How to Find Air Pressure in a Vessel Being Evacuated Over Time?

AI Thread Summary
The discussion focuses on finding the air pressure in a vessel being evacuated over time, assuming an isothermal process. The initial pressure is denoted as po, with a constant evacuation rate C. The ideal gas equation is applied, leading to a differential equation that incorporates the rate of change of moles, dn/dt, which is negative due to the decreasing amount of gas. The correct formulation results in dp/dt = -PC/V, allowing for the calculation of pressure as a function of time. The solution is confirmed as accurate by participants in the discussion.
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Homework Statement


Find the pressure of air in a vessel being evacuated as a function of evacuation time t. The vessel volume is V, the initial pressure is po. The process is assumed to be isothermal, and the evacuation rate equal to C and independent of pressure.
Note: The evacuation rate is the gas volume being evacuated per unit time, with that volume being measured under the gas pressure attained by that moment.


Homework Equations





The Attempt at a Solution


By the ideal gas equation,
pV=nRT
\frac{dp}{dt}=\frac{RT}{V}\frac{dn}{dt}
\frac{dp}{dt}=\frac{RT}{MV}\frac{dm}{dt}
where m is the mass of gas in the vessel and M is the molar mass of the gas.

Now i am stuck here, i don't know what to do next.
 
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The process removes C units of volume of gas each second. If C units of volume are removed at pressure P and temperature T, how many moles are removed?
 
TSny said:
The process removes C units of volume of gas each second. If C units of volume are removed at pressure P and temperature T, how many moles are removed?

##\frac{PC}{RT}## ?
 
Pranav-Arora said:
##\frac{PC}{RT}## ?

Yes. Use this for dn/dt in your diff eq. [Edit: think about the sign of dn/dt]
 
Last edited:
TSny said:
Yes. Use this for dn/dt in your diff eq. [Edit: think about the sign of dn/dt]

\frac{dn}{dt}=-\frac{PC}{RT}
(Negative sign because the moles keeps on decreasing)
\frac{dp}{dt}=\frac{RT}{V}\frac{-PC}{RT}
\frac{dp}{dt}=\frac{-PC}{V}
Solving this, i get the right answer. Thanks a lot TSny! :smile:
 
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