Question

One mole of a diatomic ideal gas $(\gamma =1.4)$ is taken through a cyclic process starting from point $A$. The process $A\to B$ is an adiabatic compression, $B\to C$ is an isobaric expansion, $C\to D$ is an adibatic expansion and $D\to A$ is an isochoric process. The volume ratios are $\frac{V_A}{V_B}=16,\frac{V_C}{V_B}=2$ and temperature $T_B=909\text{K},T_C=1818\text{K}$. Find the efficiency of the cycle, when work done $W_{AB}=-1522.5R\text{J},W_{BC}=909R\text{J},W_{CD}=2567.5R\text{J}$

• A. $32.84\%$
• B. $61.40\%$
• C. $28.56\%$
• D. $35.61\%$

Answer 1

The correct option is B $61.40\%$


Given, $T_B=909\text{K},T_C=1818\text{K}$

Also, $\frac{V_A}{V_B}=16$ and $\frac{V_C}{V_B}=2$

So, for adiabatic expansion process $C\to D$
$\frac{V_A}{V_C}=\frac{V_D}{V_C}=8$
Also, $T_C{V}_C^{\gamma -1}=T_D{V}_D^{\gamma -1}\therefore T_D=T_C{\left(\frac{V_C}{V_D}\right)}^{\gamma -1}=1818{\left(\frac{1}{8}\right)}^{\frac{2}{5}}\approx 791\text{K}$

Heat absorbed from $B\to C$
$Q_1=n{C}_P(T_C-T_B)=n\frac{\gamma R}{\gamma -1}(T_C-T_B)=\frac{\frac{7R}{5}}{\frac{2}{5}}(1818-909)=\frac{7R}{2}\times 909\approx 3182R\text{J}$
For process $D\to A$, heat released
$Q_2=n{C}_V(T_D-T_A)=n\frac{R}{\gamma -1}(T_D-T_A)=\frac{5R}{2}(791-300)=\frac{5R}{2}\times 491\approx 1227.5R\text{J}$

Work done during process $AB,BC,CD$ are given as
$W_{AB}=-1522.5R\text{J}$
$W_{BC}==909R\text{J}$
$W_{CD}=2567.5R\text{J}$
Therefore,
$W_{net}=W_{AB}+W_{BC}+W_{CD}=-1522.5R+909R+2567.5R$
$=1954R\text{J}$
Thus, the efficiency of cycle $\left(\eta \right)=\frac{W_{net}}{Q_1}\times 100=\frac{1954R}{3182R}\times 100\approx 61.4\%$
Hence, option (b) is the correct answer.