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JEE Main 202427 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
27 January
Shift
Evening Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2024 · 27 January · Evening Shift · Physics · Thermodynamics

During an adiabatic process, the pressure of a gas is found to be...

During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio of CpCv\dfrac{{C}_{p}}{{C}_{v}} for the gas is :

Options

  1. A53\dfrac{5}{3}
  2. B32\dfrac{3}{2}
  3. C75\dfrac{7}{5}
  4. D97\dfrac{9}{7}

Correct answer

B. 32\dfrac{3}{2}

Step-by-step solution

It is given that

P∝T3⇒PT−3=constant...(1)P ∝ {T}^{3} ⇒ P{T}^{- 3} = \text{constant} . . . \left(1\right)

For an adiabatic process,

PVγ=constant...(2)P{V}^{\gamma} = \text{constant} . . . \left(2\right)

Assuming the gas to be a perfect gas, it can be written from equation (2) that

P(nRTP)γ=constant⇒P1−γTγ=constant⇒PTγ1−γ=constantP{\left(\dfrac{nRT}{P}\right)}^{\gamma} = \text{constant} ⇒ {P}^{1 - \gamma}{T}^{\gamma} = \text{constant} ⇒ PT\dfrac{\gamma}{1 - \gamma} = \text{constant}

By comparing with equation(1), we get

⇒γ1−γ=−3⇒γ=−3+3γ⇒3=2γ⇒γ=32⇒ \dfrac{\gamma}{1 - \gamma} = - 3 ⇒ \gamma = - 3 + 3\gamma ⇒ 3 = 2\gamma ⇒ \gamma = \dfrac{3}{2}

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