What is the pressure of a gas being pumped out of a vessel over time?

[zorro]

Homework Statement

A vessel of volume V_o contains an ideal gas at pressure P_o and temperature T. Gas is continuosly 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 the pressure of the gas as a function of time.

The Attempt at a Solution

http://latex.codecogs.com/gif.latex?\int_{V_{o}}^{V}dV&space;=&space;\int_{0}^{t}-rdt\\V-V_{o}=-rt\PV=nRT\\\frac{\partial&space;V}{\partial&space;t}P+\frac{\partial&space;P}{\partial&space;t}V=0\\\--rP+\frac{\partial&space;P}{\partial&space;t}%28V_{o}-rt%29=0\\\frac{dP}{P}=\frac{rdt}{V_{o}-rt}\\&space;P=\frac{P_{o}}{rt-V_{o}}\\Correct\answer\P=P_{o}e^{-\frac{rt}{V_{o}}}

Any help appreciated.

 

[hikaru1221]

You are messing up math and physics Physics says, PV = const (assume T=const) only applies to a fixed amount of ideal gas with uniform pressure. Here you apply that to a fixed amount of gas but it has non-uniform pressure: first an amount -dV of pressure Po is taken out, then the rest (Vo+dV) is expanded to Vo so it has pressure P1 different from Po; another -dV of pressure P1 taken out results in pressure P2 of the rest, and so on. So the fixed amount there is the rest + ALL of the gas taken out. But each amount of the gas taken out each time has different pressure, and that's the problem.

I suggest that you consider the remaining amount of gas right after one pumping and right before the next pumping:
_ Right after the pumping: Vo+dV, P (dV = -rdt)
_ Right before the next pumping: Vo, P+dP
Because this amount is fixed throughout the considered time period & it has uniform pressure at any point of time during the considered time period, PV=const can be applied.

 

[zorro]

PV = const (assume T=const) only applies to a fixed amount of ideal gas with uniform pressure. Here you apply that to a fixed amount of gas but it has non-uniform pressure[\QUOTE]

Do you mean the ideal gas equation PV=nRT is valid only if pressure is uniform throughout?
It is not mentioned in the question that pressure inside the vessel is non-uniform.

In PV=nRT, P and V are variables. It is given in the question that T remains constant throughout. So its time derivative is 0.

_ Right after the pumping: Vo+dV, P (dV = -rdt)
_ Right before the next pumping: Vo, P+dP [\QUOTE]

Right after the pumping, is the volume inside Vo - dV or Vo + dV? Volume decreases right?
Right before the next pumping, how will the volume become Vo? It should be the same Vo - dV

 

[hikaru1221]

Do you mean the ideal gas equation PV=nRT is valid only if pressure is uniform throughout?

Yes. The meaning of this equation is that an amount of gas of n moles, occupying a volume V, accompanied by pressure P, has temperature T. Pressure must be uniform throughout the gas.

It is not mentioned in the question that pressure inside the vessel is non-uniform.

We have to assume that pressure inside the vessel is uniform (this is a reasonable assumption). However, the pressure of all the gas both inside and outside the vessel is non-uniform. So if you consider the gas inside the vessel, you can apply the equation, but with caution as this amount is not fixed at all time. If you consider all the gas (inside + outside), you cannot apply the equation.

_ Right after the pumping: Vo+dV, P (dV = -rdt)
_ Right before the next pumping: Vo, P+dP [\QUOTE]

Right after the pumping, is the volume inside Vo - dV or Vo + dV? Volume decreases right?
Right before the next pumping, how will the volume become Vo? It should be the same Vo - dV

It depends on how you write dV: if dV=-rdt, it's Vo+dV; if dV=rdt (less meaningful), it's Vo-dV.
When you pump some gas out, you have to open the vessel and take the amount -dV out (dV = -rdt) by some way. Then after finishing taking -dV out, you close the vessel. During that time, the remaining gas inside the vessel which initially occupies a volume Vo+dV will spread over the whole vessel and therefore expand to Vo.

 

[zorro]

However, the pressure of all the gas both inside and outside the vessel is non-uniform.If you consider all the gas (inside + outside), you cannot apply the equation.

The question says "The pressure of the gas being taken out equals the pressure inside the vessel". This clearly means that dV amount of gas taken out has the same pressure as that of the gas inside the vessel at that instant which is Po - dP
Then how come the pressure inside and outside the vessel is non-uniform?

 

[hikaru1221]

The question says "The pressure of the gas being taken out equals the pressure inside the vessel". This clearly means that dV amount of gas taken out has the same pressure as that of the gas inside the vessel at that instant which is Po - dP
Then how come the pressure inside and outside the vessel is non-uniform?

You open the vessel, and take -dV out. At this point, you can imagine that there is nothing change in the state of the gas, just that -dV is out; the gas inside is still at (Vo+dV ; P), the amount -dV has the same pressure P. That what's the question means by the same pressure. But after this, you close the vessel, and the gas inside starts to expand, and its state changes to (Vo ; P+dP).

Otherwise, the vessel will be the one that changes: its volume and thus Vo change, in order to maintain P at Po. Imagine that you first take out the amount (-dV)1, and this amount is at Po. If the pressure inside remains the same as this (-dV)1, then its pressure is Po. Then the 2nd amount (-dV)2 is taken out at the same pressure of the gas inside, i.e. Po, and again, the remaining gas has pressure Po, and so forth. That means, pressure doesn't change! That is contradictory with the answer, right? Of course, that's because a gas pump doesn't work that way.

P.S.: Opening vessel and closing vessel is just a way to understand and approach the situation. The real phenomenon is not like that. The gas is continuously pumped out, and the vessel always opens during the pumping. However because the change in state is so fast, we can assume that change is instantaneous. That means, the "opening and closing" is also instantaneous, and including them makes no disruption to the real pumping process.

 

[zorro]

Right after the pumping: Vo+dV, P (dV = -rdt)
_ Right before the next pumping: Vo, P+dP

I just have one last doubt-

Initially, the pressure and volume of the container are Po and Vo.
When volume changes to Vo + dV right after the first pumping, pressure should be Po and just before the next pumping, volume and pressure should be Vo and Po + dP resp.
Why do we take it as a variable P?

Do we consider an intermediate process and not the initial?

 

[hikaru1221]

It's the same as you move from x1 to x2. After moving a bit by dx=vdt, you are at x1+dx. Then why take x as variable?
x is variable, representing some physical property, which may change or remains constant depending on each situation, while x1 or x2 is value of x at some state. Moreover, after moving by another dx, you're at x1+2dx, and so forth. It's inconvenient to work with x1 + ...dx, right? It is not to mention that x1 + ...dx cannot present a general process, which is instead expressed by x+dx. The notation x+dx already includes x1+dx.

 

[zorro]

All your explanations are first class. If I were in your place, I would quit studying at the first place and start tutoring
Thanks alot!