Question:
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
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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
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
First law of thermodynamics: \(dQ=dU+dW\).
Step 2: Substitution
With \(dW=0\), \(dU=dQ\lt0\).
Step 3: Conclusion
For an ideal gas internal energy depends only on temperature, so a decrease in U means the temperature falls.
Final Answer:
The temperature of the gas decreases. \[ \boxed{\text{(D) }\text{T will decrease}} \]
First law of thermodynamics: \(dQ=dU+dW\).
Step 2: Substitution
With \(dW=0\), \(dU=dQ\lt0\).
Step 3: Conclusion
For an ideal gas internal energy depends only on temperature, so a decrease in U means the temperature falls.
Final Answer:
The temperature of the gas decreases. \[ \boxed{\text{(D) }\text{T will decrease}} \]
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