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 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
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Memorize specific heat ratios (\(\gamma\)) for common gases: Monoatomic \(= \frac{5}{3}\), Diatomic \(= \frac{7}{5}\), Polyatomic \(= \frac{4}{3}\). Standard powers of 2 (like \(32 = 2^5\)) frequently appear in such thermodynamic problems to simplify fractional exponents without a calculator.
Updated On: May 10, 2026
- \(100\%\)
- \(75\%\)
- \(50\%\)
- \(25\%\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
We are required to find the efficiency of a Carnot engine given the volume expansion ratio during the adiabatic expansion phase and the type of gas (diatomic).
Step 2: Key Formula or Approach:
The efficiency of a Carnot engine is given by: \[ \eta = 1 - \frac{T_2}{T_1} \] For an adiabatic process, the relationship between Temperature and Volume is: \[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \implies \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1} \] Step 3: Detailed Explanation:
For a diatomic gas, the specific heat ratio \(\gamma\) is \(\frac{7}{5}\).
Therefore, \(\gamma - 1 = \frac{7}{5} - 1 = \frac{2}{5}\).
We are given that the final volume \(V_2 = 32V_1\), meaning the volume ratio \(\frac{V_1}{V_2} = \frac{1}{32}\).
Substitute these values into the adiabatic relation: \[ \frac{T_2}{T_1} = \left(\frac{1}{32}\right)^{\frac{2}{5}} \] Recognizing that \(32 = 2^5\), we get: \[ \frac{T_2}{T_1} = \left(\frac{1}{2^5}\right)^{\frac{2}{5}} = \frac{1}{(2^5)^{\frac{2}{5}}} = \frac{1}{2^2} = \frac{1}{4} \] Now, calculate the efficiency: \[ \eta = 1 - \frac{T_2}{T_1} = 1 - \frac{1}{4} = \frac{3}{4} \] Converting to a percentage: \[ \eta = \frac{3}{4} \times 100\% = 75\% \] Step 4: Final Answer:
The efficiency of the Carnot engine is \(75\%\).
We are required to find the efficiency of a Carnot engine given the volume expansion ratio during the adiabatic expansion phase and the type of gas (diatomic).
Step 2: Key Formula or Approach:
The efficiency of a Carnot engine is given by: \[ \eta = 1 - \frac{T_2}{T_1} \] For an adiabatic process, the relationship between Temperature and Volume is: \[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \implies \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1} \] Step 3: Detailed Explanation:
For a diatomic gas, the specific heat ratio \(\gamma\) is \(\frac{7}{5}\).
Therefore, \(\gamma - 1 = \frac{7}{5} - 1 = \frac{2}{5}\).
We are given that the final volume \(V_2 = 32V_1\), meaning the volume ratio \(\frac{V_1}{V_2} = \frac{1}{32}\).
Substitute these values into the adiabatic relation: \[ \frac{T_2}{T_1} = \left(\frac{1}{32}\right)^{\frac{2}{5}} \] Recognizing that \(32 = 2^5\), we get: \[ \frac{T_2}{T_1} = \left(\frac{1}{2^5}\right)^{\frac{2}{5}} = \frac{1}{(2^5)^{\frac{2}{5}}} = \frac{1}{2^2} = \frac{1}{4} \] Now, calculate the efficiency: \[ \eta = 1 - \frac{T_2}{T_1} = 1 - \frac{1}{4} = \frac{3}{4} \] Converting to a percentage: \[ \eta = \frac{3}{4} \times 100\% = 75\% \] Step 4: Final Answer:
The efficiency of the Carnot engine is \(75\%\).
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