Question

Find the efficiency of the thermodynamic cycle shown in figure for an ideal diatomic gas.

• A. $\frac{1}{4}$
• B. $\frac{1}{9}$
• C. $\frac{1}{8}$
• D. $\frac{1}{18}$

Answer 1

The correct option is D $\frac{1}{18}$

From the $P-V$ graph,
$W_{A\to B}=\frac{1}{2}\times P_0{V}_0+P_0{V}_0=\frac{3}{2}{P}_0{V}_0$
$W_{B\to C}=0$
$W_{C\to A}=P_0(-V_0)=-P_0{V}_0$
Using ideal gas equation,
$T_A=\frac{P_0{V}_0}{nR}$, $T_B=\frac{4P_0{V}_0}{nR}$ & $T_C=\frac{2P_0{V}_0}{nR}$
$U_{A\to B}=n{C}_v(T_B-T_A)$
$=n\left(\frac{5R}{2}\right)\left(\frac{3P_0{V}_0}{nR}\right)=\frac{15}{2}{P}_0{V}_0$
$U_{B\to C}=n{C}_v(T_C-T_B)$
$=n\left(\frac{5R}{2}\right)\left(\frac{-2P_0{V}_0}{nR}\right)=-5P_0{V}_0$
$U_{C\to A}=n{C}_v(T_A-T_C)$
$=n\left(\frac{5R}{2}\right)\left(\frac{-P_0{V}_0}{nR}\right)$
$=-\frac{5}{2}{P}_0{V}_0$

$\therefore Q_{A\to B}=W_{A\to B}+U_{A\to B}$
$=\frac{3}{2}{P}_0{V}_0+\frac{15}{2}{P}_0{V}_0$
$=9P_0{V}_0$
$Q_{B\to C}=W_{B\to C}+U_{B\to C}$
$=0+(-5P_0{V}_0)$
$=-5P_0{V}_0$
$Q_{C\to A}=W_{C\to A}+U_{C\to A}$
$=-P_0{V}_0-\frac{5}{2}{P}_0{V}_0=-\frac{7}{2}{P}_0{V}_0$
Therefore,
$Q_{\text{absorbed}}=9P_0{V}_0$
Total work done $W=\frac{P_0{V}_0}{2}$
Efficiency $=\frac{W}{Q_{\text{absorbed}}}=\frac{\frac{P_0{V}_0}{2}}{9P_0{V}_0}=\frac{1}{18}$