Thermodynamics : Work done on gas inside container by the force due to pressure exerted by external atmosphere

Question

Suppose a container is fitted with a massless and frictionless piston lying on a table such that pressure due to gas is greater than pressure due to external atmosphere. Hence piston will move.

Now the molecular motion of gaseous molecules inside container exert forces that will do "work" on the piston and correspondingly by newton's third law it will be equal to the negative of the "work" done by piston on the gas.

Similarly molecular motion due to atmospheric molecules will exert forces on piston that will do "work" on the piston correspondingly by newton's third law it will be equal to the negative of the "work" done by piston on the atmosphere.

My question is that most textbook says that they calculate "work" done by external pressure( i.e. external force) due to atmosphere on the gas but actually the forces due to external atmosphere are not in direct contact with gas since our system (which includes gas only) has only forces exerted by the piston, so how can they calculate "work" done by those forces which do not directly act on the system?

Answer 1

Consider the forces acting on the piston. Since there is no friction, the only forces are $F_{\text{gas on piston}}$ and $F_{\text{atmosphere on piston}}$. So the net force on the piston is $F_{\text{piston, net}} = F_{\text{atmosphere on piston}} + F_{\text{gas on piston}}$.

Since the piston is massless, Newton's 2nd law says $F_\text{piston, net} = m_\text{piston}a = 0$. That means $F_{\text{atmosphere on piston}} = -F_{\text{gas on piston}}$.

Meanwhile, Newton's 3rd law says the force on the gas by the piston is the negative of the force on the piston by the gas, so $F_\text{on gas} = -F_\text{gas on piston}$. Putting that into the previous equation gives $F_{\text{on gas}} = F_{\text{atmosphere on piston}}$.

So, because the piston is massless and frictionless, the force exerted by the atmosphere is the force exerted on the gas. Furthermore, if the piston is rigid, the displacement associated with the force from the atmosphere is the same as the displacement associated with the force on the gas, so the work done by the atmosphere is the work done on the gas.

Answer 2

The keywords are massless and frictionless. A massless and frictionless piston is an idealization. When the atmosphere exerts a force on it, it is perfectly transferred to the gas, which isn't true with a massive piston. We can then calculate the work due to (but not from because, as you said, the atmosphere does not directly interact with the gas) atmospheric pressure on the gas.

If the piston's mass was not ignored, it wouldn't be so easy to get the work due to atmosphere on the gas. Consider an analogous and frictionless situation:

Say we have a block of mass M (analogous to the piston) and a block of mass m (analogous to the gas) to the right of it and in contact with it. If we apply a constant force $F$ (analogous to the atmospheric pressure) in the rightward direction on block M, we do not get the work done on block m due to (not from, as the force is not applied to m) the applied force by $F\Delta x$. The force on block m is actually equal to $\frac{m}{m + M}F$, so the work on m due to $F$ would be found by $\frac{m}{m + M}F\Delta x$. The force on block m is not equal to $F$. Notice though, that we can calculate the work done on m due to the applied force on M, even though the force doesn't act directly on m.

Overall, if we considered a massive piston, we couldn't just take atmospheric pressure $P$ and use $P\Delta V$ to get the work done on the gas due to the atmosphere. But if its mass is ignored, we easily obtain the work due to the atmosphere on the gas from the atmospheric work on the piston.

Answer 3

Suppose a container is fitted with a massless.......

no you cannot have a massless piston because that would mean infinite acceleration of the piston for a given force. i.e, $F=ma$ imples $a$ is infinite when $m=0$.

Now the molecular motion of gaseous molecules inside container exert forces that will do "work" on the piston and correspondingly by third law of motion......

the gases inside do work on the sorrouding not on the piston, and similarly the sorrounding works on the system as a whole (not on the piston) Again,the 3rd law is about forces(action and reaction forces) and its not about negative work done or positive work done.

I dont know why you think direct contact is necessary. The atmosphere is pushing the piston such that it changes the volume of the gas thus it has done work on the system. Its the volume of the gas that matters.