Question

A vessel of volume V₀ contains an ideal gas at pressure p₀ and temperature T. Gas is continuously pumped out of this vessel at a constant volume-rate dV/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.

Answer 1

$\begin{array}{l}\text{Let}\mathit{\text{P}}\text{be the pressure and}\mathit{\text{n}}\text{be the number of moles of gas inside the}\mathrm{vessel\; at\; any\; given\; time}t.\\ \\ \text{Suppose a small amount of gas of}\mathit{\text{dn}}\text{moles is pumped out and the decrease in pressure is}\mathit{\text{dP.}}\\ \\ \text{Applying equation of state to the gas inside the vessel, we get}\\ (P-dP)V_o=(n-dn)RT\\ \Rightarrow P{V}_o-dP{V}_o=nRT-dnRT\\ \text{But}P{V}_o=nRT\\ \Rightarrow V_{o}dP=dnRT...\left(1\right)\\ \text{The pressure of the gas taken out is equal to the inner pressure}\text{.}\\ \text{Applying equation of state, we get}\\ (P-dP)dV\text{=}dnRT\\ \Rightarrow PdV=dnRT...\left(2\right)\\ \text{From eq.}\left(1\right)\text{and eq.}\left(2\right)\text{, we get}\\ V_{o}dP=PdV\\ \Rightarrow \frac{dP}{P}=\frac{dV}{V_o}\\ \frac{dV}{dt}=r\\ \Rightarrow dV=rdt\\ \\ \Rightarrow dV=-rdt...\left(3\right)\left[\mathrm{Since}\mathrm{pressures}\mathrm{decreases},\mathrm{rate}\mathrm{is}\mathrm{negative}\right]\\ \text{Now,}\\ \frac{dP}{P}=\frac{-rdt}{V_o}\left[\mathrm{From}\mathrm{eq}.\left(3\right)\right]\\ \left(a\right)\\ {\text{Integrating the equation P = P}}_0\text{to P = P and time t = 0 to t = t, we get}\\ \underset{P_o}{\overset{P}{\int }}=\underset{0}{\overset{t}{\int }}\\ \Rightarrow \ln P-\ln P_o=-\frac{rt}{V_o}\\ \Rightarrow \ln \left(\frac{P}{P_o}\right)=-\frac{rt}{V_o}\\ \Rightarrow P=P_o{e}^{\frac{-rt}{V_o}}\end{array}$

$\begin{array}{l}\left(b\right)\\ \text{P =}\frac{P_o}{2}\\ \frac{P_o}{2}=P_o{e}^{\frac{-rt}{V_o}}\\ \Rightarrow e^{\frac{rt}{V_o}}=2\\ \Rightarrow \frac{rt}{V_o}=\ln 2\\ \Rightarrow t=\frac{V_o\ln 2}{r}\end{array}$