One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and constant volume is 5/3, the final temperature of the gas will be:
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Question

One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and constant volume is 5/3, the final temperature of the gas will be:

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G1 Physics · Accepted Answer

Step 1: First law with work done by the gas positive: Q = ΔU + W. Step 2: Adiabatic process has Q = 0, so ΔU = −W = −6R J. Step 3: For γ = 5/3, Cᵥ = R/(γ − 1) = R/(2/3) = (3/2)R. Step 4: For one mole, ΔU = Cᵥ(T₂ − T₁). Step 5: (3/2)R(T₂ − T) = −6R; cancel R to obtain T₂ − T = −4 K. Final answer: (D) (T − 4) K.