Question:
The relation obeyed by a perfect gas during an adiabatic process is \(pV^{3/2}\). The initial temperature of the gas is 'T'. When the gas is compressed to half of its initial volume, the final temperature of the gas is
The relation obeyed by a perfect gas during an adiabatic process is \(pV^{3/2}\). The initial temperature of the gas is 'T'. When the gas is compressed to half of its initial volume, the final temperature of the gas is
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Read the adiabatic index from the given p V relation, then use T times V to the power gamma minus one as constant.
Updated On: Nov 12, 2026
- \(4\text{T}\)
- \(2\sqrt{2}\text{T}\)
- \(2\text{T}\)
- \(\sqrt{2}\text{T}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
An adiabatic process obeys \(pV^{\gamma}=\text{constant}\). Comparing with \(pV^{3/2}\) gives \(\gamma=\dfrac{3}{2}\).
Step 2: Relation for temperature
Also \(TV^{\gamma-1}=\text{constant}\), so \(T\sqrt{V}=\text{constant}\).
Step 3: Halve the volume
\(T_2=T\left(\dfrac{V}{V/2}\right)^{1/2}=\sqrt{2}\,T\)
Final Answer:
The final temperature is \(\sqrt{2}\,\text{T}\). \[ \boxed{\text{(D) }\sqrt{2}\text{T}} \]
An adiabatic process obeys \(pV^{\gamma}=\text{constant}\). Comparing with \(pV^{3/2}\) gives \(\gamma=\dfrac{3}{2}\).
Step 2: Relation for temperature
Also \(TV^{\gamma-1}=\text{constant}\), so \(T\sqrt{V}=\text{constant}\).
Step 3: Halve the volume
\(T_2=T\left(\dfrac{V}{V/2}\right)^{1/2}=\sqrt{2}\,T\)
Final Answer:
The final temperature is \(\sqrt{2}\,\text{T}\). \[ \boxed{\text{(D) }\sqrt{2}\text{T}} \]
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