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JEE Advanced 2023Paper 1
Exam
JEE Advanced
Year
2023
Paper
Paper 1
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2023 · Paper 1 · Physics · Thermodynamics

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)( {T}_{A}, {V}_{0} ) to final state (Tf,5V0)( {T}_{f}, 5{V}_{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}, 5{V}_{0} ). The ratio of the specific heats at constant pressure and constant volume of this ideal gas is γ\gamma. What is the ratio TATB\dfrac{{T}_{A}}{{T}_{B}}?

One mole of an ideal gas expands adiabatically from an initial state ( T _ A , V _ 0 ) to final state ( T _ f , 5 V _ 0

Options

  1. A5γ−1{5}^{\gamma - 1}
  2. B51−γ{5}^{1 - \gamma}
  3. C5γ{5}^{\gamma}
  4. D51+γ{5}^{1 + \gamma}

Correct answer

A. 5γ−1{5}^{\gamma - 1}

Step-by-step solution

As we know for an Adiabatic process

TVγ−1=ConstantT{V}^{\gamma - 1} = \text{Constant}.

One mole of an ideal gas expands adiabatically from an initial state ( T _ A , V _ 0 ) to final state ( T _ f , 5 V _ 0</p><p></p><p>⇒ {T}_{A}{{V}_{0}}^{\gamma - 1} = {T}_{f}{\left(5{V}_{0}\right)}^{\gamma - 1} . . . \left(i\right)

For Isothermal process</p><p></p><p>{T}_{B} = {T}_{f} . . . \left(ii\right)

Dividing equation \left(i\right) by equation \left(ii\right), we get</p><p></p><p>⇒ \dfrac{{T}_{A}}{{T}_{B}} = {5}^{\gamma - 1}

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