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QJEEAdv23Ph6af1: JEE Advanced 2023 Physics - Class 11 JEEAdv23Ph6af1 Solution: One mole of an ideal gas expands adiabatically from an initial…

One mole of an ideal gas expands adiabatically from an initial state (TA,V0)\left(T_{\mathrm{A}}, V_0\right) to final state (Tf,5V0)\left(T_{\mathrm{f}}, 5 V_0\right) . Another mole of the same gas expands isothermally from a different initial state (TB,V0)\left(T_{\mathrm{B}}, V_0\right) to the same final state (Tf,5V0)\left(T_{\mathrm{f}}, 5 V_0\right) . 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_{\mathrm{A}} / T_{\mathrm{B}} ?

Multiple Choice Options

  • A 5γ−15^{\gamma-1}
  • B 51−γ5^{1-\gamma}
  • C 5γ5^\gamma
  • D 51+γ5^{1+\gamma}

Step-by-Step Verified Solution

JEE Advanced 2023 Paper 1 Online Physics - Heat and Thermodynamics Question 9 English Explanation

The expansion of an ideal gas can be analyzed using the thermodynamic relations for adiabatic and isothermal processes.

  1. Adiabatic Expansion: For an adiabatic process, we use the relation involving temperatures and volumes at the initial and final states, which is given by :

    TVγ−1=constantTV^{\gamma-1} = \text{constant}

    Therefore, for the adiabatic expansion from (TA,V0)(T_A, V_0) to (Tf,5V0)(T_f, 5V_0) , we have :

    TAV0γ−1=Tf(5V0)γ−1T_A V_0^{\gamma-1} = T_f (5V_0)^{\gamma-1}

    Simplifying, we get : TA=Tf×5γ−1T_A = T_f \times 5^{\gamma-1}

  2. Isothermal Expansion : In an isothermal process, the temperature remains constant. So, TB=TfT_B = T_f for the isothermal expansion from (TB,V0)(T_B, V_0) to (Tf,5V0)(T_f, 5V_0) .

To find the ratio TA/TBT_A / T_B , we substitute the expressions for TAT_A and TBT_B :

TATB=Tf×5γ−1Tf\frac{T_A}{T_B} = \frac{T_f \times 5^{\gamma-1}}{T_f}

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

Therefore, the ratio TA/TBT_A / T_B is 5γ−15^{\gamma-1} , which corresponds to Option A.

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