Question

A Carnot engine uses diatomic gas as a working substance. During the adiabatic expansion part of the cycle, if the volume of the gas becomes 32 times its initial volume, then the efficiency of the engine is

• A. 100%
• B. 75%
• C. 50%
• D. 25%

Hint:

For a Carnot cycle, the efficiency is determined solely by the reservoir temperatures. The adiabatic processes link the temperature ratio to the volume ratio ($T_L/T_H = (V_i/V_f)^{\gamma-1}$) or the pressure ratio ($T_L/T_H = (P_L/P_H)^{(\gamma-1)/\gamma}$). Remember the value of $\gamma$ for different gases (monatomic: 5/3, diatomic: 7/5).

Answer 1

The Correct Option is B

Step 1: State the formula for the efficiency of a Carnot engine.
The efficiency, $\eta$, of a Carnot engine is determined by the temperatures of the hot ($T_H$) and cold ($T_L$) reservoirs. \[ \eta = 1 - \frac{T_L}{T_H}. \]

Step 2: Use the equation for an adiabatic process to find the temperature ratio.
The adiabatic expansion occurs between the hot reservoir temperature $T_H$ and the cold reservoir temperature $T_L$. For an adiabatic process, the relationship between temperature and volume is $TV^{\gamma-1} = \text{constant}$. Let the initial volume be $V_i$ (at temperature $T_H$) and the final volume be $V_f$ (at temperature $T_L$). \[ T_H V_i^{\gamma-1} = T_L V_f^{\gamma-1}. \] \[ \frac{T_L}{T_H} = \left(\frac{V_i}{V_f}\right)^{\gamma-1}. \]

Step 3: Determine the value of $\gamma$ for a diatomic gas.
For a diatomic gas, the ratio of specific heats is $\gamma = \frac{C_p}{C_v} = \frac{7}{5}$. Therefore, $\gamma-1 = \frac{7}{5} - 1 = \frac{2}{5}$.

Step 4: Calculate the temperature ratio.
We are given that the volume becomes 32 times its initial volume, so $V_f = 32V_i$. \[ \frac{T_L}{T_H} = \left(\frac{V_i}{32V_i}\right)^{2/5} = \left(\frac{1}{32}\right)^{2/5}. \] Since $32 = 2^5$, we have $\frac{1}{32} = \left(\frac{1}{2}\right)^5$. \[ \frac{T_L}{T_H} = \left( \left(\frac{1}{2}\right)^5 \right)^{2/5} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}. \]

Step 5: Calculate the efficiency.
\[ \eta = 1 - \frac{T_L}{T_H} = 1 - \frac{1}{4} = \frac{3}{4}. \] In percentage terms, the efficiency is $75%$.