### Introduction:

As we know that the volume of the gas depends on the pressure as well as temperature. If the pressure of the gas is kept constant then what will be the interrelationship between the volume of the gas and its temperature? If the temperature is increased, then the average kinetic energy of the gas molecules also increases. As a result, the molecules move at a greater speed. Hence its volume increases and vice-versa.

A famous scientist, Jacques Alexandre César Charles had given a statement about the relationship between the volume of the gas and its temperature and this statement is termed as Charles’ Law.

## Statement Of Charles’ Law

At constant pressure, the volume of a given mass of a gas is increased or decreased by 1 / 273 part of its initial volume at 0 0 C on increasing or decreasing the temperature by one degree centigrade.

## Conversion of the Value of temperature from Celcius Scale to Kelvin Scale and Vice Versa

From Celcius Scale to Kelvin Scale :

Examples 1 : Find the value of 25 0 C in Kelvin scale.

Solution : 25 0 C = ( 25 + 273 ) K = 298 K

Example 2 : Express the value of 30 0 C in absolute scale.

Solution : 30 0  C = ( 30 + 273 ) K = 303 K

Example 3 : Find the value of -5 0 C in Kelvin scale.

Solution : – 5 0 C = ( – 5 + 273 ) K = 268 K

Conclusion : t 0 C = ( t + 273 ) K

From Kelvin scale to Celcius scale :

Example 1 : Find the value of 273 K in Celcius scale.

Solution : 273 K = ( 273 – 273 ) 0 C = 0 0 C

Example 2 : Find the value of 300 K in Celcius scale.

Solution : 300 K = ( 300 – 273 ) 0 C = 27 0 C

Example 3 : Find the value of 303 K in Celcius scale.

Solution : 303 K = ( 303 – 273 ) 0 C = 30 0 C

Example 4 : Find the value of 56 K in Celcius scale.

Solution : 56 K = ( 56 – 273 ) 0 C = 217 0 C

Example 5 : Find the value of 156 K in Celcius scale.

Solution : 156 K = ( 156 – 273 ) 0 C = 117 0 C

Conclusion : T K = ( T – 273 ) 0 C

## Mathematical Form Charles’ Law

Let us assume that the pressure of the gas is kept constant.

V0 = initial volume at 00 C

For 1 0 C rise in temperature, the increased volume is given by

V1  = V0  + ( 1 / 273 ) V0

For 2 0 C rise in temperature, the increased volume is given by

V2  = V0  + ( 2 / 273 ) V0

Similarly, for t 0 C rise in temperature, the increased volume is given by

V  = V0  + ( t / 273 ) V0  = V0  ( 1 + t / 273 ) = V0  ( 273 + t ) / 273 = V0  T / 273

where T = ( 273 + t ) K =Tempearture in Kelvin scale

V / T = ( V0  / 273 )

( V0  / 273 ) = constant term

V ∝ T

V1  / T1  = V2  / T2

Alternative form of Charles’ Law :

At constant pressure, the volume of the given mass of a gas is directly proportional to its temperature.

## Graphical Representation of Charles’ Law

1 ) V – T graph :

2 ) V – t graph :

## Short Questions

1 ) StateCharles’ law.

Ans : At constant pressure, the volume of a given mass of a gas is increased or decreased by 1 / 273 part of its initial volume at 0 0 C on increasing or decreasing the temperature by one-degree centigrade.

2 ) Name the constants in Charles’ Law.

Ans : Pressure and mass of the gas.

3  ) Which physical quantities are variable in Charles’ Law?

Ans : Volume and temperature of the gas.

4 ) What is the nature of the V – T graph?

Ans : Straight line.

5 ) Explain V – t grah.

Ans: V – t graph is a straight line having a negative x-intercept. This negative x-intercept explains the critical temperature of the gas that means the volume of the gas will be zero at -2730 C but this is ideally true. Therefore, in the above graph, the dotted line is shown. This concept is not true for a real gas.

6 ) Explain Charles’ Law.

Ans: We know that, if the temperature of the gas increases the average kinetic energy of the gas molecules also increases. The gas molecules tend to occupy more and more space. Hence, the volume will also increase. According to Charles’ Law, if the temperature is increased by one-degree centigrade the volume of t the fixed mass of any gas increases 1 / 273rd time of its initial volume but the temperature of the gas remains constant.

7 ) Write down the mathematical form of Charles’ Law and explain each term.

Ans : Let P = Pressure of the gas

T = temperature of the gas

M = Mass of the gas

V = Volume of the gas

Form Charles’ Law, at constant P and M

V ∝ T

V1  / T1  = V2  / T2

8 ) Give any two postulates of the kinetic theory of gas.

Ans : i ) The molecules of the gas are point mass and have the negligible volume.

ii ) The force of attraction or repulsion between the molecules is negligible.

9 ) Draw V – T graph.

Ans :

10 ) Draw the V – t graph.

Ans :

11 ) Draw the graph of Charles’ Law.

Ans : Either of the above two graphs.

12 ) What is the relation between volume of the gas and temperature of the gas?

Ans: There is a direct relationship between the volume and temperature of the fixed mass of any gas at constant pressure.

13 ) How does the volume vary with the temperature of the gas while the pressure of the gas remains constant?

Ans: The volume varies directly with the temperature of the fixed mass gas while the pressure of the gas remains constant.

14 ) How does the increment of temperature affect the volume of the gas?

Ans : The increment of temperature affects the volume of the gas. As we know from Charles’s Law that the volume of the fixed mass of any gas increases by 1 / 273rd of its initial volume on increasing the temperature by one degree centigrade at constant pressure.

15 ) What is the SI unit of temperature?

Ans : The SI unit of temperature is Kelvin.

16 ) What is the CGS unit of temperature?

Ans : The CGS unit of temperature is 0 C.

17 ) Name the instrument used to measure the temperature?

Ans : Name of the instrument used to measure the temperature is Thermometer.

18 ) What is the value of the standard temperature?

Ans : 0 0 C or 273 K

19 ) What is the value of the normal temperature ?

Ans : 20 0 C or 293 K

20 ) What is meant by ” Critical temperature?

Ans : The temperature of the gas at which its volume becomes zero, is called as the Critical temperature. Its value is – 273 K. Pratically, it is not possible. It is true only for ideal gas.