Question

Gas ‘A’ undergoes change from state ‘X’ to state ‘Y’. In this process, the heat absorbed and work done by the gas is 10 J and 18 J respectively. Now gas is brought back to state ‘X’ by another process during which 6 J of heat is evolved. In the reverse process of ‘Y’ to ‘X’, the work done is:

• A. 18 J of the work is done by the gas ‘A’.
• B. 2 J of the work is done by the gas ‘A’.
• C. 12 J of the work is done on the gas ‘A’ by the surrounding.
• D. 14 J of the work is done on the gas ‘A’ by the surrounding.

Answer 1

The Correct Option is C

Step 1: Apply the first law of thermodynamics.
The first law of thermodynamics states that: \[ \Delta Q = \Delta U + \Delta W \] where \( \Delta Q \) is the heat absorbed, \( \Delta U \) is the change in internal energy, and \( \Delta W \) is the work done by the gas.
Step 2: Understand the energy changes.
In the process from ‘X’ to ‘Y’, heat absorbed \( \Delta Q = 10 \, \text{J} \), and work done \( \Delta W = 18 \, \text{J} \). In the process from ‘Y’ to ‘X’, heat evolved \( \Delta Q = -6 \, \text{J} \).
Step 3: Work done in reverse process.
Using the first law again for the reverse process, the total energy change is zero, as the system returns to its original state: \[ \Delta Q = \Delta U + \Delta W \] Substitute the values: \[ -6 = \Delta U + \Delta W \] The change in internal energy, \( \Delta U \), remains the same in both processes, so: \[ \Delta W = -12 \, \text{J} \] This indicates that 12 J of work is done on the gas by the surrounding.
Final Answer: (C) 12 J of the work is done on the gas ‘A’ by the surrounding.