Question

One mole of an ideal gas is contained under a weightless piston of a vertical cylinder at a temperature ${27}^{\circ }\text{C}$. The space over the piston opens into the atmosphere. What work has to be performed in order to increase isothermally, the gas volume under the piston to twice its initial value, by slowly raising the piston. (Neglect friction)

• A. $750\text{J}$
• B. $775\text{J}$
• C. $765\text{J}$
• D. $790\text{J}$

Answer 1

The correct option is C $765\text{J}$


Let $A$ be the area of cross section of the piston.
Let $P$ = Inside pressure of container
$P_0$ = atmosphric pressure
$F$ = Force exerted by agent to increase volume
So, $F+PA=P_{0}A$
i.e $F=P_{0}A-PA$
Now, work done by the agent
$W={\int }_{V_1}^{V_2}Fdx={\int }_{V_1}^{V_2}(P_0-P)Adx={\int }_{V_1=V}^{V_2=2V}(P_0-P)dV$
$\Rightarrow W={\int }_V^{2V}{P}_{0}dV-{\int }_V^{V_2}PdV$
$=P_0(2V-V)-{\int }_V^{2V}nRT\frac{dV}{V}$
[isotherma, hence $T$ is constant]
$W=P_{0}V-nRT\ln 2$
Initially, since the piston is in equilibrium, $P_i=P_0$
$\Rightarrow V=\frac{nRT}{P_i}=\frac{nRT}{P_0}$
So, $W=nRT-nRT\ln 2$
$=1\times 8.314\times 300(1-\ln 2)\{\begin{array}{c}\text{From question}\\ n=1;T=27+273=300\text{K}\\ R=8.314{\text{J mol}}^{-1}{\text{K}}^{-1}\end{array}$
$=765.35\text{J}$
$\approx 765\text{J}$