Thermodynamics

JEE Main level, single correct, medium

95 of 190

Question 95 of 190

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 :

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Answer: (B) 32\dfrac{3}{2}

It is given that
P∝T3⇒PT−3=constant...(1)P\propto {T}^{3} \Rightarrow 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} \Rightarrow {P}^{1-\gamma }{T}^{\gamma }=\text{constant} \Rightarrow P{T}^{\dfrac{\gamma }{1-\gamma }}=\text{constant}
By comparing with equation(1), we get
⇒γ1−γ=−3⇒γ=−3+3γ⇒3=2γ⇒γ=32\Rightarrow \dfrac{\gamma }{1-\gamma }=-3 \Rightarrow \gamma =-3+3\gamma \Rightarrow 3=2\gamma \Rightarrow \gamma =\dfrac{3}{2}

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