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 Pressure-Temperature relation or Gay-Lussac’s law.

Gay-Lussac’s Law:

Statement:

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

Explanation:

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

Alternate Statement of Gay-Lussac’s law:

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

In general

This relation is called the pressure-temperature relation.

Graphical Representation:

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

Numerical Problems:

Example 01:

A steel tank contains air at a pressure of 15 bar at 20 ^oC. The tank is provided with a safety valve which can withstand a pressure of 35 bar. Calculate the temperature to which the tank can be safely heated.

Solution:

Given: Initial pressure P₁ = 15 bar, Initial temperature = 20 ^oC = 20 + 273 = 293 K, Final pressure P₂ = 35 bar

To Find: Temperature up to which tank can be heated = T₂ =?

By Gay-Lussac’s Law

T₂ = (P₂ x T₁)/P₁ = (35 x 293)/15 = 683.67 K

T₂ = 683.67 – 273.15 = 410.15 ^oC

Ans: The tank can be heated up to 410.15 ^oC

Example 02:

An iron tank contains helium at a pressure of 2.5 atm at 25 ^oC. The tank can withstand a maximum pressure of 10 atm. The building in which the tank has been installed catches fire. Predict whether the tank will blow up first or melt if the melting point of iron is 1535 ^oC.

Solution:

Given: Initial pressure P₁ = 2.5 atm, Initial temperature = 25 ^oC = 25 + 273 = 298 K, Melting point of iron = T₂ = 1535 ^oC = 1535 + 273 = 1808 K

To Find: Final pressure = P₂ =?

By Gay-Lussac’s Law

P₂ = (T₂ x P₁)/T₁ = (1808 x 2.5)/298 = 15.16 atm

The pressure at the melting point is 15.16 atm, which is much more than the maximum pressure that the tank can withstand 10 atm. Hence the tank will blow up before reaching the melting point.

Ans: The tank will blow up before reaching the melting point.

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A matter is defined as anything that has mass, which occupies space and may be perceived by senses. There are three states of matter, viz. (a) solid, (b) liquid, and (c) gaseous states. Whatever may be the state of matter, it is composed of particles. In this article, we shall study the particle model of matter.

Particle and Kinetic Model of Matter:

The particle model is also known as a dynamic particle model. On the basis of the particle model, different states of matter can be explained easily. Some assumptions of this model are as follows.

• All matter is made of tiny particles. However, the arrangement and the distribution of particles are different in the three states of matter.
• Empty spaces exist between these particles. These empty spaces are called voids.
• The particles exert force attraction on one another but the magnitude of these interparticle forces is different in the three states of matter.
• The particles are not stationary and have a tendency to acquire motion. In solids, they are fixed at a position and only vibrate about their mean position. In liquids and solids besides vibrational motion, the particles have translatory motion.
• With the increase in the temperature the kinetic energy of the particles hence the thermal energy increases.

Evidence of Particle Nature of Matter:

If we add potassium permanganate in water kept in a glass jar. We can observe the purple coloured particles separate from the crystals of potassium permanganate and spread in water. Ultimately whole water turns purple.

If we add crystals of salt in water, they settle at the bottom. Gradually their size starts reducing and ultimately the crystal disappear but whole water gets a uniform salty taste. Besides the volume of water does not increase. it indicates salt particles occupy inter-particulate spaces.

When a scent bottle is opened at one corner of a room the fragrance can be smelt at any corner of the room. The molecules of scent occupy the inter-particulate space.

Evidence of Kinetic Nature of the Particles of Matter:

The English Botanist Robert Brown, in 1927 observed that colloidal particles exhibit continuous random motion in all directions in a straight line.  He found such movement when pollen grains were suspended in water. 

The phenomenon of continuous zig-zag movement of colloidal particles in straight-line paths in a random direction is known as a Brownian movement. A pollen grain is placed on the surface of water taken in a beaker. It shows the Brownian movement. The pollen grain is surrounded by a large number of water molecules that constantly bombard the pollen grain. On unequal bombardment, the pollen grain gets pushed in certain directions. This experiment proves the kinetic nature of particles of matter.

Characteristics of Particles of the matter:

Particles of matter are very small.

All matter is made up of very small particles that are not visible to naked eye. It can be proved by the following experiment. Take two or three crystals of potassium permanganate and add them in 100ml of water. The solution formed is deep purple in colour. Now take 10ml of this solution and add it to another beaker containing 00ml of fresh water, again you will observe that the colour of the water will change but the solution will be faint compared to that in the first case. Repeat this procedure four or more time. In every step, we observe that the colour of the water changes but it will become fainter and fainter.

The solution remains coloured even at a very high dilution. Which shows that potassium permanganate added is broken into very very small particles exhibiting their characteristic properties. Hence we can conclude that particles of matter are very very small.

Particles have spaces between them.

If we add crystals of salt in water, they settle at the bottom. Gradually their size starts reducing and ultimately the crystal disappear but whole water gets a uniform salty taste. Besides the volume of water does not increase. it indicates salt particles occupy inter-particle spaces present between water particles.

Particles are constantly moving.

A pollen grain is placed on the surface of water taken in a beaker. It shows Brownian movement. The pollen grain is surrounded by a large number of water molecules which constantly bombard the pollen grain. On unequal bombardment, the pollen grain gets pushed in certain directions. This experiment shows that the particles of matter are constantly moving. Thus they possess kinetic energy. As the temperature rises, particles move faster. Hence with the increase in temperature the kinetic energy of the particles also increase.

Particles Attract each other.

Particles of matter have a force acting on them. This force keeps the particles together. The strength of this force of attraction varies from one kind of matter to another.  This force of attraction varies from substance to substance it can be verified by the fact that some forces can be powdered by applying small force, while some break into crystals, while some do not break. This force of attraction between the particles of the same substance is called cohesion. The attractive forces between the particles are maximum in solids and minimum or negligible in case of gases.

States of Matter on the Basis of Particle and Kinetic Model:

Solid State:

At room temperature, the particles of solids occupy definite positions. They can vibrate about their mean positions, but they cannot move from one position to another. The particles are bound to one another strongly. Hence solids have definite shape and definite volume.

Liquid State:

The particles of liquid do not have fixed positions, and they slide over one another within the bulk of the liquid. Thus particles can move from one position to another, but cannot leave the bulk. The particles are bound loosely. Hence liquids have definite volume but don’t have a definite shape. They take shape of the container in which they are kept.

Gaseous State:

In gases, the particles lie far apart and exert a very weak force of attraction on one another. Actually, the particles of gas are in random motion can change position continuously, and can move away from each other. They occupy the whole space available. Only walls of containers restrict their movement. Hence gases have neither definite shape nor definite volume.

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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 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

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In this article, we shall study the general characteristics of the gaseous state and the measurable properties of gases.

General Characteristics of Gaseous State:

• A gas has neither a definite shape nor a definite volume. They acquire the volume and shape of the container in which they are kept.
• A gas is capable of expanding to any limit and is capable of occupying whole available space.
• Gas is highly compressible. As the pressure on the gas is increased its volume decreases.
• Gases have comparatively less density w.r.t. solids and liquids.
• Gases have two specific heats viz. specific heat at constant pressure and specific heat at constant volume.
• In the gaseous state, the intermolecular forces of attraction are very weak i.e. almost zero.
• The molecules of a gas are in continuous random motion.
• Gases have the capacity to diffuse with each other irrespective of the difference in their densities.
• Gases have low viscosity.
• A gas exerts pressure on the walls of the container in which it is kept.
• All the gases obey certain laws known as gas laws. These relat5ions or laws are based on experimental results.

Note: Eleven out of all known elements exist in the gaseous state under normal atmospheric conditions of temperature (25° C) and pressure (1 bar) are as shown below.

Measurable Properties of a Gases:

The measurable properties used to describe the physical state of a gas are (a) mass, (b) volume (c) pressure and (d) temperature.

Mass:

It is given directly or in terms of moles of gas. It is measured in grams or kilograms. Generally, the mass of a gas is expressed in terms of moles.

Number of moles of a gas = Given mass of the gas/Molecular mass of the gas = m/M

Molecular masses of some gases are hydrogen (2), oxygen (32), nitrogen (28), carbon dioxide (44), ammonia (17), methane (16), and chlorine (71)

Measurement of Mass of a Gas:

The mass of the container with enclosed gas is measured. Then the container is completely emptied and the mass of the empty container is measured. The difference between the two masses gives the mass of the gas in the container.

Conversions:

Volume:

It means space occupied by the gas. It is measured in cubic centimeters, millilitres (mL), litres (L), dm³, m³. In the case of gas, the volume of the gas is equal to the volume of the container containing it.

Measurement of Volume of a gas:

The gas acquires volume and shape of the container in which they are kept. Hence the volume of the container is taken as the volume of the gas.

Conversions:

Temperature:

The temperature of a gas is a measure of the quantity of energy possessed by the gas molecules.  It is measured in °C (centigrade or celsius) or K (Kelvin).

Measurement of Temperature of a gas:

The temperature may be defined as the degree of hotness. It is measured by a device called thermometer. The common thermometer is a mercury thermometer.

Various Temperature Scales:

Celsius Scale (° C):

In this scale, the melting point of ice at one-atmosphere pressure and at mean sea level is taken as the lower reference point and consider as 0° C. While boiling point of water at one-atmosphere pressure and at mean sea level is taken as an upper reference point and consider as 100° C. The range between the two reference points is divided into 100 equal parts and each part is called 1° C (one-degree celsius). This scale is also called a centigrade scale.

A lower limit 0° C is considered arbitrary, this scale can be extended to indicate negative temperatures also. The temperature below -273.15° C is not possible.

Farenheit Scale (° F):

In this scale, the melting point of ice at one-atmosphere pressure and at mean sea level is taken as the lower reference point and consider as 32° F. While boiling point of water at one-atmosphere pressure and at mean sea level is taken as the upper reference point and consider as 212° F. The range between the two reference points is divided into 180 equal parts and each part is called 1° F (one-degree farenheit). Nowadays, this scale is not in use.

Kelvin Scale (K):

In this scale, the lowest possible temperature -273.15° C  is taken as a lower reference point. This temperature is called absolute zero. The division of 1 K is equal to 1° C. The unit of temperature in the kelvin scale is K (kelvin) and is considered as the fundamental unit in S.I. system of units.

Conversion of Temperature in Different Scales:

Celsius scale to Kelvin scale  ° C  +  273 = K

Kelvin scale to celsius scale  K  –  273 = ° C

Pressure:

It means the total force exerted by the gas molecules per unit area of the container walls due to their collision on the same walls.  It is measured In terms of the height of the mercury column in centimetres or milimetres or Nm^-2 or Pa (pascal).

The gas molecules are in random motion. During random motion, they collide with each other and with the walls of the container. During the collision with walls of the container, they exert outward force on the walls of the container. The outward force per unit area of the walls is called the pressure of the gas.

The S.I. unit of pressure is Nm^-2 (newton per square metre). It is called Pa (pascal). Other units of pressure used are atm (atmosphere), bar, and in terms of the height of the mercury column.

Atmospheric Pressure:

A thick blanket of air around the earth is called atmosphere. Molecules of various gases present in the atmosphere are under constant pull of gravitational force of the earth. Thus the weight of air column exerts pressure on the surface of the earth. The force experienced by any area of the earth exposed to the atmosphere is equal to the height of the column of air above it. This force per unit area of the earth is called the atmospheric pressure. The value of atmospheric pressure is not constant over the place. It varies with location, temperature and weather conditions. The atmospheric pressure is very large, but living beings are not getting crushed under it because the pressure inside the body of living beings is equal to that of atmospheric pressure.

Atmospheric pressure is measured by a simple device called barometer. It consists of a glass tube of about 1 m long, closed at one end filled with mercury and inverted in an open vessel. The level of mercury adjusts itself in the tube to balance the atmospheric pressure. At sea level, the height of the mercury column is approximately 76 cm above the mercury level in the open vessel.

A standard atmospheric pressure (1 atm) is the pressure exerted by exactly 76 cm of mercury column at 0° C (273.15 K) measured at sea level where standard gravity is 9.806 ms^-2, with the density of mercury being 13.596 g cm^-3. mm of Hg unit is also called torr. The empty space (vacuum) above the mercury column in the tube is called Torricelli’s space.

Measurement of Pressure of a Gas:

The pressure of a gas is measured using a simple apparatus called mercury manometer or pre-calibrated pressure gauges. Open end manometers are suited for measuring pressure equal to or greater than the atmospheric pressure. While closed end manometer are suited for measure pressure below the atmospheric pressure. In closed end manometer, the difference in the levels of mercury in the two limbs gives the pressure of the gas.

In open end manometer, the difference in the levels of mercury in the two limbs gives gauge pressure of the gas.

There are three possibilities in the measurement

• Levels of Hg is same in both the limbs

Gas Pressure = P_atm

• Level of Hg in open limb is more than that connected to the gas container

Gas Pressure = P_atm + Gauge Pressure

• Level of Hg in open limb is lower than that connected to the gas container

Gas Pressure = P_atm – Gauge Pressure.

Conversions:

Numerical Problems:

Example – 01:

A manometer is connected to a gas containing bulb. The open arm reads 43.7 cm whereas the arm connected to the bulb reads 15.6 cm. If barometric pressure is 743 mm Hg. What is the pressure of the gas in the bulb in bar.

Solution:

Difference in mercury level of two arms = 43.7 cm – 15.6 cm = 28. 1 cm

Hence gauge pressure = 28. 1 cm of Hg = 281 mm of Hg

Absolute Pressure = Gauge pressure + Atmospheric pressure = 743 mm + 281 mm = 1024 mm

= 1024 mm/ 760 mm = 1.347 atm = 1.347 × 1.013 = 1.365 bar

Example – 02:

The side arm of an open-end mercury manometer is attached to a gaseous system. The level of mercury in the arm attached to the system is 125 mm lower than the open end arm. Is it higher than the atmospheric pressure or lower? What is the pressure of the system if the atmospheric pressure is 760 mm of Hg?

Solution:

The level of mercury attached to the system is lower than the level in the open arm. Hence pressure of the gas is more than the atmospheric pressure.

Gauge Pressure = 125 mm of Hg

Pressure of system = Gauge Pressure + Atmospheric Pressure  = 125 mm   + 760 mm = 885 mm of Hg

Example – 03:

Reading in mercury manometer closed end arm is 100 mm and in the arm attached to the system is 70 mm. What is the pressure of the system?

Solution:

System Pressure = Gauge Pressure = Difference in mercury level = 100 mm – 70 mm = 30 mm of Hg

Standard Temperature and Pressure (STP) and Normal Temperature and Pressure (NTP):

The volume of a given mass of a gas depends on its pressure and temperature. Hence we have to specify the pressure and temperature of the gas when specifying its volume. When solving problems on the gaseous state, the condition of STP and NTP means

 P = 1 atm = 1.013 × 10⁵ Pa, T = 273.15 K

Science > Chemistry > States of Matter > Gaseous State

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In this article, we shall study a change in the state of a substance.

Melting (Solid → Liquid):

The process of change of solid substance into its liquid state is called melting or fusion. The constant temperature at which the solid becomes liquid upon absorption of heat at constant pressure is called the melting point of that solid at that pressure.

Generally melting point increases with the increase in pressure. Ice is the exception to this because its melting point decreases with the increase in the pressure. Melting point at standard pressure is a characteristic property of a substance. The melting point decreases with the addition of the impurity. Hence melting point can be considered as criteria for purity.

Melting points of some important substances are ice (0 °C), iron (1535 °C), aluminium (660 °C), gold (1064 °C), silver (961 °C), aluminium (660 °C), tin (232 °C), zinc (419.5 °C), copper (1084°C), etc.

Explanation on the Basis of Kinetic Model: 

When solids are heated the thermal energy of particles increases. Thus the cohesive forces between the particles weaken to such extent that the particles can have relative motion with respect to each other but cannot move out of the bulk.  Thus solid gets converted into liquid (melts).

Applications of Melting:

• Melting is very important in the production of alloys. If a binary alloy is to be produced. The element with a higher melting point is melted in a crucible and an element with a lower melting point is added to the molten metal. The second element also melts forming almost a homogeneous solution called alloy. Alloys have many applications in everyday life. Some examples of alloys are

• Substances with a high melting point are used to make high-temperature devices. For example, tungsten is used in an incandescent bulb.
• Metals are melted and they are cast (moulded) to give the solids required shape.

Factors Affecting Melting Point:

Internal factors:

• Inter-Molecular (Particle) Forces: If the attractive forces between the molecules of solid are weaker and then the solid has a low melting point. The attraction between the molecules of covalent compounds is weaker than that in ionic solids and hence covalent compounds have a lower melting point than that of the ionic compounds.
• The shape of molecules: If the shape of the molecule is such that it can have a closed packing of the molecules, then the substance has a higher melting point.
• Size of the molecule: The smaller size of molecules can have a closed packing (less void space) of the molecules, then the substance has a higher melting point.

External Factors: 

• Impurity: The melting point of a substance decreases with the presence of impurities in it, The phenomenon is called melting point depression. The particles of impurity disrupt the repeating pattern of forces that hold the solid together. Hence less energy is required to melt the part of the solid surrounding the impurity. Salt is spread on the frozen street so that the melting point decreases and the ice melt fast.
• Pressure: For the solids, those expand on heating, the melting point increases with increase in the pressure. It is due to the fact that the pressure opposes the increase in the distance between molecules (expansion). e.g. silver, gold, copper, paraffin wax, etc. For the solids, those contract on heating, the melting point decreases with increase in the pressure. It is due to the fact that the pressure supports the decrease in the distance between molecules (contraction). e.g. ice, cast iron, bismuth, brass, etc.

When two ice cubes are pressed together they form a single block of ice. The phenomenon is called regelation. When the two cubes are pressed against each other. the ice at the interface melts due to lowering of melting point. When the pressure is released the melted ice (water) at the interface solidifies again and a single block of ice is obtained.

Sublimation (Solid ⇔ Gas):

Sublimation is the process by which a heated solid directly changes into its gaseous state i.e. vapour state. These vapours on cooling directly give solid. Such substances are called sublimates. Examples are ammonium chloride, ammonia, naphthalene balls, camphor, etc.

Explanation on the Basis of Kinetic Model: 

Certain solids are heated the thermal energy of molecules increases so that the interparticle forces become negligible and the particles can move freely.  Thus such solids on heating get converted directly into gases. This phenomenon is known as sublimation. The cohesive forces between the particles in such substances are weak.

Freezing (Liquid → Solid):

The process of change of matter from a liquid state to a solid state is called freezing or solidification. The constant temperature at which a liquid changes into solid by giving out heat energy (or cooling) is called the freezing point of the liquid. The freezing point of a liquid is a characteristic property of the liquid. Hence can be considered as criteria of purity.

Freezing points of some important substances are water (0 °C), benzene (5.5 °C), mercury (- 38.87 °C), etc.

Explanation on the Basis of Kinetic Model: 

When liquids are cooled the thermal energy of particles decreases. Thus the cohesive forces between the particles strengthen to such extent that the particles can not have relative motion with each other and they occupy the fixed positions.  Thus liquid gets converted into solid (freezes).

Applications of Freezing:

• It is used for the preparation of ice creams.
• The lowering of the freezing point on the addition of solute to the solution is used to find molecular mass of the solute.

Factors Affecting Freezing Point:

For the same substance, the freezing point of the liquid is equal to the melting point of the solid. Therefore the factors those affect melting point of solid obviously affect the freezing point of the liquid.

Freezing Mixtures: 

a mixture of two or more substances (e.g. ice water and salt, or dry ice and alcohol) which can be used to produce temperatures below the freezing point of water.

A freezing mixture of 3 parts of ice and 1 part of NaCl produces a temperature of – 21 °C. A freezing mixture of 2 parts of ice and 3 parts of K₂CO₃ produces a temperature of  – 46 °C.  A freezing mixture of dry ice and alcohol or ethers can produce a temperature of – 60 °C.

In a freezing mixture, a soluble salt is added. The heat required to dissolve one mole of soluble solute in a solvent is called heat of solvation. This heat required for dissolution of solid is taken from the mixture itself and thus the freezing point decreases in steps.

Freezing mixtures ate used to preserve perishable foodstuff like meat and fishes. They are used for producing sub-zero temperatures in laboratories and industrial units.

Evaporation or Vaporization (Liquid → Gas):

The process of conversion of a substance from the liquid state to its vapour state at any temperature below boiling point is called evaporation or vaporization.

Explanation on the Basis of Kinetic Model: 

Some particles from liquid surface possess kinetic energy sufficient to overcome the attractive forces from remaining particles of the liquid and become completely free and escape out as a gas particle in the surroundings. This phenomenon is called evaporation or vaporization.

The rate of evaporation is directly proportional to the surface area and the temperature of the liquid.

During evaporation, the temperature of liquid falls. To maintain temperature balance the liquid particles absorb heat from the surroundings making the surrounding cooler. We have already seen that the molecules with higher kinetic energy leave the surface of the liquid, thus there is an overall decrease in the kinetic energy of liquid. This is one of the reasons for the decrease in the temperature of the liquid.

To increase the rate of evaporation we should increase the surface area, the temperature and the wind speed and should decrease the humidity.

Characteristics of Evaporation:

• It is a surface phenomenon as it takes place on the surface of the liquid.
• It takes place at all temperatures.
• It is a slow process
• The temperature of liquid falls.

Applications of Evaporation:

• During hot day sweat is formed on the body which evaporates. The necessary heat required for the evaporation of the sweat is taken from the body and thus the body temperature is maintained.
• Common salts are produced in shallow lagoons. The water from creek or sea is collected. Water evaporates leaving common salt behind.
• Water gets cooled in an earthen pot (matka). Water seeps through the porous earthen pot and gets on the surface of the pot. It evaporates and the necessary heat required for the evaporation of the water is taken from the water inside the pot and thus the temperature of the water inside the pot decreases.
• Drying of clothes is due to evaporation of water. We have to spread the clothes (increase in surface area), under the sun (increasing temperature) at a windy place.
• In refrigerator the cooling gas (freon) gets evaporator in tubes surrounding freezer region, The necessary heat required for the evaporation of the water is taken from the freezer region and thus the temperature of the freezer region decreases.

Boiling (Liquid → Gas):

Boiling process of change of a liquid into a vapour at a particular temperature and pressure from all part of the liquid. Boiling is a bulk process and takes place throughout the liquid.

When we supply heat energy to liquid the particles start moving faster. At a certain temperature, a point is reached when the particles have enough energy to break free from the forces of attraction of each other. At this temperature, the liquid starts changing into a gas (vapours). The temperature at which a liquid starts boiling at the atmospheric pressure is known as its boiling point. Pure liquids have fixed boiling points. It can be considered as the criteria of purity.

The constant temperature at which a liquid changes to vapour under normal atmospheric pressure is called the boiling point of the liquid. Boiling points of some important liquids are water (100 °C), Ethyl alcohol (78.3 °C), benzene (80.2 °C), chloroform (62 °C), sulphuric acid (280 °C), diethyl ether (35 °C), etc.

Explanation on the Basis of Kinetic Model: 

During boiling, not only the particles on the surface of the liquid but those near walls of the container also start leaving the liquid. It can be seen that small vapour bubbles are formed inside the liquid on walls of the container. As temperature increases the pressure of vapours in bubble increases. The bubbles start growing in size. A point is reached when the vapour pressure inside the bubble is equal to that of atmospheric pressure. At that instant, the bubble detaches from the walls of the container and rise upward. Reaching the surface it bursts giving vapours to the surroundings. Thus there is continuous agitation of the mass of liquid and we say liquid is boiling.

As the pressure increases the boiling point increases. Soluble impurities increase boiling point.

Characteristics of Boiling:

• It is a bulk phenomenon as it takes place throughout the liquid.
• It takes place at fixed temperatures.
• It is a fast process
• The temperature of liquid remains constant.

Factors Affecting Boiling Point:

• Pressure: As the external (atmospheric) pressure decreases boiling point decreases. Hence at higher altitude water boils below 100 °C. Hence higher altitude, food is not cooked properly. To avoid this problem pressure cooker is used for cooking food.

Working of Pressure Cooker:

The basic principle of a pressure cooker is that the boiling point of water increases with the increase in pressure. A pressure cooker is a steel or aluminum vessel with a lid which is airtight. There is a safety valve to release steam to decrease the excess pressure above certain designated pressure. The steam is formed from water in the pressure cooker which has no escape route gets collected in the vessel which put extra pressure on water, which leads to increase in the boiling point of water above 100 °C. Thus gradually the boiling point of water goes on increasing. When the required pressure is reached, the safety valve lifts due to steam pressure and excess of steam is blown out. The safety wall closes and the process restarts. The pressure of steam is even throughout the vessel and hence the food is cooked fast and evenly. The pressure cooker saves a lot of fuel required for cooking.

• Impurity: When a solid is dissolved in liquid the boiling point increases beyond the normal boiling point. Hence during steaming of food, some salt is added to water, so that the food cooks well.

Liquefaction or Condensation:

Liquefaction is the process in which the gaseous substance changes into a liquid state at a particular temperature.

Explanation on the Basis of Kinetic Model:

On cooling the particles of gas lose their kinetic energy and their speed decreases. The decrease in their speed reduces interparticle space and the particles come so close so that the attractive forces between them increase and the gas gets converted into a liquid.

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A matter is defined as anything that has mass, which occupies space and may be perceived by senses. There are three states of matter, viz. (a) solid, (b) liquid, and (c) gaseous states.

Historical Perspective of States of Matter:

Ancient Indian philosophers suggested that all the forms of matter are made up of five basic elements (they called it tatva) they called these five basic elements as ‘panch maha bhoota’. These elements are the sky (Akash), air (vayu), fire (teja), water (aap) and earth (soil). Ancient Greek philosophers considered that all for of matter are made of fire, water, air and the earth. Thales (640-546 BC) suggested that all things arose from the water.

The properties which decide the state of matter are the interparticle space, the force of attraction between particles, and the kinetic energy of particles due to their motion. Thus different states of matter can be explained on the basis of particle and kinetic model.

Solid State:

It has a definite shape and definite volume at a given temperature and pressure. A substance is said to be in the solid state if its melting point is above the room temperature at the atmospheric pressure e.g. Chair, chalk, desk, salt, silver, etc.

Characteristics of Solid State:

• Solids have a definite shape and volume.
• There are strong cohesive forces between the molecules of solid.
• The molecules of solid are fixed at one point.
• The melting point of the solid is above room temperature at atmospheric pressure.
• Solids have high densities.

Particle Model:

According to the particle model, in the solid state, the constituent particles are very close to each other. Hence voids between them are very small. There are strong cohesive forces between the particles of solid.

Due to small voids and strong cohesive forces, the particles are not free to change their position and thus can’t have relative motion w.r.t. each other. Thus the particles of solid are fixed at one point. Hence solids have a definite shape and definite volume at given temperature and pressure.

Kinetic Model:

Due to small interparticle space and strong cohesive forces, the particles are fixed in one position. They can only vibrate about their mean position. Hence solids have low thermal energy and thus particles cannot break away from each other by overcoming inter-particles attractive forces. Thus they have a definite spatial arrangement. Hence solids have a definite shape and definite volume at a given temperature and pressure.  When the average distance between the particles increases beyond 10-9 m, the solid melts into a liquid.

Liquid State:

It has a definite volume but has indefinite shape. It will take the shape of the container containing it. A substance is said to be in the liquid state if its boiling point is above the room temperature and melting point is above the room temperature at the atmospheric pressure e.g. water, alcohol, milk etc.

Characteristics of Liquid State:

• Liquids do not have a definite shape but have a definite volume
• In liquids, the cohesive forces are weaker compared to solid and stronger compared to gases.
• Molecules of liquid move freely anywhere but can’t leave the bulk.
• The boiling point of a liquid is above and its freezing point (melting point) is below the room temperature at the atmospheric pressure.
• Liquids have comparatively low densities compared to solids but have higher densities than gases.

Particle Model:

According to the particle model, in the liquid state, the distance between constituent particles is more compared to that between solid particles and less than that between gaseous particles. Thus voids are more compared to that in solids but less compared to that in gases. The cohesive forces between the particles of a liquid are weaker than that between solid particles and stronger than that between gaseous particles.

Hence the cohesive forces are weak enough so that the particles of liquid can have relative motion w.r.t. each other but these cohesive forces are strong enough to stop the particles of a liquid to go out of the bulk. Hence liquids have a definite volume but have indefinite shape.

Kinetic Model:

The interparticle distance between the particles is more than that in the solid state. Hence the attractive forces are weaker than that in the solid state. There is larger void space among the particles. Hence the particles can vibrate with a higher amplitude. At the same time, the particles can move in the bulk. Hence they have translational motion.

Thus particles in the solid state have more thermal energy than that in the solid state. Thus liquids can flow and have a definite volume. Due to their fluidity, they acquire the shape of the container in which they are kept.

Gaseous State:

Gas has neither a definite shape nor a definite volume. It takes the shape and volume of the container. Thus it occupies the whole available volume. A substance is said to be in the gaseous state if its boiling point is below room temperature at atmospheric pressure. e.g. air, oxygen, nitrogen, carbon dioxide.

Characteristics of Gaseous State:

• Gases have neither definite shape nor a definite volume.
• In gases, the intermolecular forces of attraction are very weak i.e. almost zero.
• Molecules of gases move freely anywhere.
• The condensation point (boiling point) of gas is below the room temperature at atmospheric pressure.
• Gases have very low densities.

Particle Model:

According to the particle model, iIn the gaseous state, the distance between constituent particles is very large compared to that between solid particles of the liquid. Voids are very large. The cohesive forces between the particles of a gas are negligible. Hence the particles of a gas can move away freely from the bulk and occupy any space available. Hence, gases have neither a definite shape nor a definite volume.

Kinetic Model:

In the gaseous state, the distance between constituent particles is very large compared to that between solid particles of the liquid. The cohesive forces between the particles of a gas are negligible. Hence the particles are free to move and free to vibrate. Hence they have the highest kinetic energy (hence thermal energy) in this state compared to the solid and liquid state.

On cooling the gas the kinetic energy of the gas particles decreases and the molecules come near to each other resulting in an increase in the cohesive forces and thus the gas condenses to form a liquid.

Notes:

• By changing the temperature or pressure or both, the state of the substance can be changed.
• Besides these three standard states of matter, there are two more states called plasma state (Exists at very high temperature) and Bose-Einstein condensate (Exists at the very very cold condition).

Comparative Study of States of Matter:

Plasma State:

This state exists at superheated gaseous state consisting of a mixture of electrons and positively charged ions with unusual properties. These particles are super energetic and are at super excited state. It is found at extremely high temperatures such as interiors of suns and stars or intense electric fields as a discharge tube. Astronomers reveal that 99% of all matter in the universe exists in the plasma state.

Bose-Einstein Condensate (1924):

This state was predicted by Einstein and proved by Satyendra Nath Bose in 1920.  It is super cooled solid in which atoms lose their separate identity. They get condensed and behave like a single super atom. This state is very useful for the modern concept of superconductivity.

Bulk Properties of Matter:

The bulk properties of matter depict the collective behaviour of a large number of particles taken together. These properties are not exhibited by the particle individually.

Volume, pressure, temperature, melting point, boiling point, vapour pressure, density, surface tension, viscosity etc. are the bulk properties of matter.

Bulk properties of matter are dependent on the state of the matter and they change with the change in the state of the matter. Similarly, these bulk properties depend on the energy of constituent particles and electrostatic attraction between them. The change in the physical state and the bulk properties of matter depend on the energy of constituent molecules and intermolecular attraction between them.

It is to be noted that the chemical properties of a substance do not change with the change in the state of the substance

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