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While learning about the different thermodynamic processes like the adiabatic, isothermal and isobaric process I stumbled upon a question which just left me with a major doubt about the role of external pressure on the work done by the system. In the question, we were given the atmospheric pressure and the change in volume of a gas on heating which was enclosed in a container with a movable piston. The work done by the gas was calculated as P(atmospheric) x ∆V, saying that it is an isobaric process, but I couldn't really understand why is it so? Since we are heating the gas there must be some increase in pressure so how can it be isobaric? And if we go by the same logic then even an adiabatic process carried out under similar conditions should ultimately become an isobaric process? Now, what is wrong and what is correct? What is an isobaric process ultimately?

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In that example, as the gas makes work, it would lose internal energy, and so the pressure would go down (even if the system absorbs heat at constant temperature, because for an idea gas $PV=nRT$. So in order to keep it isobaric, you need to heat the gas to larger temperatures as it expands, to keep the pressure constant.

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The work done by the gas was calculated as P(atmospheric) x ∆V, saying that it is an isobaric process, but I couldn't really understand why is it so? Since we are heating the gas there must be some increase in pressure so how can it be isobaric?

The work done by (or on) the gas only depends on the external pressure only. Work is force x displacement, which is equivalent to pressure x $\Delta V$ and that pressure is the external pressure. In this context "isobaric" means constant external pressure. It does not necessarily refer to the gas pressure.

In order for the gas pressure to be constant, heat has to added very slowly to the gas (reversibly) so that the gas is always in internal thermal and mechanical equilibrium, i.e. the difference between the external pressure and gas pressure is always infinitesimally small as the gas expands. If you have an ideal gas you can apply then apply the equation of state to the gas during the entire process.

If the gas is heated rapidly, then the gas will not be in internal equilibrium, i.e., there will be pressure and temperature gradients within the gas, so that the gas pressure is undefined and one cannot say it (the gas) is at constant pressure.

And if we go by the same logic then even an adiabatic process carried out under similar conditions should ultimately become an isobaric process? Now, what is wrong and what is correct?

No, an adiabatic process can not be constant pressure. The difference between and adiabatic and isobaric process is that there is no heat transfer in an adiabatic process. All of the work done in an adiabatic expansion is at the expense of internal energy. Per the first law for an adiabatic process $\Delta U=-W$ and since for an ideal gas $\Delta U=C_{V}\Delta T$ that means the temperature will drop. For an ideal gas:

$$PV=nRT$$

If $T$ drops and $V$ increases, with $nR$ constant, pressure $P=\frac {nRT}{V}$ of the gas will drop.

What is an isobaric process ultimately?

Ultimately it is at minimum a constant external pressure process, but if carried out reversibly, it is both a constant gas and external pressure process.

Hope this helps.

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Here are some questions to assist in your understanding.

Question 1: I have a massless rod, and two people are pulling on opposite ends of the rod. Even if the rod is accelerating, according to Newton's second law, the forces exerted by the two people on the opposite ends are (a) the same (b) not the same

Question 2: I have a massless frictionless piston being pushed on one side by an ideal gas contained inside a cylinder and on the other side by an external force. Even if the piston is accelerating, according to Newton's second law, the force exerted by the ideal gas on the inside face of the piston and by the external force on the outside face of the piston are (a) the same (b) not the same

Question 3: True of false For a reversible process, an ideal gas inside a cylinder satisfies PV = nRT

Question 4: True of false For an irreversible (non-quasistiatic) process, an ideal gas inside a cylinder satisfies PV=nRT

Please answer these, and we can continue.

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