A monoatomic ideal gas, initially at temperature $T_1$ is enclosed in a cylinder fitted with frictionless…

A monoatomic ideal gas, initially at temperature $T_1$ is enclosed in a cylinder fitted with 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 ratio $T_2 / T_1$ is
  1. $\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{2 / 3}$
  2. $\left[\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right]^{2 / 3}$
  3. $\left[\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right]^{1 / 2}$
  4. $\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{1 / 2}$

Solution

For an adiabatic process $\begin{aligned} & \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\ \therefore \quad & \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1} \end{aligned}$ For a monoatomic gas, $\gamma=\frac{5}{3}$ $\begin{array}{ll} & \Rightarrow \gamma-1=\frac{5}{3}-1=\frac{2}{3} \\ & \text {Volume }=\text { Area } \times \text { Length, } \\ \therefore \quad & \mathrm{V}_1=\mathrm{AL}_1 \text { and } \mathrm{V}_2=\mathrm{AL}_2 \\ \therefore \quad & \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\left[\frac{\mathrm{AL}_1}{\mathrm{AL}]_2}\right]^{2 / 3}=\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{2 / 3} \\ \therefore \quad & \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right)^{\frac{2}{3}} \end{array}$ ^

Asked in: MHT CET 2024 (15 May Shift 2)

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