A French chemist, Jacques Charles found the relationship between the volume and temperature under constant pressure. Charles law states that the volume of a given mass of gas is directly proportional to its absolute temperature or temperature in Kelvin, provided pressure remaining constant.

It means that the volume of the gas increases as temperature increases and also decreases as temperature decrease i.e (vice versa).

Mathematically,

:- \( \scriptsize V \propto T \) P (T, n constant)

:- \( \scriptsize V \propto T \) PK

:- \( \scriptsize V = KT \)

OR

K = \( \frac{V}{T} \)

Where V = Volume

T = Kelvin temperature

K = Mathematical constant

For two gases (1) and (2):

:- \( \normalsize \frac{V_1}{T_1} \scriptsize = K, \; \normalsize \frac{V_2}{T_2} \scriptsize = K \)

Therefore, \( \frac{V_1}{T_1}= \frac{V_2}{T_2} \)

V_{1} = Initial volume, T_{1} = Initial Temperature

V_{2} = Final volume, T_{2} = Final Temperature.

### Graphical Representation of Charles Law

**Note:**

The temperature at -273^{0}C is called the Absolute zero temperature. This is the temperature at which the molecules of a gas have no volume, no kinetic energy (no velocity), and practically no pressure.

The advantage of the Kelvin scale is that it eliminates the use of zero or negative temperature values when solving problems involving gases. This is because the absolute scale, -273^{0}c = 0K.

**Pictorial Representation of Charles Law**

### How kinetic theory explains Charles Law

An increase in temperature increases the average kinetic energy and the molecules of gas move more rapidly and collide with the walls of the container.

At constant pressure, the volume of the container increases with an increase in temperature. Conversely, a decrease in temperature lowers the average kinetic energy of the gas molecules and also lowers the frequency of collision of gas molecules.

### Calculations based on Charles Law

1. The volume of a given mass of gas at 30^{0}C is 600cm^{3}, if the pressure remains constant, determine the temperature when the volume is 800cm^{3}

According to Charles Law \( \frac{V_1}{T_1}= \frac{V_2}{T_2} \)

V_{1} = 600cm^{3}, T_{1} = (30^{0}C + 273)k = 303k

V_{2} = 800cm^{3}, T_{2} = ?

T_{2} = \( \frac{T_1 V_2}{V_1} \)

T_{2} = \( \frac{303 \; \times \; 800}{600} \)

T_{2} = 404K or 131^{0}C

2. If 35cm^{3} of a gas is heated from 14^{0}C to 55^{0}C, what is the new volume using Charles Law.

:- \( \frac{V_1}{T_1}= \frac{V_2}{T_2} \)

V_{1} = 35cm^{3}

T_{1} = (14 + 273)k = 287k

V_{2} = ?

T_{2} = (55 + 273)k = 328k

V_{2} = \( \frac{V_1 \; \times \; T_2}{T_1}\\ = \frac{35 \; \times \; 328}{287}\\ = \scriptsize 40cm^3 \)

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