Question:medium

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:

Updated On: Jun 6, 2026
  • 18 J of the work is done by the gas ‘A’.
  • 2 J of the work is done by the gas ‘A’.
  • 12 J of the work is done on the gas ‘A’ by the surrounding.
  • 14 J of the work is done on the gas ‘A’ by the surrounding.
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
Internal energy (\(U\)) is a state function. This means the change in internal energy for a round trip (from X to Y and back to X) is exactly zero.
We apply the First Law of Thermodynamics to both the forward and reverse processes to find the unknown work value.
Step 2: Key Formula or Approach:
First Law of Thermodynamics (IUPAC convention): \(\Delta U = q + w\).
Here, \(q\) is positive if heat is absorbed, negative if evolved.
\(w\) is positive if work is done ON the gas, negative if work is done BY the gas.
For a cyclic process: \(\Delta U_{X \to Y} = -\Delta U_{Y \to X}\).
Step 3: Detailed Explanation:
Analyze the forward process (\(X \to Y\)):
Heat absorbed, \(q_1 = +10 \text{ J}\).
Work done BY the gas means it expanded, so \(w_1 = -18 \text{ J}\).
Calculate the change in internal energy:
\[ \Delta U_{X \to Y} = q_1 + w_1 = 10 \text{ J} + (-18 \text{ J}) = -8 \text{ J} \] Analyze the reverse process (\(Y \to X\)):
Since internal energy is a state function:
\[ \Delta U_{Y \to X} = -\Delta U_{X \to Y} = -(-8 \text{ J}) = +8 \text{ J} \] Heat is evolved in this step, so \(q_2 = -6 \text{ J}\).
Apply the First Law to find the work \(w_2\):
\[ \Delta U_{Y \to X} = q_2 + w_2 \] \[ 8 \text{ J} = -6 \text{ J} + w_2 \] \[ w_2 = 8 \text{ J} + 6 \text{ J} = +14 \text{ J} \] Since \(w_2\) is positive, it means \(14 \text{ J}\) of work is done ON the gas by the surroundings.
Step 4: Final Answer:
14 J of the work is done on the gas 'A' by the surrounding.
Was this answer helpful?
0