Question

$1litre$ of an ideal gas $(\gamma =1.5)$ at $300K$ is suddenly compressed to half its original volume.
If the original pressure is $100kPa$, find the work done by the gas in the process.

Take: $\sqrt{2}=1.41$

• A. $-82.82J$
• B. $-92.47J$
• C. $-72.47J$
• D. $-42.47J$

Answer 1

Answer: (A)

$-82.82J$

As the process is done suddenly,

It is adiabatic process,$(\gamma =1.5)$

$T{V}^{\gamma -1}=\text{constant}$

$T_1=300K,$

By the above relationship, $T_2=300\sqrt{2}K$

And as $P_1=100kPa$

$n=\frac{PV}{RT}=\frac{100\times {10}^{3}Pa\times {10}^{-3}{m}^3}{300R}$

$W=-\Delta U=-\frac{nR\Delta T}{\gamma -1}=\frac{\frac{1}{3R}\times R\times 300(\sqrt{2}-1)}{0.5}$

$W=-200(\sqrt{2}-1)=-82.82J$

Option $\text{A}$ is the correct answer.