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:

  1. The Pressure inside the Container $P_t$ as a function of time $t$.
  2. 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!

  • $\begingroup$ You have been very perceptive. Have you learned yet about the open-system (control volume) version of the 1st law of thermodynamics? $\endgroup$ Jan 24, 2022 at 11:06
  • $\begingroup$ Hey @ChetMiller, I hadn't come across the Control Volumes 1st Law of Thermodynamic, but I've gone through that now and seem to understand my question better. Thankyou for you suggestion! $\endgroup$
    – Aaryan
    Jan 24, 2022 at 12:20

1 Answer 1


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: $$\frac{dP}{dt}=-\frac{r}{V_0}P$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.