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JEE Main 2020 (7 Jan, Shift 2) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2020 (7 Jan, Shift 2)

Q211JEE Main 2020 · 7 JAN · SHIFT 2Difficult

Thermodynamic Processes

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\frac{C_p}{C_v} = \gamma for this gas then a good estimate for τ2τ1\frac{\tau_2}{\tau_1} is given by:

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Answer: (a)

  1. Mean collision time τ=λvrms\tau = \frac{\lambda}{v_{rms}}, with λ∝V\lambda \propto V (fixed number of molecules) and vrms∝Tv_{rms} \propto \sqrt{T}, so τ∝VT\tau \propto \frac{V}{\sqrt{T}}.
  2. Adiabatic: T∝V1−γT \propto V^{1-\gamma}, so τ∝V⋅Vγ−12=Vγ+12\tau \propto V \cdot V^{\frac{\gamma-1}{2}} = V^{\frac{\gamma+1}{2}}.
  3. Volume doubles: τ2τ1=21+γ2\frac{\tau_2}{\tau_1} = 2^{\frac{1+\gamma}{2}}
  4. So the answer is (a).
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