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JEE Advanced 2023 · Paper 1, Physics question 6

One mole of an ideal gas expands adiabatically from an initial state (TA,V0)(T_A, V_0) to final state (Tf,5V0)(T_f, 5V_0). Another mole of the same gas expands isothermally from a different initial state (TB,V0)(T_B, V_0) to the same final state (Tf,5V0)(T_f, 5V_0). The ratio of the specific heats at constant pressure and constant volume of this ideal gas is γ\gamma. What is the ratio TA/TBT_A/T_B?

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Answer and worked solution

Official answer: A

Adiabatic gas. TVγ−1TV^{\gamma-1} is constant:
TAV0γ−1=Tf(5V0)γ−1  ⟹  TA=5γ−1Tf.T_AV_0^{\gamma-1}=T_f(5V_0)^{\gamma-1}\implies T_A=5^{\gamma-1}T_f.

Isothermal gas. The temperature does not change, so TB=TfT_B=T_f.

TATB=5γ−1.\frac{T_A}{T_B}=5^{\gamma-1}.

Topics:Thermodynamic processes
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