This question stems from the online thermodynamics notes provided by web.mit.edu. I'm having some trouble understanding the example given of air expanding into a vacuum.

Consider the transient, unsteady process of filling a tank, initially evacuated, from a surrounding atmosphere, which is at pressure $P_0$ and temperature $T_0$. The tank is insulated, so the process is adiabatic.

filling an evacuated tank

The author of the notes states that work of magnitude $P_0V_0$ is done on the system. However, I thought that because the pressure in the tank is zero, the work done ($PdV$) would be zero.

Why is it the case that the work done on the system is $P_0V_0$ and not zero?


1 Answer 1


the work done on the system is has caused it to move $V_0$ volume of gas up to the pressure of $P_0$

As if you had blown a deflated balloon to that size and that pressure, you worked moving air out of you lung to the balloon by the magnitude of $$P_0V_0$$

The fact that the tank is vacated give you the chance to do more work on the gas as opposed to what you think, else if there was some pressure left in the tank the work would be $$W=(P_0-P_{tank})V_0 < the \ emty\ tank\ case$$ Remeber we are considering the gas getting into the tank and how much work has gone into pumping it. this work is coming from the ambient gas sorrunding the system!


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