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JEE Main 20207 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
7 January
Shift
Evening Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2020 · 7 January · Evening Shift · Physics · Thermodynamics

Under an adiabatic process, the volume of an ideal gas gets...

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

Options

  1. A22
  2. B12\dfrac{1}{2}
  3. C{\left(\dfrac{1}{2}\right)}^{\gamma}
  4. D(12)γ+12{\left(\dfrac{1}{2}\right)}^{\dfrac{\gamma + 1}{2}}

Correct answer

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

Step-by-step solution

Relaxation time (τ)∝VT\left(τ\right) ∝ \dfrac{V}{\sqrt{T}}
And T∝1Vγ−1T ∝ \dfrac{1}{{V}^{\gamma - 1}}
τ∝V1+γ−12τ ∝ {V}^{1 + \dfrac{\gamma - 1}{2}}
τ∝V1+γ2τ ∝ V\dfrac{1 + \gamma}{2}
τfτi=(2VV)γ+12\dfrac{{τ}_{f}}{{τ}_{i}} = {\left(\dfrac{2V}{V}\right)}^{\dfrac{\gamma + 1}{2}}
τfτi=(2)γ+12\dfrac{{τ}_{f}}{{τ}_{i}} = {\left(2\right)}^{\dfrac{\gamma + 1}{2}}
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