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Adiabatic process: work done and change in temperature

This type of question, asked 3 times: 2 May 2023 S1, 10 May 2024 S1, 13 Apr 2026 S2. All 3 questions below, oldest first.

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  1. Question 1

    2 May 2023 S1

    30
    One mole of an ideal gas at an initial temperature of TT K does 6R6R J of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/35/3, the final temperature of gas will be (R=8.31R = 8.31 J mol−1^{-1} K−1^{-1})

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  2. Question 2

    10 May 2024 S1

    30
    Gas at temperature T is adiabatically expanded to double its volume. Work done (γ=32\gamma=\frac{3}{2}) is

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  3. Question 3

    13 Apr 2026 S2

    115
    A sample of gas at temperature TT is adiabatically expanded to double its volume. The work done by the gas in the process is (given, γ=32\gamma = \dfrac{3}{2}):

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