Initially the pressure of 1 mole of an ideal gas is $10^5 \mathrm{Nm}^{-2}$ and its volume is 16 litre. When…
Initially the pressure of 1 mole of an ideal gas is $10^5 \mathrm{Nm}^{-2}$ and its volume is 16 litre. When it is adiabatically compressed, its final volume is 2 litre. Work done on the gas is $\left[\right.$ molar specific heat at constant volume $\left.=\frac{3 R}{2}\right]$
72 kJ
7.2 kJ
720 kJ
360 kJ
Solution
$\begin{aligned}
& \text { } P_1=10^5 \mathrm{Nm}^{-2}, \mathrm{~V}_1=16 \text { lit, } \mathrm{V}_2=2 \text { lit, } \mathrm{C}_{\mathrm{v}}=\frac{3 \mathrm{R}}{2} \\
& \therefore \mathrm{C}_{\mathrm{p}}=\mathrm{C}_{\mathrm{v}}+\mathrm{R}=\left(\frac{3 \mathrm{R}}{2}+\mathrm{R}\right)=\frac{5 \mathrm{R}}{2} \\
& \therefore \gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\frac{\frac{5 \mathrm{R}}{\frac{2}{3 R}}}{\frac{3 \mathrm{R}}{2}}=\frac{5}{3}
\end{aligned}$ For adiabatic process, $P_1 V_1^\gamma=P_2 V_2^\gamma$
$\Rightarrow \quad P_2=P_1\left(\frac{V_1}{V_2}\right)^\gamma=10^5\left(\frac{16}{2}\right)^{\frac{5}{3}}=32 \times 10^5 \mathrm{Nm}^{-2}$
$\therefore \quad$ Work done, $\mathrm{W}=\frac{\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2}{\gamma-1}$
$=\frac{10^5 \times 16 \times 10^{-3}-32 \times 10^5 \times 2 \times 10^{-3}}{\left(\frac{5}{3}-1\right)}$
$=-7.2 \mathrm{~kJ}$ (by the gas)
$\therefore \quad$ Work done on the gas is 7.2 kJ