Question:
A vessel of volume \( V \) L contains an ideal gas at \( T(K) \). The vessel is partitioned into two equal parts. The volume (in L) and temperature (in K) in each part is respectively:
A vessel of volume \( V \) L contains an ideal gas at \( T(K) \). The vessel is partitioned into two equal parts. The volume (in L) and temperature (in K) in each part is respectively:
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For an ideal gas in a rigid container, partitioning the volume does not affect the temperature if no external heat transfer or work is involved.
Updated On: Mar 18, 2026
- \( V, \frac{T}{2} \)
- \( \frac{V}{2}, T \)
- \( V, T \)
- \( \frac{V}{2}, \frac{T}{2} \)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Partitioning of the Vessel
We are given that a vessel of volume \( V \) L contains an ideal gas at temperature \( T \) K. The vessel is then partitioned into two equal parts. Step 1: Understanding the effect of partitioning
When a vessel is partitioned into two equal parts, the volume of each part will be: \[ V' = \frac{V}{2}. \] Since no heat exchange occurs between the two parts after partitioning, the temperature remains the same. That is, \[ T' = T. \] Step 2: Verifying the correct answer
From the above calculations, the new volume in each partition is \( \frac{V}{2} \) and the temperature remains \( T \). Hence, the correct answer is: \[ \frac{V}{2}, T. \]
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