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

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

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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 derive an expression for pressure exerted by gas on the walls of container. We shall also derivation of different gas laws using the kinetic theory of gases.

Expression for Pressure Exerted by a Gas Using Kinetic Theory of Gases:

Let us consider a gas enclosed in a cube whose each edge is of length ‘l’. Let A be the area of each face of the cube. So A = l². Let V be the volume of the cube (or the gas). So V = l³. Let ‘m’ be the mass of each molecule of the gas and ‘N’ be the total number of molecules of the gas and ‘M’ be the total mass of the gas. So M = mN.

Suppose that the three intersecting edges of the cube are along the rectangular co-ordinate axes X, Y and Z with the origin O at one of the corners of the cube.  By the kinetic theory of gases, we know that molecules of a gas are in a state of random motion so it can be imagined that on an average N/3 molecule are constantly moving parallel to each edge of the cube i.e. along the co-ordinate axes. Let velocities of N/3 molecules moving parallel to X-axis be C₁, C₂, C₃, …. , C_N/3 respectively.

Consider a molecule moving with the velocity C₁in the positive direction of the X-axis. The initial momentum of the molecule,   

p₁ = mC₁,

It will collide normally with the wall ABCD and as the collision is perfectly elastic rebounds with the same velocity. Hence, momentum of molecule  after  collision, 

p₂ =  – mC₁

Change in the momentum of the molecule due to one collision with ABCD

Δp =  p₂  –   p₁

Δp = – mC₁  – mC₁     =  – 2 – mC₁

Before the next collision with the wall ABCD the molecule will travel a distance 2l with velocity C₁. So time interval between two successive collisions of the molecule with ABCD = 2l /C₁

Number of collisions of the molecule per unit with wall ABCD =  C₁/ 2l

Change in the momentum of the molecule per   unit   time

= –  2mC1  ×  C₁/ 2l   = –  mC₁² /l

But by Newton’s second law we know that the rate of change of momentum is equal to the impressed force.

So force exerted on the molecule by wall ABCD =- mC₁² /l

From    Newton’s third law of motion, action and reaction are equal and opposite. So force exerted on the molecule by wall ABCD   = mC₁² /l. Thus every molecule will exert a force on the wall ABCD. So total force exerted on the wall ABCD due to molecules moving in the positive X-axis direction is given by

Let P be the pressure of the gas. The pressure exerted on the wall ABCD is given by

The pressure exerted on each wall will be the same and i.e. equal to the pressure of the gas. By definition of r.m.s. velocity

This is an expression for pressure exerted by a gas on the walls of the container.

Boyle’s Law from Kinetic Theory of Gases:

Statement: 

The temperature remaining constant the pressure exerted by a certain mass of gas is inversely proportional to its volume.

Explanation: 

If P is the pressure and V is the volume of a certain mass of enclosed gas, then

P ∝1 / V      ∴ P V = constant

Proof:

From kinetic theory of gases, the pressure exerted by a gas is given by

Where M = Total mass of the gas = Nm

V = Volume of the gas

ρ = Density of the gas

C̅ = r.m.s. velocity of gas molecules.

M = Molecular mass of the gas

N = Number of molecules of a gas

m = Mass of each molecule of a gas.

But by assumptions of the kinetic theory of gases the average kinetic energy of a molecule is constant at a constant temperature. Thus the right-hand side of the equation is constant.

Thus,         PV = constant i.e.  P ∝ 1 / V .   This is Boyle’s Law.

Thus Boyle’s law is deduced from the kinetic theory of gases.

Relation Between r.m.s. Velocity of Gas Molecule and the Absolute Temperature of the Gas:

From kinetic theory of gases, the pressure exerted by a gas is given by

Where M = Total mass of the gas = Nm

V = Volume of the gas

ρ = Density of the gas

C̅ = r.m.s. velocity of gas molecules.

M = Molecular mass of the gas

For one mole of a gas, the total mass of the gas can be taken as molecular weight

This is an expression for r.m.s. velocity of gas molecules in terms of its molecular weight. Now R is the universal gas constant, molecular mass M for a particular gas is constant.

Thus, the R.M.S. velocity of a gas is directly proportional to the square root of the absolute temperature of the gas.

Kinetic Energy of Gas:

Kinetic Energy Per Unit Volume of a Gas:

From kinetic theory of gases, the pressure exerted by a gas is given by

Where M = Total mass of the gas = Nm

V = Volume of the gas

ρ = Density of the gas

C̅ = r.m.s. velocity of gas molecules.

M = Molecular mass of the gas

N = Number of molecules of a gas

m = Mass of each molecule of a gas.

This is an expression for the kinetic energy of gas molecules per unit volume of the gas.

Kinetic Energy of a Gas:

From kinetic theory of gases, the pressure exerted by a gas is given by

Where M = Total mass of the gas = Nm

V = Volume of the gas

ρ = Density of the gas

C̅ = r.m.s. velocity of gas molecules.

M = Molecular mass of the gas

N = Number of molecules of a gas

m = Mass of each molecule of a gas.

This is an expression for the kinetic energy of gas molecules.

Average Kinetic Energy Per Mole of a Gas:

From kinetic theory of gases, the pressure exerted by a gas is given by

Where M = Total mass of the gas = Nm

V = Volume of the gas

ρ = Density of the gas

C̅ = r.m.s. velocity of gas molecules.

M = Molecular mass of the gas

For one mole of a gas, the total mass of the gas can be taken as molecular weight

This is an expression for r.m.s. velocity of gas molecules in terms of its molecular weight. Now R is the universal gas constant, molecular mass M for a particular gas is constant.

This is an expression for kinetic energy per mole of a gas

Average Kinetic Energy Per Molecule of a Gas:

From kinetic theory of gases, the pressure exerted by a gas is given by

Where M = Total mass of the gas = Nm

V = Volume of the gas

ρ = Density of the gas

C̅ = r.m.s. velocity of gas molecules.

M = Molecular mass of the gas

For one mole of a gas, the total mass of the gas can be taken as molecular weight

This is an expression for r.m.s. velocity of gas molecules in terms of its molecular weight. Now R is the universal gas constant, molecular mass M for a particular gas is constant.

This is an expression for kinetic energy per mole of a gas

Where N = Avogadro’s number

This is an expression for average kinetic energy per molecule of a gas.

Average Kinetic Energy Per Unit Mass of a Gas:

From kinetic theory of gases, the pressure exerted by a gas is given by

Where M = Total mass of the gas = Nm

V = Volume of the gas

ρ = Density of the gas

C̅ = r.m.s. velocity of gas molecules.

M = Molecular mass of the gas

For one mole of a gas, the total mass of the gas can be taken as molecular weight

This is an expression for r.m.s. velocity of gas molecules in terms of its molecular weight. Now R is the universal gas constant, molecular mass M for a particular gas is constant.

This is an expression for kinetic energy per mole of a gas

This is an expression for average kinetic energy per unit mass of a gas.

Charle’s Law:

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

Dalton’s Law of Partial Pressure:

At constant temperature, the pressure exerted by a mixture of two or more non-reacting gases in enclosed in a definite volume, is equal to the sum of the individual pressure which each gas exerts, if present alone in the same volume.

Explanation: Let P₁, P₂, P₃, …. be the partial pressure of the mixture of non-reacting gases enclosed in a definite volume and at temperature T. Let P be the pressure of a mixture of these gases. Then by the law of partial pressure,

P = P₁ + P₂ + P₃, ….

Maxwell Distribution of Molecular Speeds:

For a given mass of a gas, the velocities of all molecules are not the same, even when bulk parameters like pressure, volume and temperature are fixed. Collisions change the direction and speed of the molecules, but in a state of equilibrium, the distribution of speed is constant.

It is observed that the molecular speed distribution gives the number of molecules dN(v) between speeds v and v + dv, which is proportional to dv (difference in velocities) is called Maxwell distribution.

Following graph shows the distribution of velocity at different temperatures.

The fraction of the molecules with speed v and v + dv is equal to the area of the strip shown.

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