Verified Solution • 2025–26 Session moderate

QPages 34-37: Q37: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q37 Solution: A vessel of volume contains an ideal gas at pressure and…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the given constant volume‑rate $r=\,dV/dt$ as a flow at a fixed pressure (e.g., atmospheric) and write $dn/dt=-rP_0/(RT)$ or assume the vessel volume changes. The correct approach is to recognise that $r$ is the volume of gas removed per unit time at the instantaneous internal pressure $P(t)$, so the mole loss is $dn/dt=-\frac{rP(t)}{RT}$, leading to $dP/dt=-(r/V_0)P$ and the exponential decay.

A vessel of volume V0V_0 contains an ideal gas at pressure p0p_0 and temperature TT. Gas is continuously pumped out of this vessel at a constant volume-rate dV/dt=rdV/dt = r keeping the temperature constant. The pressure of the gas being taken out equals the pressure inside the vessel. Find a) the pressure of the gas as a function of time, b) the time taken before half the original gas is pumped out. Use R=8.3 J K−1 mol−1R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}

Step-by-Step Verified Solution

**Answer:** (a) P=P0e−rt/V0P = P_0 e^{-rt/V_0} (b) t=V0rln⁡2t = \frac{V_0}{r} \ln 2 **Step-by-Step Explanation:** (a) Gas is pumped out at constant volume rate r=dV/dtr = dV/dt. At any instant, the amount of gas in the vessel is n=PV0/(RT)n = PV_0/(RT). The gas being removed carries away moles at rate dn/dt=−r⋅P/(RT)dn/dt = -r \cdot P/(RT) (since volume rr at pressure PP contains rP/(RT)rP/(RT) moles). But n=PV0/(RT)n = PV_0/(RT), so dn/dt=(V0/RT)dP/dtdn/dt = (V_0/RT) dP/dt. Equating:
V0RTdPdt=−rPRT⇒dPdt=−rV0P\frac{V_0}{RT}\frac{dP}{dt} = -\frac{rP}{RT} \Rightarrow \frac{dP}{dt} = -\frac{r}{V_0}P
Solving: P=P0e−rt/V0P = P_0 e^{-rt/V_0}. (b) For half the gas pumped out: P=P0/2P = P_0/2 (since TT constant).
e−rt/V0=1/2⇒t=V0rln⁡2e^{-rt/V_0} = 1/2 \Rightarrow t = \frac{V_0}{r}\ln 2

All Questions in this Chapter

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