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?

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.

  • Thanks for your answer. You proved mathematically that in massless piston the force due to atmospheric pressure is directly equal to the force exerted by the piston on the gas. Hence the work done is same. Thank you very much! – Kartoos Jul 27 '15 at 7:24
  • Is this the reason why Pressure remains constant in this case where atmospheric displacement is equal to gas displacement so pressure becomes equal? – Hydrous Caperilla Apr 2 at 1:44
  • @HydrousCaperilla I'm not sure I understand your question. What pressure remains constant? But assuming the external reservoir is very large and the piston moves slowly enough, you would not expect the atmospheric pressure to change as the piston moves. – pwf Apr 2 at 21:52

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.

  • No,massless piston cannot even defined because that would mean infinite acceleration for a given force. but thats not the key misunderstanding though. – Paul Mar 4 '15 at 17:12
  • But it's not really defined as massless, its mass is just ignored. I think it is the key issue. The asker would have a valid concern, as I pointed out, if the mass of the piston was not ignored. – Tim Clark Mar 4 '15 at 17:27
  • From defination of work in thermodynamics the initial volume and final volume of the gas would matter and also the process ofcourse .I understand your point,you are saying it would be very hard for the gas to push to piston if it is massive.But thats not what we want,we just need the change in volume. – Paul Mar 4 '15 at 17:50
  • "We just need the change in volume." Agreed, but why is this the case? The work done on the gas by atmospheric pressure is: $P\Delta V$ = $PA\Delta x$ = $F\Delta x$ where F is the total force on the piston from atmospheric pressure. If the piston's mass is not ignored, this is not going to equal to the work done by atmospheric pressure on the gas for the reason explained in my answer. So one reason that we just need the change in volume is that we ignore the mass of this piston. – Tim Clark Mar 4 '15 at 18:44
  • In your explanation you said that we consider work done by force F even though it is actually done by contact forces on m due to M which is equal to (m/(m+M))*F. Still it is not correct and rigorous. I am still confused. – Kartoos Mar 5 '15 at 9:08

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.

  • You are correct that it should be "newton's third law" instead of "third law of motion" in my question but my point is that in physics we choose first of all a system , then we try to find out the forces acting on the system and then evaluate total external force acting on it and then work done by those forces. If we take the gaseous molecules as the system then since atmospheric molecules do not even interact with it that is why I have a confusion whether work is actually done by forces due to atmospheric molecules forces or due to the forces exerted by piston's atoms – Kartoos Mar 5 '15 at 8:43
  • Actually your point is that volume change matters in thermodynamic work but I want an explanation using newton's laws and with the help of basic definition of work involving force and displacement – Kartoos Mar 5 '15 at 9:19
  • @user74571:grc.nasa.gov/WWW/k-12/airplane/Images/work2.gif – Paul Mar 7 '15 at 2:57

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