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: Understanding the first process \( X \to Y \).
For the process \( X \to Y \), the work done by the gas is 10 J and the heat absorbed is 2 J. According to the first law of thermodynamics: \[ \Delta U = Q - W \] where: - \( \Delta U \) is the change in internal energy, - \( Q \) is the heat absorbed, - \( W \) is the work done by the gas. For the process \( X \to Y \): \[ \Delta U = 2 \, \text{J} - 10 \, \text{J} = -8 \, \text{J} \]
Step 2: Understanding the reverse process \( Y \to X \).
In the reverse process, heat is evolved (i.e., heat is released) and the amount of heat evolved is given as 6 J. The change in internal energy for the reverse process will be the same as for the forward process (since internal energy depends only on the initial and final states, not the path), so: \[ \Delta U = -8 \, \text{J} \] For the reverse process, we know that heat is released, so \( Q = -6 \, \text{J} \) (negative because heat is evolved). Using the first law of thermodynamics for the reverse process: \[ \Delta U = Q - W \] \[ -8 \, \text{J} = -6 \, \text{J} - W \] Solving for \( W \): \[ W = -6 \, \text{J} + 8 \, \text{J} = 2 \, \text{J} \] Thus, the work done on the gas is: \[ W = 14 \, \text{J} \] Final Answer: Work done on the gas is 14 J.