# If work done on a system is taken as work of external pressure, then where am I going wrong?

The following problem is taken from "APTITUDE TEST PROBLEMS IN PHYSICS" by SS KROTOV.
'A vertical thermally insulated cylinder of volume V contains $$n$$ moles of an ideal monatmoic gas under a weightless piston. A load of mass M is placed on the piston, as a result of which the piston is displaced by a distance h. Determine the final temperature $$T_f$$ of the gas established after the piston has been displaced if the area of the piston is S and the atmospheric pressure is $$p_0$$.'
In the above question as the piston is thermally insulated, so work done on it is totally converted into it's internal energy. $$\Delta U=q-W$$ $$nC_v\Delta T=-W$$ Now work can be calculated using work of external pressure, but we can see external pressure is sum of atmospheric pressure and pressure due of load of mass M.(It is also constant) So
$$p_{ext}=p_0+\frac{Mg}S$$ And hence $$W=p_{ext}\Delta V=(p_0+\frac{Mg}S)Sh$$ where 'Sh' is change in volume. Now after solving the equations I am ending up at a completely different answer, here's why.
The author in the solution has taken work done on the system as work done by gravity i.e W=Mgh, and proceeded. He has ignored the work done by the already existing external atmospheric pressure $$p_0$$.
This is where I always screw up problems in thermodynamics, because my chemistry lecturer tells that work done is defined as $$W=-\int p_{ext}dV$$ whereas my physics lecturer tells that work done is defined as $$W=\int pdV$$ where $$p$$ is the pressure of the gas contained in the cylinder, and is so defined in different books which I use.Hence I always messes up.
If any one can help with how actually is work defined then I will be very grateful to you.
Following is the link to the book where solution is also available at the end.
https://ia801603.us.archive.org/8/items/ProblemsInPhysicsSSKrotov/Problems%20In%20Physics%20SS%20Krotov.pdf
Problem 2.9

## 2 Answers

The ideal gas law is valid only if the gas is in thermodynamic equilibrium, and, if not, the ideal gas law produces an incorrect result for the for the force exerted by the gas on the piston. In this irreversible compression, the gas passes through a sequence of intermediate thermodynamic states that are not at thermodynamic equilibrium, so the ideal gas law can't be used. So, in such a case, it may be necessary, if possible, to use the external pressure to calculate the work.

By Newton's 3rd law, at the interface between the gas and its surroundings, the force that the gas exerts on the surroundings must be equal to the force exerted by the surroundings on the gas. And, by Newton's 2nd law, for the massless piston in this problem, we must have: $$F_g=P_{ext}S=p_0S+mg$$If we multiply this by the piston velocity and integrate with respect to time, we obtain: $$W_{gas}=P_{ext}\Delta V=p_0\Delta V+\frac{mg}{S}\Delta V=-p_oSh-mgh$$where $$W_{gas}$$ is the work done by the gas in the process.

What this shows is that your assessment of this problem was completely correct, and the author (as well as @Vid) are incorrect.

• Hi Chet. Agree. On an even more basic level can we also say that the pressure used to calculate work (be it the gas pressure for a reversible process or the external pressure for any process) is always absolute pressure? To ignore atmospheric pressure is tantamount to calculating work based on gauge pressure. – Bob D May 5 at 12:18
• @BobD I guess that's one way of saying it. – Chet Miller May 5 at 12:26
• @BobD If the fluid is incompressible, you can use gauge pressure. Otherwise, no. – Chet Miller May 5 at 12:42
• @chet miller. Thanks a lot for clarifying my doubt. – sameed hussain May 5 at 14:56
• @ChetMiller If the fluid is incompressible, then there is no pdV work right? – Bob D May 7 at 12:55

Definition $$dW=-pdV$$ is valid for so called p,V,T system, in general gas. You can also use it for any system with intensive variable $$Y$$ and extensive variable $$X$$. Then $$dW=-YdX$$. In thermodynamics you are generally observing only one system, and don't think much on technicalities (how exactly is the force applied, how does apparatus look like, etc..).

For your case, $$dV$$ is change of volume of observed system (gas inside a cylinder) and $$p$$ is pressure of your system, this means pressure of gas inside cylinder. But bear in mind, that during compressing of your air, $$p$$ isn't constant, so $$W$$ is calculated as change of potential energy. External reassure doesn't do any work, since it is already in equilibirum at the begining, eg. it can't compressed anything, becouse there is no $$\Delta p$$.

• In my judgment, this is incorrect, and the OP's assessment is right on target. – Chet Miller May 5 at 10:48