Question

Two moles of Helium gas undergo a reversible cyclic process as shown in figure. Assuming gas to be ideal. What is the work for the process C to D?

• A. -800 Rln2
• B. zero
• C. +200 Rln2
• D. -600 Rln2

Answer 1

The correct option is A -800 Rln2

Solution:- (A) $-800R\left(\ln 2\right)$

Given:- No. of moles $\left(n\right)=2$

As we know that work done in an isothermal reversible expansion is given by-

$W=-nRT\ln \frac{V_f}{V_i}=-nRT\ln \frac{P_i}{P_f}$

For work done from $C$ to $D$,

$T=400K$

$P_i=2atm$

$P_f=1atm$

$\therefore W_{C\to D}=-2\times R\times 400\times \ln \frac{2}{1}$

$\Rightarrow W_{C\to D}=-800R\left(\ln 2\right)$

Hence the work done for the process $C$ to $D$ is $-800R\left(\ln 2\right)$.