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

Answer: (D)

$\frac{1}{18}$

$\text{Solution}:$

Let the number of moles of gas be n and the temperature be $T_0$ in the state A.

Now, work done during the cycle

$W=\frac{1}{2}\times (2V_0-V_0)(2P_0-P_0)=\frac{1}{2}{P}_0{V}_0$

For the heat $\Delta Q_1$ given during the process $A\to B$, we have

$\Delta Q_1=\Delta W_{AB}+\Delta U_{AB}\Delta W_{AB}$= area under the straight line AB =$\frac{1}{2}(P_0+2P_0)(2V_0-V_0)=\frac{3P_0{V}_0}{2}$

Applying equation of state for the gas in the state A and B.

$\frac{P_0{V}_0}{T_0}=\frac{\left(2P_0\right)\left(2V_0\right)}{T_B}$

$⟹T_B=4T_0$

$\therefore U_{AB}$=$n{C}_v\Delta T$ = $n\frac{5R}{2}(4T_0-T_0)$ = $\frac{15nR{T}_2}{2}$=$\frac{15P_0{V}_0}{2}$

$\therefore \Delta Q_1=\frac{3}{2}{P}_0{V}_0+\frac{15}{2}{P}_0{V}_0=9P_0{V}_0$

Obviously, the processes $B\to C$ and $C\to A$ involve the abstraction of heat from the gas.

Efficiency=$\frac{Workdonepercycle}{Totalheatsuppliedpercycle}$

i.e., $\eta =\frac{\frac{1}{2}{P}_0{V}_0}{9P_0{V}_0}=\frac{1}{18}$

$\text{Hence D is the correct option}$