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A conducting and closed container of capacity 100 litre contains an ideal gas at a high pressure. Now using a pump, the gas is taken out at a constant rate of 5 litre $\mathrm{s}^{-1}$. Find the time taken in which the pressure will decrease to $\frac{P_{\text {initial }}}{100}$. (Assume isothermal condition)
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$92 \mathrm{~s}$
The gas is taken out at the rate of 5 lit s$^{-1}$. The volume of gas ejected in ' $d t$ ' time is $d(\mathrm{vol})=5 d t$
Moles of gas ejected $=\frac{n}{V}(5 d t)$
$\begin{aligned} & P V=n R T \Rightarrow(d p) V=(d n) R T \\ & \Rightarrow(d p) V=-\left(\frac{n}{V} 5 d t\right) R T \\ & \Rightarrow(d P) V=-5 d t \frac{n R T}{V}=-(5 d t) P \\ & \Rightarrow \frac{-d P}{P}=\frac{5}{V} d t \Rightarrow \frac{d P}{P}=-\frac{5}{100} d t=-\frac{1}{20} d t \\ & \int_{p_i}^{p_f} \frac{d P}{P}=-\frac{1}{20} \int_{t=0}^{t=t} d t \Rightarrow P_f=P_i e^{\frac{-t}{20}} \\ & \Rightarrow \frac{P_i}{100}=P_i e^{\frac{-t}{20}} \\ & t=20 \ln 100=92 \mathrm{sec} .\end{aligned}$
Moles of gas ejected $=\frac{n}{V}(5 d t)$
$\begin{aligned} & P V=n R T \Rightarrow(d p) V=(d n) R T \\ & \Rightarrow(d p) V=-\left(\frac{n}{V} 5 d t\right) R T \\ & \Rightarrow(d P) V=-5 d t \frac{n R T}{V}=-(5 d t) P \\ & \Rightarrow \frac{-d P}{P}=\frac{5}{V} d t \Rightarrow \frac{d P}{P}=-\frac{5}{100} d t=-\frac{1}{20} d t \\ & \int_{p_i}^{p_f} \frac{d P}{P}=-\frac{1}{20} \int_{t=0}^{t=t} d t \Rightarrow P_f=P_i e^{\frac{-t}{20}} \\ & \Rightarrow \frac{P_i}{100}=P_i e^{\frac{-t}{20}} \\ & t=20 \ln 100=92 \mathrm{sec} .\end{aligned}$
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