Question

State and explain Charle’s law.

Answer 1

At constant pressure, the volume of a given mass of a gas increases or decreases by 1/273 of its
original volume at 0°C for every 1°C increase or decrease in temperature, respectively. If V₀ is the
volume of gas at 0°C and V, is the volume of gas at t°C, (P = constant), then

$$\begin{gathered} {\mathrm{V}}_{\mathrm{t}}={\mathrm{V}}_0\left[1+\frac{\mathrm{t}}{273}\right] \\ {\mathrm{V}}_{\mathrm{t}}={\mathrm{V}}_0\left[\frac{273+\mathrm{t}}{273}\right] \\ {\mathrm{V}}_{\mathrm{t}}={\mathrm{V}}_0\left[\frac{\mathrm{T}}{273}\right]\mathrm{T}=\mathrm{absolute}\mathrm{temperature} \\ \frac{{\mathrm{V}}_0}{273}=\mathrm{Constant} \\ \therefore \mathrm{V}\propto \mathrm{T} \\ \mathrm{If}{\mathrm{V}}_1\mathrm{and}{\mathrm{V}}_2\mathrm{are}\mathrm{the}\mathrm{volume}\mathrm{occupied}\mathrm{by}\mathrm{a}\mathrm{given}\mathrm{mass}\mathrm{of}\mathrm{gas}\mathrm{at}\mathrm{tempreature}{\mathrm{T}}_1\mathrm{and}{\mathrm{T}}_2,\mathrm{then} \\ \frac{{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{V}}_2}{{\mathrm{T}}_2}(\mathrm{P}=\mathrm{Constant}) \end{gathered}$$

Charles' law can otherwise be stated as the volume occupied by a given mass of a gas is directly
proportional to the absolute temperature of the gas at constant pressure.