Thermodynamics

JEE Main level, single correct, medium

140 of 190

Question 140 of 190

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently, the mean collision time between the gas molecule changes from τ1{\tau }_{1} to τ2{\tau }_{2} . If CPCv=γ\dfrac{{C}_{P}}{{C}_{v}}=\gamma for this gas then a good estimate for τ2τ1\dfrac{{\tau }_{2}}{{\tau }_{1}} is given by

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Answer: (D) (12)γ+12{\left(\dfrac{1}{2}\right)}^{\dfrac{\gamma +1}{2}}

Relaxation time (τ)∝VT\left(\tau \right)\propto \dfrac{V}{\sqrt{T}}
And
T∝1Vγ−1T\propto \dfrac{1}{{V}^{\gamma -1}}
τ∝V1+γ−12\tau \propto {V}^{1+\dfrac{\gamma -1}{2}}
τ∝V1+γ2\tau \propto {V}^{\dfrac{1+\gamma }{2}}
τfτi=(2VV)γ+12\dfrac{{\tau }_{f}}{{\tau }_{i}}={\left(\dfrac{2V}{V}\right)}^{\dfrac{\gamma +1}{2}}
τfτi=(2)γ+12\dfrac{{\tau }_{f}}{{\tau }_{i}}={\left(2\right)}^{\dfrac{\gamma +1}{2}}

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