Newton's third law in thermodynamics

Question

In my physics textbook, the foundation for work is derived using newton's third law, where F_surr = - F_gas, where surrounding represents a piston-cylinder device and gas is pushing against the inner surface of the piston towards the right.

My questions are:

1. Given this information, it is obvious that W_surr = - W_gas. So, shouldn't the terms cancel each other and result in zero acceleration of the piston, i.e., the piston remains stationary?
2. I understand that newton's third law is applicable for forces acting on different objects in contrast to second law which is used for analysing forces acting on a single object. So using second law, if i were to dissect the system and piston into two bodies and assuming no friction, we will have F_surr acting towards the left on the gas and F_gas acting on the piston towards the right. So, I am confused here, shouldn't F_surr being the only force on the gas cause it to compress and F_gas on the piston cause the piston to move towards the right?
3. Finally, during a quasi-equilibrium process, will there be any accelerations of the piston at all at each equilibrium states or is the equilibrium state like a point at which the system and surrounding are in complete mechanical equilibrium and are stationary?

Essentially, I am having a hard time applying newton's second and third law to derive work equations.

Answer 1

For question 1, when we do a force balance on a body, we include only the forces exerted by other bodies on that body, and not forces which it exerts on other bodies. So the two action-reaction forces you refer to do not cancel.

For question 2, in the first law of thermodynamics energy balance, $\Delta U=Q-W$, W is the work that the gas does on the piston. So $W=F_{gas}\delta$, where $\delta$ is the displacement to the right. The magnitude of the surroundings force on the gas $F_{surr}$ is equal to the magnitude of the gas force on the surroundings $F_{gas}$ (although it is pointing to the left). Based only on the magnitude of the surrounding force, the work that the gas does on the surroundings is $W=F_{surr}\delta$.

For question 3, in a quasi-static process, the piston is not accelerating and gaining kinetic energy; the system and surroundings are essentially in mechanical equilibrium (except for a slight, quantitatively insignificant, difference).

Answer 2

Newton's 3rd law only describes what I consider to be the same force that acts on two objects, not two different forces. And forces not acting on the same object can never sum or cancel each other.

The 3rd law says that the Earth pulls on the Moon, and the Moon pulls on the Earth. It's just as if there is a rope stretched between them pulling on both equally. But there is only one rope. The Earth has an unbalanced force on it and the Moon has an unbalanced force on it.

On your example, look only at the piston and you'll see that it has a force $F_{gas} = P_{gas} A$ on one end, and $F_{atm} = P_{atm} A$ on its other end. The net force on the piston is the difference between the two, and this is the force that is doing work on the piston.

Separately, the high pressure gas has a force equal to $F_{gas}$ being exerted on it by the piston, and doing negative work on it. But this has nothing to do with the previous paragraph about forces and work on the piston.