Question

A fixed mass of oxygen gas performs a cyclic process ABCA as shown in the figure. Efficiency of the cycle will be

• A. $\frac{3\ln (3)-2}{5+3\ln (3)}$
• B. $\frac{3\ln (4)-2}{4+5\ln (3)}$
• C. $\frac{3\ln (4)-3}{4+5\ln (3)}$
• D. $\frac{3\ln (3)-1}{6+3\ln (4)}$

Answer 1

Answer: (A)

$A\to B$ isochoric as $P$ is directly proportional to $T$ $W_{AB}=0$
$Q_{AB}=n{C}_v(3T-T)$
$=\left(\frac{5}{2}\right)nR2T=5nRT$
$B\to C$ isothermal and thus $Q=W$ as there won't
be any change in internal energy. $W_{BC}=nR3T\ln \left(\frac{P_B}{P_C}\right)=3nRTln3$
$=3nRT\ln 3=Q_{BC}$
$C\to$ Aisobaric process $W_{CA}=P\left(V_A-V_C\right)=nR\left(T_A-3T\right)=-2nRT$
Efficiency of a cyclic process is given by $\eta =\frac{\text{ Work done in cyclic process }}{\text{ Heat supplied to the system }}=\frac{W}{Q_{\text{in }}}$
Efficiency $\eta =\frac{3nR{T}_A\ln 3-2nR{T}_A}{5nR{T}_A+3nR{T}_A\ln 3}$
$=\frac{3\ln 3-2}{5+3\ln 3}$