Question:

Carnot cycle of an engine is given below. What is the total work done by the gas in one cycle?

Show Hint

The total work in a Carnot cycle depends only on the isothermal steps. Remember: work is done by the gas in expansion, and on the gas in compression.
Updated On: Apr 23, 2025
  • \( \mu R T_2 \log\frac{V_2}{V_1} - \mu R T_1 \log\frac{V_3}{V_4} \)
  • \( \mu R T_1 \log\frac{V_2}{V_1} - \mu R T_2 \log\frac{V_3}{V_4} \)
  • \( \mu R T_1 \log\frac{V_2}{V_1} + \mu R T_2 \log\frac{V_3}{V_4} \)
  • Zero
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation


The Carnot cycle consists of two isothermal and two adiabatic processes: - \( P \rightarrow Q \): Isothermal expansion at \( T_1 \)\ - \( Q \rightarrow R \): Adiabatic expansion\ - \( R \rightarrow S \): Isothermal compression at \( T_2 \)\ - \( S \rightarrow P \): Adiabatic compression The total work done in the Carnot cycle is the net work done during the two isothermal processes, because no work is done in the adiabatic processes in terms of heat exchange: - Work done during isothermal expansion: \[ W_1 = \mu R T_1 \log\left( \frac{V_2}{V_1} \right) \] - Work done during isothermal compression: \[ W_2 = \mu R T_2 \log\left( \frac{V_3}{V_4} \right) \] Total work done: \[ W = W_1 - W_2 = \mu R T_1 \log\left( \frac{V_2}{V_1} \right) - \mu R T_2 \log\left( \frac{V_3}{V_4} \right) \]
Was this answer helpful?
0
0