Question

The cycle is shown in figure is made of one mole of perfect gas in a cylinder with a piston. The processes A to B and C to D are isochoric whereas process B to C and D to A are adiabatic, the work done in one cycle is ($V_A$=$V_B=V$, $V_C$=$V_D=2V$ and ${\rm Y}$=5/3)

• A. $\left[{1-4}^{3/2}\right](P_B-P_A)V$
• B. $\frac{3}{2}\left[{1-3}^{2/3}\right](P_B-P_A)V$
• C. $\frac{3}{2}[1-2^{-2/3}](P_B-P_A)V$
• D. $\frac{5}{2}[1-2^{-2/3}](P_B-P_A)V$

Answer 1

The correct option is C $\frac{3}{2}[1-2^{-2/3}](P_B-P_A)V$

The work integral $W={\int }_{v_i}^{v_f}PdV$.

Using adiabatic condition $P{V}^{\gamma }=constant=k$

$\Rightarrow W=k{\int }_{v_i}^{v_f}\frac{dV}{v^{\gamma }}$

intregal yield $W=\frac{k(V_f^{1-\gamma }-V_i^{1-\gamma })}{1-\gamma }$...eq(1)

Work done in process B to C is $W_{BC}$

where $k=P_B{V}^{\gamma }$ and volume at B and C is V and 2V respectively

$W_{BC}=\frac{k(2V{)}^{1-\frac{5}{3}}-V^{1-\frac{5}{3}}}{1-\frac{5}{3}}=-\frac{3}{2}(2^{-\frac{2}{3}}-1)P_{B}V$

Work done in process D to A is $W_{DA}$

where $k=P_A{V}^{\gamma }$ and volume at B and C is V and 2V respectively

$W_{DA}=\frac{k(V{)}^{1-\frac{5}{3}}-(2V{)}^{1-\frac{5}{3}}}{1-\frac{5}{3}}=-\frac{3}{2}(1-2^{-\frac{2}{3}})P_{A}V$

total work done is $W=W_{BC}+W_{DA}=-\frac{3}{2}(2^{-\frac{2}{5}}-1)P_{B}V+(-\frac{3}{2}(1-2^{-\frac{2}{5}}\left)P_{A}V\right)=\frac{3}{2}(1-2^{-\frac{2}{3}})(P_B-P_A)V$

Hence C option is correct.