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: Process identification
Zero work for a gas means no volume change, so the process is isochoric. Heat is removed from the gas.

Step 2: Energy
All heat removed comes out of the internal energy, so the molecules lose kinetic energy and the temperature drops.

Step 3: Pressure
At constant volume $P/T = \text{constant}$, so $P$ also drops.

Step 4: Answer
Option (C): $T$ decreases.

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