly proportional to its pressure.
• Atmospheric pressure is low at high altitudes, so air is less dense. Hence, a lesser quantity of oxygen is available for breathing. This is the reason why mountaineers have to carry oxygen cylinders with them.

Numerical Problems:

Example – 01:

5 dm³ volume of a gas exerts a pressure of 2.02 × 10⁵ kPa. This gas is completely pumped into another tank where it exerts a pressure of 1.01 × 10⁵ kPa at the same temperature, calculate the volume of the tank

Given: Initial volume = V₁ = 5 dm³, Initial pressure = P₁ = 2.02 × 10⁵ kPa, Final pressure P₂ = 1.01 ₁ kPa, Temperature constant

To Find: Final volume = V₂ =?

Solution:

At constant temperature by Boyle’s law

P₁V₁ = P₂V₂

∴ V₂ = P₁V₁ /P₂

∴  V₂ = 2.02 × 10⁵ kPa × 5 dm³ / 1.01 × 10⁵ kPa

∴ V₂ = 10 dm³

Hence the volume of the tank is 10 dm³.

Example – 02:

Given the mass of a gas occupies a volume of 2.5 dm³ at NTP. Calculate the change in volume of gas at same temperature if the pressure of the gas is changed to 1.04 × 10⁵ Nm^-2.

Given: Initial volume = V₁ = 2.5 dm3, Initial pressure = P₁ = 1.013 × 10⁵ Nm^-2 (Normal pressure), Final pressure P₂ = 1.04 × 10⁵ Nm^-2, Temperature constant

To Find: Final volume = V₂ =?

Solution:

At constant temperature by Boyle’s law

P₁V₁ = P₂V₂

∴ V₂ = P₁V₁ /P₂

∴  V₂ = 1.013 × 10⁵ Nm^-2 × 2.5 dm³ / 1.04 × 10⁵ Nm^-2

∴   V₂ = 2.435 dm³

∴  Change in volume = V₁ – V₂ = 2.5 dm³ – 2.435 dm³ = 0.065 dm³

Hence the change in volume of the gas is 0.065 dm³.

Example – 03:

A balloon is inflated with helium gas at room temperature of 25 ^oC and at 1 bar pressure when its initial volume is 2.27 L and allowed to rise in the air. As it rises the external pressure decreases and volume of the gas increases till finally, it bursts when external pressure is 0.3 bar. What is the limit to which volume of the balloon can be inflated?

Given: Initial volume = V₁ = 2.27 L, Initial pressure = P₁ = 1 bar, Final pressure P₂ = 0.3 bar, Temperature constant

To Find: Final volume = V₂ =?

Solution:

At constant temperature by Boyle’s law

P₁V₁ = P₂V₂

∴ V₂ = P₁V₁ /P₂

∴  V₂ = 1 bar × 2.27 L/ 0.3 bar

∴ V₂ = 7.567 L

Thus balloon can be inflated to the maximum volume of 7.567 L.

Example – 04:

The volume of given mass of a gas is 0.6 dm³ at a pressure of 101.325 kPa. Calculate the volume of the gas if its pressure is ballooned to 142.860 kPa at the same temperature.

Given: Initial volume = V₁ = 0.6 dm³, Initial pressure = P₁ = 101.325 kPa, Final pressure P₂ = 142.860 kPa, Temperature constant

To Find: Final volume = V₂ =?

Solution:

At constant temperature by Boyle’s law

P₁V₁ = P₂V₂

∴ V₂ = P₁V₁ /P₂

∴  V₂ = 101.325 kPa × 0.6 dm³ / 142.860 kPa

∴ V₂ = 0.426 dm³

Thus final volume of the gas is 0.426 dm³.

Example – 05:

In a J tube partially filled with mercury, the volume of the air column is 4.2 mL and the mercury level in the two limbs is the same. Some mercury is now added to the tube so that the volume of air enclosed in the shorter limb is now 2.8 mL. What is the difference in the level of mercury in this case? Atmospheric pressure is 1 bar.

Given: Initial volume = V₁ = 4.2 mL, Initial pressure = P₁ = 1 bar, Final volume V₂ = 2.8 mL Temperature constant

To Find: Difference in mercury levels =?

Solution:

At constant temperature by Boyle’s law

P₁V₁ = P₂V₂

∴ P₂ = P₁V₁ /V₂

∴  P₂ = 1 bar × 4.2 mL / 2.8 mL

∴ P₂ = 1.5 bar

Difference in pressure = P2 – P1 = 1.5 bar – 1 bar = 0.5 bar

Now 1 bar corresponds to 750.12 mm of mercury

Hence 0.5 bar corresponds to 750.12 x 0.5 =375.06 mm of mercury

Hence the difference in mercury levels is 375.06 mm or 37.51 cm

Example – 06:

A thin glass bulb of 100 mL capacity is evacuated and kept in 2.0 L container at 27 °C and 800 mm pressure. If the bulb implodes isothermally, calculate the new pressure in the container in kPa.

Given: Initial Volume = 2000 mL – 100 mL = 1900 mL

To Find: Final pressure =?

Solution:

At constant temperature by Boyle’s law

P₁V₁ = P₂V₂

∴ P₂ = P₁V₁ /V₂

∴  P₂ = 800 mm × 1900 L / 2000 mL

∴  P₂ = 76o mm = 760 mm/760 mm = 1 atm = 101.325 kPa

Hence the new pressure is 101.325 kPa

Example – 07:

Certain bulb A containing gas at 1.5 bar pressure was put into communication with an evacuated vessel of 1.0 dm³ capacity through the stop cock. The final pressure of the system dropped to 920 mbar, at the same temperature. What is the volume of container A.

Given: Initial pressure P1 = 1.5 bar, Final Pressure = 920 mbar = 0.920 bar. Let V dm³ be the volume of the container A. Initial Volume = V1 = V dm³, Final Volume = (1.0 + V) dm³

To Find: Volume of container A = V =?

Solution:

At constant temperature by Boyle’s law

P₁V₁ = P₂V₂

∴ 1.5 bar × V dm³ = 0.920 bar × (100 – V) dm³

∴ 1.5 V = 92  – 0.920V

∴ 1.5 V + 0.920V = 92

∴ 2.42 V = 92

∴ V = 92 / 2.42 = 39.02

Hence the volume of container A is 38.02 dm³.

Example – 08:

What will be the minimum pressure required to compress 500 dm³ of air at 1 bar to 200 dm³ at 30 °C

Given: Initial volume = V₁ = 500 dm³, Initial pressure = P₁ = 1 bar, Final volume V₂ = 200 dm³ Temperature constant

To Find: Final Pressure = P₂ =?

At constant temperature by Boyle’s law

P₁V₁ = P₂V₂

∴ P₂ = P₁V₁ /V₂

∴ P₂ = 1 bar × 500 dm³ / 200 dm³

∴ P₂ = 2.5 bar

Hence minimum pressure required = 2.5 bar

Example – 09:

A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What would be the pressure?

Given: Initial volume = V₁ = 120 mL, Initial pressure = P₁ = 1.2 bar, Final volume V₂ = 180 mL, Temperature constant

To Find: Final Pressure = P₂ =?

At constant temperature by Boyle’s law

P₁V₁ = P₂V₂

∴ P₂ = P₁V₁ /V₂

∴ P₂ = 1.2 bar × 120 dm³ / 180 dm³

∴ P₂ = 0.8 bar

Hence minimum pressure required = 0.8 bar

Science > Chemistry > States of Matter > Boyle’s Law

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In this article, we shall study to calculate the change in internal energy and change in enthalpy in a chemical reaction.

Example – 01:

For a particular reaction, the system absorbs 6 kJ of heat and does 1.5 kJ of work on its surroundings. What are the change in internal energy and enthalpy change of the system?

Given: q = + 6 kJ (Heat absorbed), W = -1.5 kJ (Work done on the surroundings).

To Find: Change in internal energy = ΔU =? Enthalpy change = ΔH =?

Solution:

By the first law of thermodynamics

Δ U = q + W

Δ U = 6 k J – 1.5 kJ = 4.5 kJ

ΔH = q_p = Heat supplied at constant pressure = + 6 kJ

Ans: The change in internal energy is 4.5 kJ and enthalpy change is 6 kJ

Example – 02:

An ideal gas expands from a volume of 6 dm³ to 16 dm³ against constant external pressure of 2.026 x 10⁵ Nm^-2. Find Enthalpy change if ΔU is 418 J.

Given:  Initial volume = V₁ = 6 dm³ = 6 × 10^-3 m³, Final volume = V₂ = 16 dm³ = 16 × 10^-3 m³, P_ext = 2.026 x 10⁵ Nm^-2, ΔU = 418 J.

To Find: Enthalpy change = ΔH =?

Solution:

Work done in the process is given by W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 2.026 x 10⁵ Nm^-2 × (16 × 10^-3 m³ – 6 × 10^-3 m³)

∴ W = – 2.026 x 10⁵ × (10 × 10^-3) = – 2026 J

By the first law of thermodynamics

Change in internal energy

Δ U = q_p   + W

∴   418 J = q_p – 2026 J

∴   q_p =   418 J +  2026 J = 2444 J

ΔH = q_p = Heat supplied at constant pressure = 2444 kJ

Ans: enthalpy change is 2444 J

Example – 03:

A sample of gas absorbs 4000 kJ of heat. a) If volume remains constant, what is ΔU? b) Suppose that in addition to absorption of heat by the sample, the surrounding does 2000 kJ of work on the sample. What is ΔU? c) Suppose that as the original sample absorbs heat, it expands against atmospheric pressure and does 600 kJ of work on its surroundings. What is ΔU?

Solution:

Part – a

Given: q = + 4000 kJ (Heat absorbed by sample), ΔV = 0 (Volume remains the same)

To Find: ΔU =?

Work done in the process is given by   W = – P_ext × ΔV = – P_ext × 0 = 0

By the first law of thermodynamics

Δ U = q   + W

∴ Δ U = + 4000 kJ   + 0 kJ = + 4000 kJ

Part – b

Given: q = + 4000 kJ (Heat absorbed by sample), W = + 2000 kJ (Work done by surroundings)

To Find: ΔU =?

By the first law of thermodynamics

Δ U = q + W

∴ Δ U = + 4000 kJ   + 2000 kJ = + 6000 kJ

Part – c

Given: q = + 4000 kJ (Heat absorbed by sample), W = – 600 kJ (Work done on the surroundings)

To Find: ΔU =?

By the first law of thermodynamics

Δ U = q   + W

∴ Δ U = + 4000 kJ   –   600 kJ = + 3400 kJ

Example – 04:

Calculate the work done in the following reaction when 2 moles of HCl are used at constant pressure and 423 K. State whether work is on the system or by the system.

4 HCl_(g)  + O_2(g)  →  2 Cl_2(g)   +  2 H₂O_(g)

Given: R = 8.314 J K^-1 mol^-1, T = 423 K

To Find: Work done = W =?

Solution:

The reaction is 4 HCl_(g)  + O_2(g)  →  2 Cl_2(g)   +  2 H₂O_(g)

Given 2 moles of HCl are used, hence dividing equation by 2 to get 2 HCl, we get

2 HCl_(g)  + ½O_2(g)  →   Cl_2(g)   +  H₂O_(g)

Δn = n_{product (g)}   – n_{reactant (g)} = (1 + 1) – (2 + ½) = 2 – 5/2 = – ½

Work done in chemical reaction is given by

∴ W = – Δn RT = – (-½) mol × 8.314 J K^-1 mol^-1 × 423 K = 1758 J

∴ W = + 1758 J

Positive sign indicates that work is done by the surroundings on the system

Ans: Work done by the surroundings on the system in the reaction is 1758 J

Example – 05:

Calculate the work done in the following reaction when 1 mol of SO₂ is oxidised at constant pressure at 5o °C. State whether work is on the system or by the system.

2SO_2(g) + O_2(g)  →  2 SO_3(g)

Given:  Temperature = T = 5o °C = 50 + 273 = 323 K, R = 8.314 J K^-1 mol^-1,

To Find: Work done = W =?

Solution:

The reaction is 2SO_2(g) + O_2(g)  →  2 SO_3(g)

Given 1 mole of SO₂ is used, hence dividing equation by 2 to get 1 mol of SO₂

SO_2(g) + ½O_2(g)  → SO_3(g)

Δn = n_{product (g)}   – n_{reactant (g)} = (1 ) -(1 + ½) = 1 – 3/2 = – ½

Work done in chemical reaction is given by

∴ W = – Δn RT = – (-½) mol × 8.314 J K^-1 mol^-1 × 323 K = 1343 J

∴ W = + 1343 J

Positive sign indicates that work is done by the surroundings on the system

Ans: Work done by the surroundings on the system in the reaction is 1343 J

Example – 06:

Calculate the work done in the following reaction when 2 mol of NH₄NO₃ decomposes at constant pressure at 10o °C. State whether work is on the system or by the system.

NH₄NO_3(s) → N₂O_(g)  +  2 H₂O_(g)

Given:  Temperature = T = 10o °C = 100 + 273 = 373 K, R = 8.314 J K^-1 mol^-1 ,

To Find: Work done = W =?

Solution:

The reaction is NH₄NO_3(s) → N₂O_(g)  +  2 H₂O_(g)

Given 2 mol of NH₄NO₃ decomposes, hence multiplying equation by 2

2 NH₄NO_3(s) → 2 N₂O_(g)  +  4 H₂O_(g)

Δn = n_{product (g)}   – n_{reactant (g)} = (2 + 4) -(0) =6

Work done in chemical reaction is given by

∴ W = – Δn RT = – (6) mol × 8.314 J K^-1 mol^-1 × 373 K = – 18607 J

∴ W = – 18.61 kJ

Negative sign indicates that work is done by the system on the surroundings

Ans: Work done by the surroundings on the system in the reaction is – 18.61 kJ.

Example – 07:

CO reacts with O2 according to the following reaction. How much pressure volume work is done and what is the value of ΔU for the reaction of 7.0 g of CO at 1 atm pressure, if the volume change is – 2.8 L.

2CO_(g)  +  O_2(g) → 2CO_2(g)      Enthalpy change = Δ H = – 566 kJ

Given: ΔV = – 2.8 L

To Find: Work done = W =? ΔU = ?

Solution:

Work done in the process is given by

W =  – P_ext × ΔV  = – 1 atm  ×( -2.8) L = 2.8 L atm =  2.8 L atm  × 101.3 J L^-1atm^-1 = 283.6 J

∴ W = 0.2836 kJ

Positive sign indicates the work is done on the system.

From given reaction 2 x (12 + 16) = 56 g of CO on oxidation liberates  566 kJ energy

Hence heat liberated on oxidation of 7.0 g of CO = (7.0/56) × 566 = 70.75 KJ

Hence ΔH = – 70.75 kJ (negative sign as heat is liberated)

By the first law of thermodynamics

Δ U = q   + W

∴ Δ U = 0.2836 kJ – 70.75 kJ =  -70.47 kJ

Ans: The work done on the system is 0.2836 kJ and Δ U = -70.47 kJ

Example – 08:

The enthalpy change for the following reaction is – 620 J, when 100 mL of ethylene and 100 mL of H₂ react at 1 atm pressure. Calculate pressure-volume type work and ΔU.

C₂H_4(g)  +  H_2(g) → C₂H_6(g)

Given: ΔH = -620 J

To Find: Work done = W =? ΔU = ?

Solution:

The reaction is C₂H_4(g)  +  H_2(g) → C₂H_6(g)

Thus 1 vol of C₂H_4(g) reacts with 1 vol of H_2(g) to give 1 vol of C₂H_6(g)

Thus 100 mL of C₂H_4(g) reacts with 100 mL of H_2(g) to give 100 mL of C₂H_6(g)

Change in volume during the reaction = ΔV = V_product – V_reactant = (100) – (100 + 100) = – 100 mL = – 0.1 L

Work done in the process is given by

W = – P_ext × ΔV = – 1 atm  ×( -0.1) L = 0.1 L atm =  0.1 L atm  × 101.3 J L^-1atm^-1 = 10.13 J

∴ W = + 10.13 J

Positive sign indicates the work is done on the system.

By the first law of thermodynamics

Δ U = q   + W

∴ Δ U =   + 10.13 J – 620 J =  – 609.9 J

Ans: the work done on system is 10.13 J and Δ U = – 609.9 J

Previous Topic: The Concept of Enthalpy of a System

Next Topic: Concept of Enthalpy of a Reaction, Thermochemistry

Science > Chemistry > Chemical Thermodynamics and Energetics > Change in Internal Energy and Enthalpy

The post Problems on Internal Energy Change and Enthalpy Change appeared first on The Fact Factor.

In this article, we shall study change in enthalpy for different chemical processes.

Enthalpy of Formation (Δ_fH° or Δ_formationH°):

The change in enthalpy of a chemical reaction at a given temperature and pressure, when one mole of the substance is formed from its constituent elements in their standard states is called the heat of formation.

Explanation: Consider the following reaction

C_(s) + O_2(g)  → CO_2(g)   ,  Δ_formationH°  =  -395.39 kJ mol^-1

Since one mole of carbon dioxide gas is formed we can say that the heat of formation of carbon dioxide gas is -395.39 kJ.

Notes:

• The standard state of the element is that stable state of the element in which it exists at 1 atm. Pressure and 298 K
• Enthalpies of elements in their standard states are arbitrarily taken as zero.
• Enthalpy of a compound is equal to its heat of formation.
• When solving problems on the heat of formation make sure that the product side of the thermochemical equation has one mole of the substance whose heat of formation is to be calculated.

Enthalpy of Dissociation (Δ_dissociationH°):

Change in enthalpy of a chemical reaction at a given temperature and pressure, when one mole of a substance is dissociated into its constituent elements is called the heat of dissociation.

Explanation: Consider the following reaction

H_2(g) →  2H_(g),    Δ_dissociationH° = + 435.136 kJ mol^-1

Since one mole of hydrogen gas is dissociated the heat of dissociation of hydrogen gas is + 435.136 kJ

Enthalpy of Combustion (Δ_cH° or Δ_combustionH°):

Change in the enthalpy of a chemical reaction at a given temperature and pressure when one mole of a substance is combusted (burn) completely in excess of oxygen is called the heat of combustion.

Explanation: Consider the following reaction

C_(s) + O_2(g)  → CO_2(g)   , ΔH  =  -395.39 kJ mol^-1

Since one mole of carbon is combusted completely the heat of combustion of carbon is – 395.39 kJ.

Enthalpy of Neutralization (Δ_{neutralization}H°):

Change in the enthalpy of a chemical reaction at a given temperature and pressure when one gram equivalent weight of acid is completely neutralized by one gram equivalent weight of the base is called the heat of neutralization.

Explanation: Consider following reaction

HCl_(aq) + NaOH_(aq)  →   NaCl  +    H₂O     Δ_{neutralization}H°  =  -56.9 kJ

The heat of neutralization of HCI by NaOH is -56.9 KJ.

For all strong acids and bases, the heat of neutralization is the same because, in their neutralization reaction, there is a combination of H+ ions of an acid with OH- ions of the base to produce un-dissociated water.

Change of Phase:

Change of phase involves the change in the physical state of matter. During the phase change, the chemical properties of the substance do not change but physical properties change. The following are the types of phase changes.

Fusion: This is the process in which the matter changes from solid-state to liquid state. It is endothermic process.

e.g. melting of ice H₂O_(s) → H₂O_(l)

Vapourization: This is the process in which the matter changes from liquid state to gaseous state. It is an endothermic process.

e.g. boiling of water H₂O_(l) → H₂O_(g)

Sublimation: This is the process in which the matter changes from the solid-state into a gaseous state directly. It is an endothermic process.

e.g. heating of camphor Camphor_(s) → Camphor_(g)

Characteristics of Change of Phase:

• The phase change always takes place at constant pressure and temperature.
• During the phase transition, there is an equilibrium between the two phases. Thus both the phases exist simultaneously.
• The Change in temperature takes place only when completion of phase transition.

Enthalpy of Fusion (Δ_fusH°):

The enthalpy-change that accompanies the fusion of one mole of a solid without the change in temperature at constant pressure is called its enthalpy of fusion.

Explanation: Consider the following reaction

H₂O_(s) → H₂O_(l),               Δ_fusionH  = +6.01  kJ mol^-1

Thus, the equation indicates that when one mole of ice melts at 0° C at 1 atm pressure, the enthalpy-change is 6.01 kJ. i.e. 6.01 kJ of energy is absorbed.

Enthalpy of Freezing (Δ_freezeH°):

The enthalpy change that accompanies the freezing of one mole of a liquid without a change in temperature at constant pressure is called its enthalpy of freezing.

Explanation: Consider the following reaction

H₂O_(l) → H₂O_(s),               Δ_freezeH  = +6.01  kJ mol^-1

Thus, the equation indicates that when one mole of water freezes at 0° C at 1 atm pressure, the change in enthalpy is -6.01 kJ. i.e. 6.01 kJ of energy is released.

Enthalpy of Vaporization (Δ_vaporizationH°):

The enthalpy change that accompanies the vaporization of one mole of a liquid without a change in temperature at constant pressure is called its enthalpy of vaporization.

Explanation: Consider the following reaction

H₂O_(l) → H₂O_(g),   Δ_{vapourization}H  = + 40.7  kJ mol^-1

Thus, the equation indicates that when one mole of water vaporizes at 100° C at 1 atm pressure, the change in enthalpy is + 40.7 kJ. i.e. 40.7 kJ of energy is absorbed.

Enthalpy of Condensation (Δ_condensationH°):

The enthalpy change that accompanies the condensation of one mole of a liquid without a change in temperature at constant pressure is called its enthalpy of condensation.

Explanation: Consider the following reaction

H₂O_(l) → H₂O_(g),   Δ_condensationH  = – 40.7  kJ mol^-1

Thus, the equation indicates that when one mole of water vapours condense at 100° C at 1 atm pressure, the enthalpy-change is – 40.7 kJ. i.e. 40.7 kJ of energy is released.

Enthalpy of Sublimation (Δ_sublimationH°):

The direct conversion of solid to vapour without going through the liquid state is called sublimation. The enthalpy-change that accompanies the condensation of one mole of a solid directly into vapours at a constant temperature at constant pressure is called its enthalpy of sublimation.

Explanation: Consider the following reaction

H₂O_(s) → H₂O_(g),       Δ_sublimationH  = +51.08  kJ mol^-1

Thus, the equation indicates that when one mole of ice sublimes at O° C at 1 atm pressure, the enthalpy change is + 51.08 kJ. i.e. 51.8 kJ of energy is absorbed.

It is to be noted that Δ_sublimationH = Δ_fusionH + Δ_{vapourization}H

Atomic or Molecular Changes:

Enthalpy of Ionization (Δ_ionizationH°):

The enthalpy change that accompanies the removal of an electron from each atom or ion in one mole of gaseous atoms or ions is called its enthalpy of ionization.

Explanation: Consider the following reaction

Na_(g)   →   Na⁺_(g)+ e^– ,    Δ_ionizationH  = 494  kJ  mol^-1

Thus, the equation indicates that when one mole of gaseous sodium atom ionizes to Na⁺_(g) ion, the change in enthalpy is + 494 kJ. i.e. 494 kJ of energy is absorbed.

Enthalpy of Atomization (Δ_atomizationH°):

The enthalpy change that accompanies the dissociation of all the molecules in one mole of gas-phase substance into gaseous atoms is called its enthalpy of atomization.

Explanation: Consider the following reaction

Cl_2(g) → Cl_(g)+ Cl_(g),          Δ_atomizationH  = 242  kJ mol^-1

Thus, the equation indicates that when one mole of gaseous chlorine molecule dissociates completely into its atomic form in the gaseous state then the change in enthalpy is + 242 kJ. i.e. 242 kJ of energy is absorbed.

Enthalpy of Solution (Δ_solutionH°):

Change in enthalpy of chemical reaction at a given temperature and pressure, when one mole of a solution is dissolved in a specified quantity of solvent so as to form a solution of particular concentration is called as enthalpy of a solution.

Explanation: Consider the following reaction

KOH_(s)   +   H₂O_(l) → KOH_(aq)        Δ_solutionH   = -58.57 KJ mol^-1

Thus, the equation indicates that when one mole of potassium hydroxide (solute) dissolves in one mole of water (solvent) to form one mole of potassium hydroxide solution in water then the change in enthalpy is -58.57 kJ. i.e. 58.57 kJ of energy is released

Bond Enthalpy (Bond Energy):

Chemical reactions involve the breaking and making of chemical bonds. Energy is required to break a bond and energy is released when a bond is formed. It is possible to relate the heat of reaction to changes in energy associated with the breaking and making of chemical bonds. With reference to the enthalpy changes associated with chemical bonds, two different terms are used in thermodynamics. (i) Bond dissociation enthalpy (ii) Mean bond enthalpy. Let us discuss these terms with reference to diatomic and polyatomic molecules.

Diatomic Molecules:

Consider the following process in which the bonds in one mole of dihydrogen gas (H2) are broken:

H_2(g) →  2H_(g) ;   ΔH–HHO = 435.0 kJ mol^-1

The enthalpy change involved in this process is the bond dissociation enthalpy of H–H bond.

The bond dissociation enthalpy is the change in enthalpy when one mole of covalent bonds of a gaseous covalent compound is broken to form products in the gas phase. Note that it is the same as the enthalpy of atomization of dihydrogen. This is true for all diatomic molecules.

Polyatomic Molecule:

In the case of polyatomic molecules, bond dissociation enthalpy is different for different bonds within the same molecule. Let us consider a polyatomic molecule like methane, CH4.  The overall thermochemical equation for its atomization reaction is given below:

CH_4(g) →  C_(g) + 4H_(g) , ΔH = +1665 KJ.

In methane, all the four C – H bonds are identical in bond length and energy. However, the energies required to break the individual C – H bonds in each successive step differ. In such cases, we use mean bond enthalpy

CH_4(g) →  CH_3(g) + H_(g) , ΔH = +427 KJ.

CH_3(g) → CH_2(g) + H_(g), ΔH = +439 KJ

CH_2(g) → CH_(g) + H_(g), ΔH = +452 KJ

CH_(g) → C (g) + H_(g),   ΔH = +347 KJ

Adding above reactions we get

CH_4(g) →  C_(g) + 4H_(g) , ΔH = +1665 KJ

We find that mean C–H bond enthalpy in methane as 1664/4 = 416 kJ/mol. Using Hess’s law, bond enthalpies can be calculated.

Enthalpy of Reaction from Bond Enthalpy:

The reaction enthalpies are very important quantities as these arise from the changes that accompany the breaking of old bonds and the formation of the new bonds. We can predict the enthalpy of a reaction in the gas phase if we know different bond enthalpies. The standard enthalpy of reaction is related to bond enthalpies of the reactants and products in gas phase reactions as:

ΔH = ∑ Bond enthalpies _reactants  –     ∑ Bond enthalpies _products

Remember that this relationship is approximate and is valid when all substances (reactants and products) in the reaction are in a gaseous state.

The values of given bond enthalpy can be used to calculate bond enthalpies of specific bonds in the molecule.

Crystal Lattice Energy (Δ_LatticeH°):

Crystal lattice energy is defined as the enthalpy change or energy released accompanying formation of 1 mole of crystalline solid from its constituent ions in the gaseous state at a constant temperature.

Explanation:

M⁺_(g)  + X^–_(g) → M⁺X^–_{(s) ,}  ΔH = – x KJ  mol^-1

Thus, the equation indicates that when one mole of ionic compound M+X-(s) is formed from its constituent ions the change in enthalpy is – x kJ i.e. x kJ of energy is evolved. Crystal lattice energy is always negative.

The sequence of actions involved in the formation of 1 mole of an ionic compound in its standard state from its constituent elements in their states at constant temperature and pressure is called Born-Haber cycle.

Factors Affecting Crystal Lattice Energy:

Crystal lattice energy depends on the interionic distance in the crystalline solid. As the distance decreases, the crystal lattice energy increases. Crystal lattice energy depends on the charge of constituent cations and anions.

Enthalpy of Hydration (Δ_HydrationH°):

It is defined as the heat evolved when one mole of gaseous ions dissolve in water by hydration to give infinitely dilute solution at constant temperature and pressure.

Explanation: Consider the following reaction

Na⁺_(g) + aq → Na⁺_(ag) ,     ΔH = – 390 KJ mol^-1

Thus, the equation indicates that when one mole of sodium ion in the gaseous state is dissolved in water the change in enthalpy is – 390 kJ. i.e. 390 kJ of energy is released.

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Science > Chemistry > Chemical Thermodynamics and Energetics > Heat of Reaction Of Different Processes

The post Enthalpy Of Different Processes appeared first on The Fact Factor.

The branch of chemistry which deals with the quantitative study of thermal or heat changes in various chemical reactions is known as thermochemistry. In this article, We shall discuss very important concept of chemistry i.e. heat of reaction.

Thermochemical Equation:

An equation which indicates the heat changes in a chemical reaction at a certain temperature and pressure with an indication of states of reactants and products is called a thermochemical equation

Example:

C_(s)  + O_2(g)  → CO_2(g)   , ΔH  =  -395.39 kJ

This thermochemical equation indicates that when one mole of solid carbon reacts with one mole of gaseous oxygen at constant pressure one mole of gaseous carbon dioxide is obtained. During this reaction, 395.39 kJ of heat is evolved.

Guidelines for Writing Thermochemical Equation:

• The equation must be automatically balanced like a chemical equation.
• The value of ΔH must be indicated. ΔH is negative for exothermic reaction and positive for an endothermic reaction.
• The physical state of each reactant and each product must be indicated. Symbols such as (s) for solid-state, (l) for the liquid state, (g) for gaseous state and (aq.) for an aqueous solution should be indicated.
• The thermochemical equation can be reversed. During the reversal of the sign of ΔH must be changed.
• The temperature of the reaction can be written as a suffix to ΔH

Example:

C_(s)  + O_2(g)  → CO_2(g)   , ΔH  =  -395.39 kJ

This thermochemical equation indicates that when one mole of solid carbon reacts with one mole of gaseous oxygen at constant pressure one mole of gaseous carbon dioxide is obtained. During this reaction, 395.39 kJ of heat is evolved.

The necessity of Mentioning of State of a Substance:

It is important to mention the physical state of substances in the thermochemical equation because the change of physical state is also accompanied by the enthalpy change.

Example: Consider the following thermochemical equations

H_2(g) +    1/2O_2(g)  →    H₂O_(l) , ΔH =  – 286 kJ

Thus, when 1 mole of hydrogen gas reacts with half a mole of oxygen gas to form one mole of liquid water, 286 kJ of heat is produced.

H_2(g) +    1/2O_2(g)  →    H₂O_(g) , ΔH =  – 249 kJ

Thus, when 1 mole of hydrogen gas reacts with half a mole of oxygen gas to form one mole of water vapours, 249 kJ of heat is produced. Hence, whenever there is a state change, there is a change in enthalpy of reaction. Hence in thermochemical reaction, the state of each substance involved should be mentioned.

Heat of Reaction OR Enthalpy of Chemical Reaction:

The difference between the sum of enthalpies of products and the sum of enthalpies of reactants at a given temperature is called as the heat of reaction. It is denoted by ΔH or ΔU

Explanation:

Consider a general chemical reaction 

A    +    B →  C     +   D.

Let H_A , H_B, H_C, and H_D be the enthalpies of A, B, C and D respectively then the heat of reaction is given by

ΔH  =   (H_C +  H_D)  –  (H_A +  H_B)

The heat of reaction can be determined either at constant pressure or at constant volume.

Heat of Reaction at Constant Pressure:

The difference between the sum of enthalpies of products and the sum of enthalpies of reactants at a given temperature and constant pressure is called the heat of reaction at the constant pressure at a given temperature.

It is denoted by ΔH. Normally heat of reaction at constant pressure is specified at 298 K and 1 atm. Pressure. This heat of reaction is called the standard heat of reaction.

Thus at constant pressure heat of reaction is given by

ΔH = ∑ ΔH_Products  –     ∑ ΔH_Reactants

ΔH is negative for exothermic reaction and positive for an endothermic reaction.

Heat of Reaction at Constant Volume:

The difference between the sum of internal energies of products and the sum of internal energies of reactants at a given temperature and constant volume is called the heat of reaction at the constant volume at a given temperature. It is denoted by ΔE.

Thus at constant pressure heat of reaction is given by

ΔU = ∑ ΔU_Products  –     ∑ ΔU_Reactants

Factors Affecting Heat of Reaction:

• Physical states of substances involved.
• The amount of substance involved.
• Way of carrying out the reaction i.e. in a case of gaseous reactions heat of reaction depends on whether the reaction is carried out at constant pressure or at constant volume.
• The pressure of reactants and products.
• The temperature (as explained by Kirchhoff’s equations)

Different Types of Chemical Reactions on the Basis of Change in the Enthalpy:

On the basis of change in the enthalpy chemical reactions are classified into exothermic reactions and endothermic reactions

Exothermic Reactions:

The chemical reactions which involve the evolution of heat are called as exothermic reactions.

Example:

C_(s)+ O_2(g)    →  CO_2(g) , ΔH = -395.39 kJ

In this case, the enthalpy of products is less than the enthalpy of reactants. For such reactions, the change in enthalpy is always negative.

Characteristics of Exothermic Reactions:

• The chemical reactions which involve the evolution of heat are called exothermic reactions.
• For such reactions, the change in enthalpy is always negative.
• In this case, the enthalpy of products is less than the enthalpy of reactants
• Products are more stable than the reactants.

Endothermic Reactions:

The chemical reactions which involve absorption of heat are called endothermic reactions.

Example :

2C_(s)+ H_2(g)    →  C₂H_2(g) , ΔH = + 225.94kJ

In this case, the enthalpy of products is more than the enthalpy of reactants. For such reactions, the change in enthalpy is always positive.

Characteristics of Endothermic Reactions:

• The chemical reactions which involve absorption of heat are called endothermic reactions.
• For such reactions, the change in enthalpy is always positive.
• In this case, the enthalpy of products is more than the enthalpy of reactants.
• Products are less stable than the reactants.

Previous Topic: Numerical Problems on Enthalpy and Internal Energy Changes

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Science > Chemistry > Chemical Thermodynamics and Energetics > Heat of Reaction

The post Heat of Reaction appeared first on The Fact Factor.

The enthalpy of system is defined as the sum of the internal energy of the system and energy that arises due to pressure and volume. It is denoted by letter H. Mathematically,

H = U + PV

Where, H = Enthalpy U = Internal Energy P = Pressure V = Volume

As the internal energy E, Pressure P and Volume V are state functions, the enthalpy of the system is a state function.

Change in Enthalpy of System:

Let H₁ and H₂ be the enthalpies of the system in the initial state and final state respectively. Let P₁, V_1, and U1 be the pressure, volume and internal energy of the system in the initial state. Let P2, V2, and U₂ be the pressure, volume and internal energy of the system in the final state.

The change in enthalpy is given by ΔH = H₁ – H₂         ……………….    (1)      

By definition of enthalpy   H = U + PV

For initial state,       H1 = U₁ + P ₁V₁

For final state,     H₂ = U₂ + P₂ V₂

Substituting these values in equation (1)

Δ H =   (U₂ + P₂ V₂) – (U₁ + P₁V₁)   

∴ Δ H =   (U₂ – U₁) + (P₂ V₂ – P₁ V₁)

At constant pressure P₁    =   P₂ =   P 

∴ Δ H =   (U2 – U₁) + (P V₂ – P V₁) 

∴ Δ H =    (U₂ – U₁) + P (V₂ – V₁)

∴ Δ H   = ΔU + PΔ V

This is mathematical expression for change in enthalpy of system.

Thus the change in enthalpy of the system is equal to the sum of the increase in the internal energy of the system and the mechanical work due to expansion.

If Δ H is positive then the reaction is endothermic and when ΔH is negative the reaction is exothermic.

Change in Enthalpy of System at Constant Pressure (Isobaric Process):

The expression for the change in enthalpy of a system at constant pressure is

Δ H = Δ U   + PΔ V     …………  (1)

Where, ΔH = Change in enthalpy ΔU = Change in internal energy P = Pressure ΔV = Change in volume

The mathematical statement of the first law of thermodynamics is  

ΔU = q + PΔ V 

Where,   q = Heat supplied to the system ΔU = Change in internal energy P = Pressure ΔV = Change in volume

Let us denote q = q_p = heat absorbed by the system at constant pressure.

q_p = ΔU + PΔV    ……….   (2)  

From equations (1) and (2) we have

ΔH = q_p

Thus at constant pressure, the change in enthalpy is equal to heat absorbed at constant pressure.

Change in enthalpy of System at Constant Volume (Isochoric Process):

The expression for the change in enthalpy of a system is

Δ H = Δ U + PΔ V

Where, ΔH = Change in enthalpy ΔU = Change in internal energy P = Pressure ΔV = Change in volume

In isochoric process ΔV = 0

∴   Δ H = Δ U +  P(0)

Δ H = Δ U    ………….. (1)

The mathematical statement of the first law of thermodynamics is

Δ U = q + PΔV ………….. (2)

Where,   q = Heat supplied to the system ΔU = Change in internal energy P = Pressure ΔV = Change in volume

Substituting ΔV = 0 in equation (2)

q = ΔU

let us denote q = q_v = heat absorbed by system at constant volume.

q_v = ΔE   …………..  (3)

From equations (1) and (3) we have

ΔH = ΔU = q_v

Thus at constant volume, the change in enthalpy is equal to the heat absorbed at constant volume and also equal to change in internal energy. Thus in the isochoric process heat supplied to the system is used for increasing the internal energy of the system.

The Relation between ΔH and ΔU:

Change in the enthalpy of a system is given by

Δ H = ΔU   + PΔV ……(1)

Where, ΔU = Change in internal energy P = Pressure of the system ΔV = Change in volume

But for reactions involving gases

PV = nRT

Where P = Pressure of a gas V = Volume of the gas n = Number of moles of the gas

R = Universal gas constant T = Absolute temperature of the gas

For initial state,    P V₁     = n₁RT

For Final state,    P V₂       = n₂RT

PV₂ – PV₁   =    n₂RT – n₁RT

∴ P( V₂ –  V₁)  =   (n₂ –  n₁ )RT

∴   PΔV = ΔnRT ….. (2)

Substituting in equation (1)

Δ H = ΔU   +  ΔnRT

WhereΔH =   Change in enthalpy or heat of reaction at constant pressure.

ΔU =   Change in internal energy or heat

Δn = Difference in the number of moles of gaseous products and reactants.

R   = Universal gas constant 8.314 J/ mol / k,      T   =   Absolute temperature.

Work Done in a Chemical Reaction:

The work done by a system at constant pressure and temperature is given by

W =   – P_extΔV

Assuming P_ext = P

W = – P( V₂  –  V₁)

∴ W =   – PV₂ + PV₁

But for reactions involving gases    PV = nRT

Where, P = Pressure of a gas V = Volume of the gas n = Number of moles of the gas

R = Universal gas constant T = Absolute temperature of the gas

For initial state,    P V₁ = n₁RT

For Final state,    P V₂       = n₂RT

W = – PV₂ + PV₁ =    – n₂RT + n₁RT

∴ W =   – (n₂ –  n₁ )RT

∴   W = – ΔnRT

This is an expression for work done in chemical reaction.

Conditions Under Which Change in Enthalpy of System (ΔH) is Equal to Change in Internal Energy of the System (ΔU):

When the reaction is carried out in a closed vessel, there is no change in volume. ΔV = 0, hence Δ H = ΔU + PΔV gives Δ H = ΔU.

When the reaction involves only solids and liquids, then change in volume is negligible. ΔV = 0, hence Δ H = ΔU + PΔV gives Δ H = ΔU.

In a chemical reaction in which the number of gaseous reactants consumed is equal to the number of moles of gaseous product formed then

Δn = 0, hence ΔH =ΔU + ΔnRT gives ΔH = ΔU.

Notes: 

• Work has significance only when the number of gaseous reactants consumed and the number of moles of gaseous products formed is different and there is a change in volume.
• Heat Supplied q is not thermodynamic function but heat supplied at constant pressure (qp) and heat supplied at constant volume (qv) are thermodynamic functions:
• By the first law of thermodynamics, ΔU = q + W.  In this relation ΔU is a state function as well as thermodynamic function. Work is path function as well as non-thermodynamic function. For algebraic addition q on the right-hand side of the equation, heat supplied is path function as well as non-thermodynamic function.
• For a constant pressure process, ΔH = qp Where, ΔH is the enthalpy of a system which is state function as well as thermodynamic function. Hence heat supplied at constant pressure qp is a thermodynamic function.
• For a constant volume process, ΔU = q_p Where, ΔU is the internal energy of a system which is state function as well as thermodynamic function. Hence heat supplied at constant volume q_v is a thermodynamic function.

Previous Topic: First Law of Thermodynamics

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Science > Chemistry > Chemical Thermodynamics and Energetics > Enthalpy of a System

The post Enthalpy of a System appeared first on The Fact Factor.

In this article, we shall study the first law of thermodynamics and its application to different chemical processes.

Zeroth Law of Thermodynamics:

Statement:

If two bodies (say A and B) are in thermal equilibrium of the third body (say C) then body A and B will also be in thermal equilibrium with each other

Significance of the Law:

The common use of the thermometer in comparing the temperature of any two or more systems is based on the zeroth law of thermodynamics.

First Law of Thermodynamics:

Statement:

Different forms of the first law of thermodynamics are as follows

• Energy can’t be created nor destroyed but it can be converted from one form into the other (or forms) or into work.
• When a quantity of energy of one kind disappears, then an equivalent amount of energy of another kind makes its appearance.
• It is impossible to make a perpetual motion machine which would produce work without consuming energy.
• The total amount of energy of an isolated system remains constant, it may change from one form to another.
• The energy of the universe remains constant.
• For a system in contact with the surroundings, the sum of energies of the system and its surroundings remains constant however differently it may be shared between the two.

Mathematical Statement of the First Law:

Consider the system in state I (initial state) with internal energy U₁. It is converted into state II (final state) with internal energy U₂ by supplying ‘q’ amount of heat to it. During this process, some work “W” is done by the system on the surroundings. Heat absorbed by the system is used for a) increasing internal energy of a system and b) to do some mechanical work “W”. 

Now, Change in the internal energy of the system is equal to the heat supplied plus Work done.

Thus, Final internal energy = U₂ = U₁ +   q +

∴   (U₂ – U₁)   =      q + W

∴   ΔU = q   + W

In the pressure-volume type of work W = PΔV

∴   ΔU = q +  PΔV

This is the mathematical equation of the first law of thermodynamics.

Sign Conventions for q, W, and ΔU:

• When heat is absorbed by the system q is positive.
• When heat is rejected or given out by the system q is negative.
• When the work is done on the system by surroundings (Work of compression) then W is positive.
• When the work is done by the system on the surroundings (Work of expansion) then, W is negative.
• When there is an increase in the internal energy of the system (increase in temperature) ΔU is positive.
• When there is a decrease in the internal energy of the system (decrease in temperature) ΔU is negative.

Application of First Law of Thermodynamics to Different Chemical Processes:

Isothermal Process:

Internal energy is a function of temperature. As the temperature is constant, the internal energy is also constant. Hence there is no change in internal energy.     ΔU = 0

By the first law of thermodynamics,

ΔU = q   + W

∴ 0 = q   + W

∴ q = -W or W = – q

Thus in the isothermal process heat absorbed is entirely used for doing work on the surroundings or the work done by the surrounding at constant pressure results in the release of the heat (energy) by the system.

Adiabatic Process:

In an adiabatic process, there is no exchange of heat      q = 0

By the first law of thermodynamics

ΔU = q   + W

ΔU = 0   + W

∴   W = q

Thus the increase in internal energy of a system is due to work done by the surroundings on the system or work done by the system on the surroundings is due to the expense of internal energy of the system.

Isochoric Process:

In isochoric process there is no change in volume   ΔV = 0, Thus the work done W = P Δ V = 0

By the first law of thermodynamics,

ΔU = q   + W

∴ ΔU = q   + 0

∴ ΔU = q

Thus the increase in internal energy of a system is due to the absorption of the heat from the surroundings or the decrease in internal energy of a system is due to the release of the heat from the system to the surroundings.

Isobaric Process:

In isobaric process, there is no change in pressure   ΔP = 0 Thus work done W = – P_ext ΔV

By the first law of thermodynamics,

ΔU = q   + W

∴ ΔU = q   – P_ext ΔV

Thus the increase in internal energy of a system is due to the absorption of the heat from the surroundings or the decrease in internal energy of a system is due to release of the heat from the system to the surroundings.

Note:

Most of the chemical reactions take place at constant pressure.

Second Law of Thermodynamics:

We observe that heat flows from the hot end of a metal rod flows to its cold end. The opposite flow of heat is not taking place. The first law (law of conservation of energy) allows heat flow from cold end to hot end. It is possible when heat lost by the cold end is equal to the heat gained by hot end. Thus energy is conserved. But such heat transfer is not possible. Thus the first law of thermodynamics is insufficient to put a restriction on the direction of the heat flow.  To overcome the deficiency of the first law, the second law of thermodynamics is proposed using human experience.

Statement:

Heat cannot be completely converted into an equivalent amount of work without producing permanent changes either in the system or its surroundings. or The spontaneous flow of heat is always unidirectional, from higher temperature to lower temperature.

Previous Topic: Concept of Maximum Work

Next Topic: Enthalpy of a System

Science > Chemistry > Chemical Thermodynamics and Energetics > First Law of Thermodynamics

The post Laws of Thermodynamics appeared first on The Fact Factor.

According to the first law of thermodynamics, ΔU = q  +  W    In an isothermal process, ΔU  =   0, ∴   q   =   – W

Therefore, all the heat absorbed by the system is utilized to do work. (i.e. maximum utilization of the energy takes place. )

The work of expansion is given by the product of external pressure and the volume change. W = – P_ext ΔV

In any expansion, the external pressure must be less than the pressure of the gas. If the external pressure is zero, the work done is also zero as the gas expands into the vacuum. If the external pressure is increased gradually, more and more work will be done by the gas during expansion. If the external pressure becomes equal to the pressure of the gas, there will be no change in the volume and thus ΔV = 0. The work done is also zero. If P_ext is more than the pressure of the gas cannot expand. Therefore, when Pext becomes P then ΔV will be maximum. In the reversible process, P_ext is always less than the pressure of the gas, by an infinitesimally small quantity.

W =  (P –  dp) dV

In the equation W tends to the maximum as (P – dp) tends to P or dp tends to zero. Therefore work done in an isothermal reversible expansion of an ideal gas is maximum work.

Conditions for Maximum Work:

All the changes taking place in a system during the process are reversible. All the changes taking place in a system during the process should take place in infinitesimally small infinite steps. During the change, the driving and opposing forces should defer by an infinitesimally small amount. The system remains in mechanical equilibrium with the surroundings.

Expression for Maximum Work:

Work done in an isothermal reversible expansion is maximum work.

Consider  ‘n ‘ moles of an ideal gas enclosed in a cylinder fitted with a weightless, frictionless, airtight movable piston. Let the pressure of the gas be P which is equal to external atmospheric pressure P. Let the external pressure be reduced by an infinitesimally small amount dp and the corresponding small increase in volume be dV. So the small work done in the expansion process

dW  =  – P_ext . dV

∴ dW =  – (P  –  dp). dV

∴ dW =   – (P.dV  –  dp.dV)  …..  (1)

Since both dp and dV are very small, the product dp.dV is very small and can be neglected in comparison with P.dV. Then the above equation becomes

dW  =   – P.dV                ……….  (2)

When the expansion of the gas is carried out reversibly then there will be series of such p.dV terms. The total maximum work W_max can be obtained by integrating above equation between the limits V₁ to  V₂. Where V₁ is initial volume and V₂ is final volume.

For an isothermal expansion, Boyle’s law is applicable.
Hence P₁V₁ = P₂V₂ i.e. V₂ / V₁ = P₁ / P₂

Where P₁  and P₂  are initial and final pressure respectively.

Notes:

• As the ratio of volumes or pressure is used in the above equation when work done is to be calculated the volumes and the pressures may be expressed in any unit, provided both the quantities are expressed in the same unit.
• The work obtained using the above equation will be in joule if R is taken in the S.I. unit.
• The number of moles  = Wt. in gram / Molecular wt.in grams
• Absolute temperature T k   = to C   + 273

Work is Path Function:

The work done in isothermal constant pressure process is given by

W   =    – Pext ΔV i.e. W   =    – Pext  ( V₂  –  V₁) 

Where, Pext = External opposing pressure

V₁ = Initial volume V₂ = Final volume.

The work done in the isothermal reversible process is given by

Where, n = Number of moles of the gas R = Universal gas constant

T = Absolute temperature of the gas

V₁ = Initial volume V₂ = Final volume

We can see that the work done depends on the manner or the conditions under which the process carried out. Thus it is not dependent on initial and final conditions of the system but on the path followed by the system. Hence work is path function or it is not a state function.

Numerical Problems on Maximum Work:

Example – 01:

3 moles of an ideal gas are expanded isothermally and reversibly from volume of 10 m³ to the volume 20 m³ at 300 K. Calculate the work done.

Given: n = 3 moles, V₁ = 10 m³, V₂ = 20 m³, T = 300 K, R = 8.314 J K^-1 mol^-1.

To Find: Work done =?

Solution:

Work done in isothermal reversible process is given by

W_max = -2.303 nRT log₁₀(V₂/V₁)

∴ W_max = -2.303 × 3 mol × 8.314 J K^-1 mol^-1 × 300 K × log₁₀( 20 m³/ 10 m³)

∴ W_max = -2.303 × 3 × 8.314 × 300 × log₁₀(2)

∴ W_max = -2.303 × 3 × 8.314 × 300 × 0.3010

∴ W_max = – 5187 J = – 5.187 kJ

Ans: maximum work done =  – 5187 J = – 5.187 kJ

Negative sign indicates the work is done by the system on the surroundings

Example – 02:

24 g of oxygen are expanded isothermally and reversibly from 1.6 × 10⁵ Pa pressure to 100 kPa at 298 K. Calculate the work done.

Given: m = 24 g, P₁ = 1.6 × 10⁵ Pa, P₂ = 100 kPa = 100 × 10³ Pa = 1 × 10⁵ Pa, T = 298 K, R = 8.314 J K^-1 mol^-1.

To Find: Work done =?

Solution:

Number of moles of oxygen = Given mass of oxygen / molecular mass of oxygen

Number of moles of oxygen = 24 g / 32 g mol^-1 = 0.75 mol

Work done in isothermal reversible process is given by

W_max =  -2.303 nRT log₁₀(P₁/P₂)

∴ W_max = -2.303 × 0.75 mol × 8.314 J K^-1 mol^-1 × 298 K × log₁₀( 1.6 × 10⁵ Pa / 1 × 10⁵ Pa)

∴ W_max = -2.303 × 0.75 × 8.314× 298 × log₁₀( 1.6)

∴ W_max = -2.303 × 0.75 × 8.314 × 298  × 0.2041

∴ W_max = – 873.4 J

Ans: maximum work done =  – 873.4 J

Negative sign indicates the work is done by the system on the surroundings

Example – 03:

4.4 × 10^-2 kg of CO₂ is compressed isothermally and reversibly at 293 K from the initial pressure of 150 kPa when work obtained is 1.245 kJ. Find the final pressure.

Given: m = 4.4 × 10^-2 kg, P₁ = 150 kPa, Maximum work of compression = W = + 1.245 kJ = 1245 J, T = 293 K, R = 8.314 J K^-1 mol^-1, P₂ =?

To Find: Final pressure =?

Solution:

Number of moles of CO₂ = Given mass of CO₂ / molecular mass of CO₂

Number of moles of CO₂ = n = 4.4 × 10^-2 kg / 44 × 10^-3 kg mol^-1 = 1 mol

Work done in isothermal reversible process is given by

W_max = -2.303 nRT log₁₀(P₁/P₂)

∴ 1245 J = -2.303 × 1 mol × 8.314 J K^-1 mol^-1 × 293 K × log₁₀(P₁/P₂)

∴ 1245 J = -2.303 × 1 mol × 8.314 J K^-1 mol^-1 × 293 K × log₁₀(P₂/P₁)

∴ log₁₀(P₂/P₁) = (1245) J / (2.303 × 1 × 8.314 × 293)  J

∴ log₁₀(P₂/P₁) = 0.2219

∴ (P₂/P₁) = Antilog (0.2219) = 1.667

∴ P₂ = 1.667 ×  P₁  = 1.667 × 150 kPa = 250 kPa

Ans: Final Pressure is 250 kPa

Example – 04:

2.8 × 10^-2 kg of nitrogen is expanded isothermally and reversibly at 300 K from the initial pressure of 15.15 × 10⁵ Nm^-2 when work obtained is 17.33 kJ. Find the final pressure.

Given: m = 42.8 × 10^-2 kg, P₁ = 15.15 × 10⁵ Nm^-2, Maximum work of expansion = W = – 17.33 kJ = – 17.33 × 10³ J, T = 293 K, R = 8.314 J K^-1 mol^-1, P₂ =?

To Find: Final Pressure =?

Solution:

Number of moles of Nitrogen = Given mass of Nitrogen / molecular mass of Nitrogen

Number of moles of Nitrogen = n = 2.8 × 10^-2 kg / 28 kg mol^-1 = 1 mol

Work done in isothermal reversible process is given by

W_max = -2.303 nRT log₁₀(P₁/P₂)

∴ – 17.33 × 10³ J = -2.303 × 1 mol × 8.314 J K^-1 mol^-1 × 300 K × log₁₀(P₁/P₂)

∴ 17.33 × 10³ J = 2.303 × 1 mol × 8.314 J K^-1 mol^-1 × 300 K × log₁₀(P₁/P₂)

∴ log₁₀(P₁/P₂) = (17.33 × 10³) J / (2.303 × 1 × 8.314 × 300)  J

∴  log₁₀(P₁/P₂) = 3.0170

∴ (P₁/P₂) = Antilog (3.0170) = 1040

∴ P₂ =  P₁ / 1040 = 15.15 × 10⁵ Nm^-2 / 1040 =  14056.7 Nm^-2

Ans: Final Pressure is 14056.7 Nm^-2

Example – 05:

3 moles of an ideal gas are compressed isothermally and reversibly at 22° C to a volume 2L when work done is 2.983 kJ. Find the initial volume.

Given: n = 3 mol, V₂ = 2 L, Maximum work of compression = W = + 2.983 kJ = 2.983 × 10³ J , T = 22° C = 22 + 273 = 295 K, R = 8.314 J K^-1 mol^-1, V₁ = ?,

To Find: Initial volume =?

Solution:

Work done in isothermal reversible process is given by

W_max =  -2.303 nRT log₁₀(V₂/V₁)

∴ 2.983 × 10³ J =  -2.303 ×3 mol × 8.314 J K^-1 mol^-1 × 295 K × log₁₀(V₂/V₁)

∴ 2.983 × 10³ J = 2.303 ×3 mol × 8.314 J K^-1 mol^-1 × 295 K × log₁₀(V₁/V₂)

∴ log₁₀(V₁/V₂) = 2.983 × 10³ J / (2.303 × 3 × 8.314 × 295)  J

∴ log₁₀(V₁/V₂) = 0.1760

∴ (V₁/V₂) = Antilog (0.1760) = 1.5

∴ V₁= 1.5  × V₂  = 1.5  ×  2 L  = 4 L

Ans: Initial Volume = 4 L

Example – 06:

280 mmol of an ideal gas occupy 12.7 L at 310 K. Calculate the work done when gas expands a) isothermally against constant external pressure 0.25 atm, b) isothermally and reversibly c) in a vacuum until its volume becomes 3.3 L.

Given: n = 280 mmoles = 0.280 mol, V₁ = 12.7 L, ΔV = 3.3 L, T = 310 K, R = 8.314 J K^-1 mol^-1. External Pressure = 0.25 atm

To Find: Work done =?

Solution:

a)  The gas expands isothermally against constant external pressure 0.25 atm

W = – P_ext × ΔV = – 0.25 atm × 3.3 L = – 0.825 L atm

W = – 0.825 L atm × 101.3  J L^-1 atm^-1 = – 83.6 J

Hence work done = – 83.6 J

Negative sign indicates the work is done by the system on the surroundings

b) Gas expands isothermally and reversibly

ΔV = (V₂ – V₁) = (V₂ –  12.7 L) =  3.3 L

V₂ = 3.3 L + 12.7 L = 16 L

Work done in isothermal reversible process is given by

W_max = -2.303 nRT log₁₀(V₂/V₁)

∴ W_max = -2.303 × 0.280 mol × 8.314 J K^-1 mol^-1 × 310 K × log₁₀( 16 L/ 12.7 L)

∴ W_max = -2.303 × 0.280 × 8.314 × 310 × log₁₀(1.260)

∴ W_max = -2.303 × 0.280 × 8.314 × 310 × 0.1004

∴ W_max = – 166.9  J

Maximum work = 166.9 J

Negative sign indicates the work is done by the system on the surroundings

c) Gas expands in vacuum

In this case, it is a free expansion, there is no opposing pressure. Hence P_ext = 0

W = – P_ext × ΔV = 0 atm × 3.3 L  = 0

Work done = 0

Example – 07:

 300 mmol of a perfect gas occupies 13 L at 320 K. Calculate the work done in joules when the gas expands a) isothermally against the constant external pressure of 0.20 atm. b) Isothermal and reversible process c) Into vacuum until the volume of the gas is increased by 3 L.

Given: n = 300 mmoles = 0.3 mol, V₁ = 13 L, ΔV = 3 L, T = 320 K, R = 8.314 J K^-1 mol^-1. External Pressure = 0.20 atm

To Find: Work done =?

Solution:

a)  The gas expands isothermally against constant external pressure 0.20 atm

W = – P_ext × ΔV = – 0.20 atm × 3 L = – 0.6 L atm

W = – 0.6 L atm × 101.3 J L^-1 atm^-1 = – 60.78 J

Hence work done = – 60.78 J

Negative sign indicates the work is done by the system on the surroundings

b) Gas expands isothermally and reversibly

ΔV = (V₂ –  V₁)  =  (V₂ –  13 L) =  3 L

V₂ = 3 L + 13 L = 16 L

Work done in isothermal reversible process is given by

W_max = -2.303 nRT log₁₀(V₂/V₁)

∴ W_max = -2.303 × 0.3 mol × 8.314 J K^-1 mol^-1 × 320 K × log₁₀( 16 L/ 13 L)

∴ W_max = -2.303 × 0.3 × 8.314 × 320 × log₁₀(1.231)

∴ W_max = -2.303 × 0.3 × 8.314 × 320 × 0.0902

∴ W_max = – 165.8  J

Maximum work = – 165.8  J

Negative sign indicates the work is done by the system on the surroundings

c) Gas expands in vacuum

In this case, it is the free expansion, there is no opposing pressure. Hence P_ext = 0

W = – P_ext × ΔV = 0 atm × 3 L  = 0

Work done = 0

Previous Topic: Pressure-Volume Type Work

Next Topic: First Law of Thermodynamics

Science > Chemistry > Chemical Thermodynamics and Energetics > Concept of Maximum Work

The post Concept of Maximum Work appeared first on The Fact Factor.

Pressure Volume Work:

Consider an ideal gas having definite mass (say n moles) be enclosed in a cylinder fitted with weightless, frictionless, tightly fitted, movable piston. Let ‘A’ be the area of the cross-section of the cylinder. Let the gas expand from volume V₁ to V₂ against constant external pressure P_ext which is exerted on the piston. Due to expansion, the piston moves upward by a distance ‘d’. In this process, the system loses energy to the surroundings. Hence the work done by the system is negative.

Work done = – Opposing Force x displacement

∴   W = – F x displacement    …….  (1)

Now, pressure is force per unit area.

∴ Pressure   = Force /  Area

∴ Force = Pressure x  Area

∴       F        =      P   x    A   ……..  (2)

Substituting this value in equation (1)

∴   W   =     –  P   x   A    x   dx     ………. (3)

But,  A   x  dx  = Volume through gas expands

=   Change in volume =  ( V₂  –  V₁) = Δ V

Substituting this value in equation (3)

W = – P_ext ΔV

This is an expression for pressure volume work done in terms of pressure and volume in the isothermal expansion of an ideal gas against constant external pressure.

Notes:

• It is the external pressure against gas expands i.e. P_ext and not the pressure of the gas that is used in evaluating the pressure volume work done.
• The final equation shows that work  W  depends only on the change in volume (ΔV) and opposing pressure (P_ext) and not on the quantity of the gas (i.e. it is independent of the number of moles) and Temperature of the gas (T).
• The work done by the system in a process depends on the way or manner in which it is carried out. Work, therefore, is not a state function. It is a path function.

Free Expansion:

The free expansion is also known as work of expansion in a vacuum. When a gas expands in a vacuum there is no opposing force i.e.  P_ext = 0.

For getting work done opposing force is necessary, hence no work is done when a gas expands in a vacuum. Such an expansion is called as free expansion.

W   =    – P_ext ΔV

∴ W   =  0 × ΔV = 0

Work Done in a Cyclic Process:

A cyclic process is one which consists of a series of intermediate steps, at the end of which the system returns to its initial state.

Since in a cycle, initial and final states are the same.

∴ Δ(state function)   =   0

∴ ΔU =  0 (i.e. No change in internal energy)

Now,   ΔU   = q + W

∴ 0 = q  +   W

∴ q = -W

Thus in a cyclic process heat absorbed by the system from the surrounding (q) is equal to work done (W) by the system on the surrounding.

Difference Between Heat and Work: 

When heat is added to gas the random motion of molecules of gas increases, thus heat is a stimulus to gas which increases the random motion of the gas molecules. Thus heat is a random form of energy. Now let us consider work done on a system by compressing the gas in a cylinder. The movement of the piston makes the gas molecules to move in the direction of applied force. Thus work is a stimulus which increases the organized motion of the molecules of the gas. Thus work is an organized form of energy.

Sign Convention Used in Thermodynamics:

• During the expansion, V₂ > V₁, Hence ΔV is positive.    Hence work done W is negative. Thus when the work is done by the system on the surrounding i.e. Work of expansion is taken as negative (- W).
• During compression, V₂ < V₁, Hence ΔV is negative. Hence work done W is positive. Thus when the work done on the system by the surrounding i.e. Work of compression is taken as positive (+W).

Numerical Problems on Pressure Volume Work:

Example – 01:

2 moles of an ideal gas are expanded isothermally from volume of 15.5 L to the volume 20 L against a constant external pressure of 1 atm. Calculate the pressure-volume work in L atm and J.

Given: n = 2 moles, V₁ = 15.5 L, V₂ = 20 L, P_ext = 1 atm

To Find: Work Done =?

Solution:

Work done in an isothermal process is given by   W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 1 atm × (20 L – 15.5 L)

∴ W = – 1 atm × (4.5 L) = – 4.5 L atm

W = – 4.5 L atm × 101.3 J L^-1 atm^-1 = – 455.8 J

Ans: work done = – 4.5 L atm or – 455.8 J

The negative sign indicates the work is done by the system on the surroundings

Example – 02:

2 moles of an ideal gas are compressed isothermally from volume of 10 dm³ to the volume 2 dm³ against a constant external pressure of 1.01 × 10⁵ Nm^-2. Calculate the pressure-volume work done.

Given: n = 2 moles, V₁ = 10 dm³ = 10 × 10^-3 m³, V₂ = 2 dm³ = 2 × 10^-3 m³, P_ext = 1 atm.

To Find: Work Done =?

Solution:

Work done in an isothermal process is given by   W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 1.01 × 10⁵ Nm^-2 × (2 × 10^-3 m³ – 10 × 10^-3 m³)

∴ W = – P_ext × (V₂ – V₁) = – 1.01 × 10⁵ Nm^-2 × (- 8 × 10^-3 m³)

∴ W = + 8.08 × 10² J = + 808 J

Ans: work done = + 808 J

Positive sign indicates the work is done by the surrounding on the system

Example – 03:

3 moles of an ideal gas are expanded isothermally from volume of 300 cm³ to the volume 2.5 L against a constant external pressure of 1.9 atm at 300 K. Calculate the pressure volume work in L atm and J.

Given: n = 3 moles, V₁ = 300 cm³= 0.3 L, V₂ = 2.5 L, P_ext = 1.9 atm

To Find: Work Done =?

Solution:

Work done in an isothermal process is given by   W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 1.9 atm × (2.5 L – 0.3 L)

∴ W = – 1.9 atm × (2.2 L) = – 4.18 L atm

W = – 4.18 L atm × 101.3 J L^-1 atm^-1 = – 423.4 J

Ans: work done = – 4.5 L atm or – 423.4 J

The negative sign indicates the work is done by the system on the surroundings

Example – 04:

1 mole of an ideal gas is compressed isothermally from a volume of 500 cm³ against a constant external pressure of 1.216 × 10⁵ Pa. The pressure-volume work involved in the process is 36.5 J. calculate the final volume.

Given: n = 1 mole, V₁ = 500 cm³= 500 × 10^-6 m³, P_ext = 1.216 × 10⁵ Pa = = 1.216 × 10⁵ Nm^-2, Work of compression = + 36. 5 J, V₂ =?

To Find: Final volume = V₂ =?

Solution:

Work done in isothermal process is given by W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁)

∴ 36.5 J= – 1.216 × 10⁵ Nm^-2 × (V₂ – 500 × 10^-6 m³)

∴ 36.5 J/ 1.216 × 10⁵ Nm^-2 = – (V₂ – 500 × 10^-6 m³)

∴300 × 10^-6 m³ = (500 × 10^-6 m³ – V₂)

∴    V₂    = (500 × 10^-6 m³ – 300 × 10^-6 m³)

∴    V₂    = 200 × 10^-6 m³ = 200 cm³

Ans: final volume = 200 cm³

Example – 05:

1 mole of an ideal gas is compressed isothermally from volume of 20 L to 8 L against constant external pressure, when pressure volume work obtained is 44.9 L atm. Find the constant external pressure.

Given: n = 1 mol, V₁ = 20 L, V₂ = 8 L, Work of compression W = + 44.9 L atm, P_ext =?

To Find: External pressure =?

Solution:

Work done in isothermal process is given by   W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁)

∴ 44.9 L atm = – P_ext × (8 L – 20 L)

∴ 44.9 L atm = – P_ext × (- 12 L)

∴ P_ext   =44.9 L atm / 12 L)

∴ P_ext   = 3.74 atm

Ans: constant external pressure is 3.74 atm

Example – 06:

One mole of a gas expands by 3 L against a constant pressure of 3 atmosphere. Calculate the pressure volume work done in a) L-atm b) joules and c) calories.

Given: n = 1 mole, Δ V = 3 L, P_ext = 3 atm

To Find: Work Done =?

Solution:

Work done in isothermal process is given by   W = – P_ext × ΔV

∴ W = – 3 atm × 3 L = – 9 L atm

∴ W = – 9 L atm × 101.3 J L^-1 atm^-1 = – 911.7 J

∴ W = – 911.7 J / 4.184 J cal^-1 = – 217.9 cal

Ans: work done = – 9 L atm or – 911.7 J  or -217.9 cal

Negative sign indicates the work is done by the system on the surroundings

Example – 07:

100 mL of ethylene(g) and 100 mL of HCl (g) are allowed to react at  2 atm pressure as per the reaction given below. Calculate pressure volume type of work in it in joules.

C₂H_4(g)    +  HCl_(g)    →     C₂H₅Cl_(g)

Solution: 

Given: P_ext = 2 atm and the given reaction is

C₂H_4(g)    +  HCl_(g)    →     C₂H₅Cl_(g)

1 Vol               1Vol                      1Vol

100 mL            100 mL                  100 mL

Thus 100 ml of C₂H_4(g) reacts with 100 mL of HCl_(g)  to give 100 mL of C₂H₅Cl_(g).

Initial volume = Volume of reactants = 100 mL + 100 mL = 200 mL = 0.2 L

Final volume = Volume of products = 100 mL = 0.1 L

Work done in isothermal process is given by   W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 2 atm × (0.1 L – 0.2 L)

∴ W = – 2 atm × (- 0.1 L) = 0.2  L atm

W = 0.2 L atm × 101.3 J L^-1 atm^-1 = 20.26 J

Ans: work done = + 20.26 J

Positive sign indicates the work is done by the surroundings on the system

Example – 08:

A gas cylinder of 5 L capacity containing 4 kg of helium gas at 27 °C developed a leakage leading to the escape of the gas into atmosphere. If atmospheric pressure is 1.0 atm. calculate the pressure volume work done by the gas assuming ideal behaviour.

Given: V₁ = 5 L, mass of gas m = 4 kg = 4 × 10³ g, T = 27 °C = 27 +273 = 300 K,  P_ext = P = 1.0 atm, R = 0.0821 L atm K^-1 mol^-1.

To Find: Work Done =?

Solution:

Number of moles = Given mass of He / Molecular mass of He = 4 × 10³ g / 4 g = 10³

By ideal gas equation, we have P V₂ = nRT

∴   V₂ = nRT / P = 10³ × 0.0821 × 300 / 1   = 2.463 × 10⁴ L = 24630 L

Work done in isothermal process is given by   W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 1 atm × (24630 L – 5 L)

∴ W = – 1 atm × (24625 L) = – 24625 L atm

W = – 24625 L atm × 101.3 J L^-1 atm^-1 = – 2.494 × 10⁶ J = – 2494 kJ

Ans: work done = – 2494 kJ

Negative sign indicates the work is done by the system on the surroundings

Example – 09:

1.6 mol of water evaporates at 373 K against atmospheric pressure of 1 atm. Assuming ideal behaviour of water vapours calculate the work done.

Given: n = 1.6 mole, T = 373 K, P_ext = P = 1.0 atm, R = 0.0821 L atm K^-1 mol^-1.

To Find: Work Done =?

Solution:

The molecular mass of water 18 g. and the density of water is 1 g per cc

Hence initial volume of water = V₁ = 18 x 1.6 x 10^-3 L = 0.0288 L

By ideal gas equation, we have P V₂ = nRT

∴   V₂ = nRT / P = 1.6 × 0.0821 × 373 / 1   = 48.93 L

Work done in isothermal process is given by   W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 1 atm × (48.93 L – 0.0288 L)

∴ W = – 1 atm × (48.90 L) = – 48.90  L atm

W = – 48.90 L atm × 101.3 J L^-1 atm^-1 = – 4954.8 J

Hence work done = – 4954.8 J

Negative sign indicates the work is done by the system on the surroundings

Example – 10:

1.0 mol of water evaporates at 373 K against atmospheric pressure of 749.8 mm of Hg. Assuming ideal behaviour of water vapours calculate the pressure volume work done.

Given: n = 1.0 mole, T = 373 K, P_ext = P = 749.8 mm of H = 749.8 / 760 = 0.987 atm, R = 0.0821 L atm K^-1 mol^-1.

To Find: Work Done =?

Solution:

The molecular mass of water 18 g. and the density of water is 1 g per cc

Hence initial volume of water = V₁ = 18 x 1.0 x 10^-3 L = 0.018 L

By ideal gas equation, we have P V₂ = nRT

∴   V₂ = nRT / P = 1.0 × 0.0821 × 373 / 0.987   = 31.03 L

Work done in isothermal process is given by   W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 0.987 atm × (31.03 L – 0.018 L)

∴ W =  –  0.987 atm × (31.01 L) = – 30.61  L atm

W = – 30.61 L atm × 101.3 J L^-1 atm^-1 = – 3101 J

Ans: work done = – 3101 J

Negative sign indicates the work is done by the system on the surroundings

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In this article, we shall discuss what is meant by process and different types of chemical processes.

Process:

It is the path or the operation by which a system changes from one state to another. A process causes a change in the value of at least one of the state functions.

Types of Chemical Processes:

Isothermal process:

If a process is carried out at a constant temperature, the process is called an isothermal process. For isothermal process ΔT = 0, ΔU = 0. Internal energy (U) of a system remains constant during isothermal process provided there is no change of phase. e.g. Fusion of ice.

Characteristics of Isothermal Chemical Processes:

• If a process is carried out at a constant temperature, the process is called an isothermal process.
• The exchange of heat takes place with the surroundings.   ( q ≠ 0)
• Internal energy remains constant. Δ U = 0    (provided there is no change in a phase).
• Total heat content (enthalpy) of the system varies.(Δ H ≠ 0)
• The system is not thermally isolated from the surroundings.
• Expansion occurs with the absorption of heat, while compression occurs with the evolution of heat.
• W  =  q
• An isothermal process can be made reversible.

Adiabatic process:   

A process carried out in such a manner that the system, undergoing the change, does not exchange heat with the surroundings is called an adiabatic process. The temperature of the system changes during the adiabatic process. For adiabatic process Q = 0. e.g. Expansion of a gas in a vacuum.

Characteristics of Adiabatic Chemical Processes:

• If a process is carried out in such a manner that the system, undergoing the change, does not exchange with the surroundings is called an adiabatic process.
• The exchange of heat with the surrounding does not take place.  ( q = 0)
• Internal energy varies. (ΔE ≠ 0)
• Total heat content (enthalpy) of the system remains constant.(Δ H = 0)
• The system is thermally isolated from the surroundings.
• In expansion temperature and internal energy decreases, while in compression temperature and internal energy increase.
• W  =  E₂  –  E₁  = Δ E
• The adiabatic process cannot be made reversible.

Isobaric process:

If a process is carried out at constant pressure, the process is called an isobaric process. For isobaric process ΔP = 0.

Example: The reaction carried out an open vessel i.e. at constant atmospheric pressure.

Isochoric process:

If a process is carried out at constant volume, the process is called the isochoric process. For isochoric process ΔV = 0, W = 0.

Example: A gaseous reaction carried out in a tightly closed container.

Reversible process:

A thermodynamically reversible process is one which takes place infinitesimally slowly in a large number of stages so that at every stage driving force is only infinitesimally greater than the opposing force and direction of the process at any stage can be reversed by an infinitesimal increase in opposing force.

At every stage of the process, the system remains in thermodynamic equilibrium with the surroundings.

Characteristics of Reversible Chemical Processes:

• A reversible process is one whose direction can be reversed by an infinitesimal change in the conditions.
• It is a hypothetical (imaginary) reaction. It is a non-spontaneous process. It is to be arranged artificially.
• This process is infinitesimally slow
• There is an equilibrium at every stage of the process.
• The direction of the process can be reversed at any stage by an infinitesimal change in one of the state functions.
• The driving and opposing forces differ by an infinitesimally small amount.
• Maximum work can be obtained.

Irreversible process:

In an irreversible process, the system does not remain in thermodynamic equilibrium with surrounding during the process. Equilibrium is reached at the end of the process.

In an irreversible process driving and opposing forces differ by a finite amount and process proceeds in a definite direction, its direction cannot be reversed by very small changes of conditions.

An irreversible process is a natural spontaneous process like the flow of water from a higher level to a lower level.

Characteristics of Irreversible Chemical Processes:

• An irreversible process is one whose direction cannot be reversed by changing the conditions slightly.
• It is a natural and real process. It is a spontaneous process. It takes place naturally.
• This process is comparatively fast.
• Equilibrium is reached at the end of the process
• The direction of the process cannot be reversed by small changes in state functions.
• The driving forces are quite larger than opposing forces.
• Maximum work cannot be obtained.

Cyclic process:

A cyclic process is one which consists of a series of intermediate steps at the end of which the system returns to the initial state. Since in a cycle, initial and final states are the same, Thus, Δ (state function)  =  0

Representation of Various Chemical Processes on PV Diagram:

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