A monoatomic ideal gas, initially at temperature ' $\mathrm{T}_1$ ' is enclosed in a cylinder fitted with a…

A monoatomic ideal gas, initially at temperature ' $\mathrm{T}_1$ ' is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature ' $\mathrm{T}_2$ ' by releasing the piston suddenly $L_1$ and $L_2$ are the lengths of the gas columns before and after the expansion respectively. The $\frac{\mathrm{T}_2}{\mathrm{~T}_1}$ is
  1. $\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{2 / 3}$
  2. $\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{1 / 2}$
  3. $\left[\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right]^{1 / 2}$
  4. $\left[\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right]^{2 / 3}$

Solution

In an adiabatic process $\mathrm{T}_1 \mathrm{~V}_1^{-1}=\mathrm{T}_2 \mathrm{~V}_2^{-1}$ For an ideal mono atomic gas the no of degree of freedom is 3 $\begin{aligned} & \mathrm{Y}=1+\frac{2}{\mathrm{f}}=1+\frac{2}{3} \\ & \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{\mathrm{Y}-1}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{2 / 3}=\left(\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right)^{2 / 3} \end{aligned}$ Since volume is proportional to length due to area being constant.

Asked in: MHT CET 2021 (23 Sep Shift 1)

Practice more Chemical Thermodynamics questions on Aicharya