Question

$n$ moles of a gas filled in a container is in thermodynamic equilibrium initially at temperature $T$. If the gas is compressed quasi-statically and isothermally to half its initial volume, the work done by the atmosphere on the piston is :

• A. $\frac{nRT}{2}$
• B. $-\frac{nRT}{2}$
• C. $nRT(\text{ln}2-\frac{1}{2})$
• D. $nRT\text{ln}2$

Answer 1

The correct option is D $nRT\text{ln}2$

Work done by the gas filled in a container in a isothermal process is given by
$W=nRT\ln \left(\frac{V_f}{V_i}\right)......\left(1\right)$
Let the initial volume of gas $\left(V_i\right)$ be $V$
$\therefore$ from the data given in the question we can say that,
Final volume of gas $\left(V_f\right)=\frac{V}{2}$
From $\left(1\right)$,
$W=nRT\ln \left(\frac{V/2}{V}\right)$
Thus we get, work done by the gas
$W=nRT\ln \left(\frac{1}{2}\right)$
$W=-nRT\ln 2$
So, work done on the gas is $W=nRT\ln 2$
Since the gas is in thermodynamic equilibrium initially, we say that,
work done on the gas = work done by the atmosphere
Hence, work done by the atmosphere $W=nRT\ln 2$
Thus, option (d) is the correct answer.