JEE Challenger
More from Thermodynamics
PhyMedAdv 2023 • Paper 1

Ratio of Initial Temperatures in Adiabatic and Isothermal Expansions

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

Options

A

5γ−15^{\gamma-1}

Correct
B

51−γ5^{1-\gamma}

C

5γ5^{\gamma}

D

51+γ5^{1+\gamma}

Topics & Concepts

ThermodynamicsFirst Law of Thermodynamics

Step-by-Step Solution

To find the ratio of the initial temperatures TA/TBT_{\text{A}} / T_{\text{B}}, we analyze the two thermodynamic processes separately:

1. Adiabatic Expansion: The gas expands adiabatically from the initial state (TA,V0)(T_{\text{A}}, V_0) to the final state (Tf,5V0)(T_{\text{f}}, 5V_0). For a reversible adiabatic process of an ideal gas, the relation between temperature and volume is given by: TVγ−1=constantT V^{\gamma-1} = \text{constant}

Applying this between the initial and final states: TAV0γ−1=Tf(5V0)γ−1T_{\text{A}} V_0^{\gamma-1} = T_{\text{f}} (5V_0)^{\gamma-1}

Solving for TAT_{\text{A}}: TA=Tf(5V0V0)γ−1=Tf⋅5γ−1T_{\text{A}} = T_{\text{f}} \left(\frac{5V_0}{V_0}\right)^{\gamma-1} = T_{\text{f}} \cdot 5^{\gamma-1}

2. Isothermal Expansion: The second gas expands isothermally from the initial state (TB,V0)(T_{\text{B}}, V_0) to the final state (Tf,5V0)(T_{\text{f}}, 5V_0). Since the process is isothermal, the temperature remains constant throughout: TB=TfT_{\text{B}} = T_{\text{f}}

3. Ratio of Initial Temperatures: Taking the ratio of TAT_{\text{A}} to TBT_{\text{B}}: TATB=Tf⋅5γ−1Tf=5γ−1\frac{T_{\text{A}}}{T_{\text{B}}} = \frac{T_{\text{f}} \cdot 5^{\gamma-1}}{T_{\text{f}}} = 5^{\gamma-1}

Therefore, the correct option is (A).

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