I had a question posed to us in Physics Class about determining the pressure inside a container when a volume of gas is expelled out of it at a contant rate, the question was as follows:
A Vessel of Volume $V_0$ contains an Ideal gas at pressure $P_0$ at temperature $T$. Gas is continuously pumped out of this vessel at a constant volume-rate with respect to time as $\frac{dV}{dt} = r$, under Isothermal conditions.The Pressure outside the container is also $P_0$ (The initial pressure of the container) and doesn't change. (Note: The gas is forcefully removed, the container doesn't change its shape.) Find:
- The Pressure inside the Container $P_t$ as a function of time $t$.
- The time taken for half the original Gas to be pumped out.
I had tried taking $PV = constant$ for the gas inside the container and having differentiated with respect to time and further solved by substituting $\frac{dV}{dt} = r$, integrating under the limits of $P_0$ to $P_t$ and taking $V = V_0$ as the volume of the container doesn't change, I had gotten my answer to be: $$P_t= P_0e^{(\frac{-rt}{V_0})} – [Equation:1]$$ Which happens to match the answer of the question in class, but then I had a doubt after some thought over my method:
My solution seems wrong to me since I have taken $PV = constant$ but the number of moles of Gas inside the container also changes so infact it would have been $PV/n = constant$ but I amn't sure how to relate the change in Moles and Volume to solve the differential equation I would get if I differentiated $PV/n = constant$ with respect to time. I didn't proceed with the second half since I felt the solution as $[Equation:1]$ to the question must then have been wrong.
It would helpful if I could get to know how to approach this question the right way and solve it. Thank you!
Best Answer
The molar volume of the gas in the container is RT/P, so the molar density is P/RT. If r is the volume rate of flow out of the tank, then the rate of change in the number of moles n in the tank is $$\frac{dn}{dt}=-r\frac{P}{RT}$$From the ideal gas law, the rate of change of pressure is $$\frac{dP}{dt}=\frac{RT}{V_0}\frac{dn}{dt}$$where $V_0$ is the tank volume (a constant). If we combine the above two equations, we obtain: $$\frac{dP}{dt}=-\frac{r}{V_0}P$$