A monatomic gas at $630 \mathrm{~K}$ expands adiabatically to 27 times its initial volume. The final…

A monatomic gas at $630 \mathrm{~K}$ expands adiabatically to 27 times its initial volume. The final temperature of the gas is
  1. $30 \mathrm{~K}$
  2. $130 \mathrm{~K}$
  3. $170 \mathrm{~K}$
  4. $70 \mathrm{~K}$

Solution

For mono-atomic, $\mathrm{f}=3$ $\gamma=1+\frac{2}{\mathrm{f}}=\frac{5}{3}$ For adiabatically $\mathrm{PV}^\gamma=$ constant $\begin{aligned} & (\therefore \mathrm{PV}=\mathrm{n} \mathrm{RT}) \\ & \left(\frac{\mathrm{nRT}}{\mathrm{V}}\right) \mathrm{V}^\gamma=\mathrm{constant} \\ & \mathrm{T}_1 \mathrm{~V}_1{ }^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2 \gamma-1 \\ & \frac{630(\mathrm{~V})^{\gamma-1}}{(27 \mathrm{~V})^{\gamma-1}}=\mathrm{T}_2\end{aligned}$ Final temperature of gas, $T_2=\frac{630}{(27)^{\gamma-1}}$ $=\frac{630}{(3)^{3 \times \frac{5}{3}-1}}=70 \mathrm{~K}$

Asked in: AP EAMCET 2023 (16 May Shift 1)

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