Question

For process \( X \to Y \), work done by the gas is 10 J and heat absorbed in the process is 2 J. For the reverse process, heat evolved is 6 J. Find the work done for the reverse process.

• A. Work done on the gas is 14 J
• B. Work done by the gas is 2 J
• C. Work done on the gas is 20 J
• D. Work done by the gas is 12 J

Hint:

In thermodynamic processes, work done and heat transfer are related by the first law of thermodynamics. Pay attention to the sign convention: heat absorbed is positive, and work done by the gas is positive.

Answer 1

The Correct Option is A

Step 1: Analyze the process \( X \to Y \).
For the process \( X \to Y \), the gas performs 10 J of work and absorbs 2 J of heat. According to the first law of thermodynamics: \[ \Delta U = Q - W \] where \( Q \) is the heat absorbed by the system and \( W \) is the work done by the gas. Substituting the given values: \[ \Delta U = 2 \, \text{J} - 10 \, \text{J} \] \[ \Delta U = -8 \, \text{J} \] Thus, the change in internal energy during the process \( X \to Y \) is \( -8 \, \text{J} \).
Step 2: Consider the reverse process \( Y \to X \).
Internal energy is a state function; therefore, the change in internal energy for the reverse process will be equal in magnitude but opposite in sign: \[ \Delta U_{Y \to X} = +8 \, \text{J} \] During the reverse process, 6 J of heat is evolved, which means heat is released by the system. Hence: \[ Q = -6 \, \text{J} \] Applying the first law again: \[ \Delta U = Q - W \] \[ 8 \, \text{J} = -6 \, \text{J} - W \] Solving for \( W \): \[ W = -14 \, \text{J} \] A negative value of work indicates that work is done on the gas.
Step 3: Interpret the result.
Since the sign is negative, the magnitude of the work done on the gas is: \[ \text{Work done on the gas} = 14 \, \text{J} \]
Final Answer: Work done on the gas is \( 14 \, \text{J} \).