Question

An ideal gas taken through process ABCA. If net heat supplied is 5 J, find work done in process $C \to A$.

• A. -5 J
• B. -10 J
• C. -15 J
• D. -20 J

Hint:

In a cycle, $\Delta U = 0$, so $Q = W$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
In a cyclic process, the net internal energy change is zero. Therefore, \( \Delta Q = \Delta W \).

Step 2: Key Formula or Approach:
1. First Law for Cycle: \( Q_{\text{net}} = W_{AB} + W_{BC} + W_{CA} \).
2. Work \( W = \int P dV \).

Step 3: Detailed Explanation:
Looking at the graph (V on y-axis, P on x-axis):
- Process AB: Isobaric expansion at \( P = 10 \).
\( W_{AB} = P(V_B - V_A) = 10(2 - 1) = 10 \text{ J} \).
- Process BC: Isochoric cooling (Volume is constant at \( V=2 \)).
\( W_{BC} = 0 \text{ J} \).
- Net Heat supplied \( Q_{\text{net}} = 5 \text{ J} \).
From First Law: \( 5 = 10 + 0 + W_{CA} \).
\[ W_{CA} = 5 - 10 = -5 \text{ J} \]

Step 4: Final Answer:
The work done in process C to A is -5 J.