Question

Helium gas goes through a cycle ABCDA (consisting of two isochoric and isobaric lines) as shown in figure. Assuming the gas to be an ideal gas. Efficiency of this cycle is nearly:

 

• A. \(9.1%\)
• B. \(12.5%\)
• C. \(10.5%\)
• D. \(15.4%\)

Hint:

For cyclic processes involving ideal gases, consider paths on the P-V diagram to estimate work and heat interactions.

Answer 1

The Correct Option is D

The cycle ABCDA consists of: 
A $\rightarrow$ B: Isochoric process (Volume constant, $V = V_0$). Pressure increases from $P_0$ to $2P_0$. 
B $\rightarrow$ 
C: Isobaric process (Pressure constant, $P = 2P_0$). Volume increases from $V_0$ to $2V_0$. C $\rightarrow$ D: Isochoric process (Volume constant, $V = 2V_0$). Pressure decreases from $2P_0$ to $P_0$. 
D $\rightarrow$ A: Isobaric process (Pressure constant, $P = P_0$). Volume decreases from $2V_0$ to $V_0$. 
1. Work done in each process: 
$W_{AB} = 0$ (Isochoric process) 
$W_{BC} = P_{BC} (V_C - V_B) = 2P_0 (2V_0 - V_0) = 2P_0V_0$ 
$W_{CD} = 0$ (Isochoric process) 
$W_{DA} = P_{DA} (V_A - V_D) = P_0 (V_0 - 2V_0) = -P_0V_0$
Net work done, $W_{net} = W_{AB} + W_{BC} + W_{CD} + W_{DA} = 0 + 2P_0V_0 + 0 - P_0V_0 = P_0V_0$. 
2. Heat input in each process: For Helium (monatomic gas), $C_v = \frac{3}{2}R$ and $C_p = \frac{5}{2}R$. 
Process A $\rightarrow$ B (Isochoric): $Q_{AB} = nC_v \Delta T_{AB} = nC_v (T_B - T_A)$. Using ideal gas law $PV = nRT$, $T = \frac{PV}{nR}$. $T_A = \frac{P_0V_0}{nR}$, $T_B = \frac{2P_0V_0}{nR}$. $Q_{AB} = nC_v \left( \frac{2P_0V_0}{nR} - \frac{P_0V_0}{nR} \right) = \frac{C_v}{R} P_0V_0 = \frac{3}{2} P_0V_0$. 
Process B $\rightarrow$ C (Isobaric): $Q_{BC} = nC_p \Delta T_{BC} = nC_p (T_C - T_B)$. $T_C = \frac{2P_0 \cdot 2V_0}{nR} = \frac{4P_0V_0}{nR}$, $T_B = \frac{2P_0V_0}{nR}$. $Q_{BC} = nC_p \left( \frac{4P_0V_0}{nR} - \frac{2P_0V_0}{nR} \right) = \frac{2C_p}{R} P_0V_0 = 2 \cdot \frac{5}{2} P_0V_0 = 5 P_0V_0$. 
Process C $\rightarrow$ D (Isochoric): $Q_{CD} = nC_v \Delta T_{CD}<0$ (Heat rejected). 
Process D $\rightarrow$ A (Isobaric): $Q_{DA} = nC_p \Delta T_{DA}<0$ (Heat rejected). 
Total heat input $Q_{in} = Q_{AB} + Q_{BC} = \frac{3}{2} P_0V_0 + 5 P_0V_0 = \frac{13}{2} P_0V_0$. 
3. Efficiency of the cycle: Efficiency $\eta = \frac{W_{net}}{Q_{in}} = \frac{P_0V_0}{\frac{13}{2} P_0V_0} = \frac{2}{13} \approx 0.1538 \approx 15.4%$. 
Correct Answer: (4) 15.4%