Thermodynamics

JEE Main level, single correct, medium

161 of 190

Question 161 of 190

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as Vq{V}^{q} , where VV is the volume of the gas. The value of qq is:
(γ=CPCv)\left(\gamma =\dfrac{{C}_{P}}{{C}_{v}}\right)

Show answer and solution

Answer: (D) γ+12\dfrac{\gamma +1}{2}

For an adiabatic process TVγ−1=T{V}^{\gamma -1}= constant.
We know that average time of collision between molecules
τ=1nπ2vrmsd2\tau =\dfrac{1}{n\pi \sqrt{2}{v}_{rms}{d}^{2}}
where, n=n= number of molecules per unit volume Vrms={V}_{rms}= rms velocity of molecules
As n∝1Vn\propto \dfrac{1}{V} and vrms∝T{v}_{rms}\propto \sqrt{T}
τ∝VT\tau \propto \dfrac{V}{\sqrt{T}}
Thus, we can write
n=K1V−1n={K}_{1}{V}^{-1} and Vrms=K2T12{V}_{rms}={K}_{2}{T}^{\dfrac{1}{2}}.
where, K1{K}_{1} and K2{K}_{2} are constants.
For adiabatic process, TVγ−1=T{V}^{\gamma -1}= constant. Thus, we can write
τ∝VT−12∝V(V1−γ)−12\tau \propto V{T}^{-\dfrac{1}{2}}\propto V{\left({V}^{1-\gamma }\right)}^{\dfrac{-1}{2}}
or τ∝Vγ+12\tau \propto {V}^{\dfrac{\gamma +1}{2}}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Thermodynamics (Thermodynamic Processes), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.