Question

An ideal gas having pressure P, volume V and temperature T undergoes a thermodynamic process in which \(dW = 0\) and \(dQ < 0\). Then, for the gas

• A. T will increase.
• B. V will increase.
• C. T will decrease.
• D. P may increase or decrease.

Hint:

With no work done, the first law gives dU = dQ, and dU depends on temperature change.

Answer 1

The Correct Option is C

Step 1: Understand the concept
The first law of thermodynamics says \(dQ = dU + dW\). For an ideal gas, the internal energy depends only on temperature: \(dU = nC_v\,dT\).

Step 2: Apply the conditions
Since \(dW = 0\), we have \(dU = dQ\). Since \(dQ < 0\), we get \(dU < 0\), so \(nC_v\,dT < 0\).

Step 3: Conclusion on temperature
\(dT < 0\), so the temperature \(T\) decreases, option (C).

Step 4: Check the other options
No work is done, so the volume does not change (an isochoric process), which rules out (B). With constant \(V\), \(P \propto T\), so the pressure must fall along with \(T\), and it cannot rise, which rules out (D). Option (A) has the wrong sign for temperature change.

Final Answer:
Temperature decreases. This is option (C). \[ \boxed{\text{(C) }T\ \text{will decrease}} \]