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- 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 0oC 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

- 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.

#### Example – 1:

- At 300 K a certain mass of a gas occupies 1 x 10
^{-4}dm^{3}by volume. Calculate the volume of the gas at 450 K at the same pressure. **Solution:****Given:**Initial temperature = T_{1}= 300 K, Initial volume = V_{1}= 1 x 10^{-4}dm^{3}, Final temperature = T_{2}= 450 K**To Find:**Final volume = V_{2}= ?

By Charle’s Law

∴ V_{2} =(T_{2}/T_{1}) x V_{1}

∴ V_{2} =(450/300) x 1 x 10^{-4} dm^{3} = (1.5) x 1 x 10^{-4} dm^{3} = 1.5 x 10^{-4} dm^{3}

**Ans:** The volume of the gas at 450 K is 1.5 x 10^{-4} dm^{3}

#### Example – 2:

- The volume of given mass of a gas at 0 °c is 2 dm
^{3}. 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 = T_{1}= 0 °C= 0 + 273.15 = 273.15 K, Initial volume = V_{1}= 2 dm^{3}, Final temperature = T_{2}= 0 °C + 10 °C = 10 °C = 10 + 273.15 = 283.15 K**To Find:**Final volume = V_{2}= ?

By Charle’s Law

∴ V_{2} =(T_{2}/T_{1}) x V_{1}

∴ V_{2} =(283.15/273.15) x 2 dm^{3} = 2.07 dm^{3}

**Part – II:**the temperature is decreased by 10 °C**Given:**Initial temperature = T_{1}= 0 °C= 0 + 273.15 = 273.15 K, Initial volume = V_{1}= 2 dm^{3}, Final temperature = T_{2}= 0 °C – 10 °C = – 10 °C = – 10 + 273.15 = 263.15 K**To Find:**Final volume = V_{2}= ?

By Charle’s Law

∴ V_{2} =(T_{2}/T_{1}) x V_{1}

∴ V_{2} =(263.15/273.15) x 2 dm^{3} = 1.93 dm^{3}

**Ans:** When the temperature is increased by 10 °C, the new volume of gas is 2.07 dm^{3}

When the temperature is decreased by 10 °C, the new volume of gas is 1.93 dm^{3}

#### Example – 3:

- A certain mass of a gas occupies a volume of 0.2 dm
^{3}at 273 K. Calculate the volume of the gas if the absolute temperature is doubled at the same pressure. - Solution:
**Given:**Initial temperature = T_{1}= 273 K, Initial volume = V_{1}= 0.2 dm^{3}, Final temperature = T_{2}= 2 x 273 = 546 K**To Find:**Final volume = V_{2}= ?

By Charle’s Law

∴ V_{2} =(T_{2}/T_{1}) x V_{1}

∴ V_{2} =(546/273) x 0.2 dm^{3} = 0.4 dm^{3}

**Ans:** When the temperature is doubled, the new volume of gas is 0.4 dm^{3}

Science > Chemistry > States of Matter > You are Here |

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