Question:easy
An ideal gas having pressure P, volume V and temperature T undergoes a thermodynamic process in which \(dW=0\) and \(dQ\lt 0\). Then, for the gas
An ideal gas having pressure P, volume V and temperature T undergoes a thermodynamic process in which \(dW=0\) and \(dQ\lt 0\). Then, for the gas
Show Hint
Apply the first law with dW equal to zero.
Updated On: Nov 12, 2026
- V will increase
- T will increase
- P first decreases and then increases
- T will decrease
Show Solution
The Correct Option is D
Solution and Explanation
Step 1: Relation
$dU=nC_{v}dT$ for an ideal gas.
Step 2: Sign
Since $dU=dQ\lt0$, $dT\lt0$.
Step 3: Check
No work means constant volume, so V does not increase and T cannot rise.
Final Answer:
The gas cools, so T decreases. \[ \boxed{\text{(D) }\text{T will decrease}} \]
$dU=nC_{v}dT$ for an ideal gas.
Step 2: Sign
Since $dU=dQ\lt0$, $dT\lt0$.
Step 3: Check
No work means constant volume, so V does not increase and T cannot rise.
Final Answer:
The gas cools, so T decreases. \[ \boxed{\text{(D) }\text{T will decrease}} \]
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