If there is container with a moveable piston. If heat is supplied to it the gas in container expands right raising the volume. The gas expands raises the piston against gravity (the gas does positive work on the weight). according to Newton's third law the gas exerts force PA on the piston but the piston also exerts and equal and opposite force. My question is how will piston even move if they exerts equal and opposite forces on one another.

The second question is that in above case if we calculate work done on the gas it comes out to be $W=\int( -PA)\;dx$. How? In my physics book they are saying that minus sign is due to the fact that force exerted by the piston on the gas is in opposite direction to displacement. But in many other places i have seen that work done formula is given by $W = \int(PA)\;dx$. How?

  • $\begingroup$ Your first question is one that has been answered many times in Introductory Physics forums. This question would more properly be submitted under the Newtonian Mecahnics tag. Regarding your 2nd question, there are two sign conventions. One is for the work done by the system on the surroundings and the other is for work done by the surroundings on the system. Both conventions are used in thermodynamics and physical chemistry texts, and you just have to get used to determining which specific convention is being used in the material you are reading. $\endgroup$ Commented Oct 29, 2016 at 11:19
  • $\begingroup$ sir but in my text book of physics HRK VOLUME 2 they are taking the first one formula of work and then they are saying that according to this formula work done in expansion is negative and work done in contraction is positive. there is nothing about work done on the gas or by the gas. $\endgroup$
    – Mehwish
    Commented Oct 29, 2016 at 11:27
  • $\begingroup$ They are using the convention that work done on the system is positive and work done by the system is negative. This is consistent with the first law of thermodynamics being written as $\Delta U=Q+W$. $\endgroup$ Commented Oct 29, 2016 at 11:35
  • $\begingroup$ Your question about how the piston moves has been asked and answered here several times. For example, this post. I'm not sure what the confusion is in your statement "there is nothing about work done on the gas or by the gas" In all cases the system in question is the gas. $\endgroup$
    – garyp
    Commented Oct 29, 2016 at 13:09

1 Answer 1


I will answer the second question first. Work is a dot product. $W=\vec{F}\cdot \vec{x}$. This is works out nicely if $\vec{x}$ is fixed and $\vec{F}$ is not dependent on $x$. In your problem, $\vec{x}$ is indeed fixed but technically $\vec{F}$ is not. For the solution we must now do an integral $\int_{x_o}^{x_f} \vec{F}(x)\cdot d\vec{x}$. Let's look at your problem outside with physically what is happening. The work here is done by the gas. As the gas expands, a force is applied to the movable piston. The piston is moved in the direction of the force. So the dot product is positive and there is work done by the gas. The negative sign in the math comes in because the asked question is the work done on the gas.

Think of work as that thing which transforms energy from one form to another. In this example you increase the temperature of the gas. This gives the gas more energy. This increased energy is shown in the system as in increase of pressure of the gas. This is a force applied to the piston which has the force of gravity also applied. These two forces are for the moment not balanced. So work is done by the gas on the piston so that the forces to cancel out. And therefore some energy will move from the gas(in the form of pressure) into the gravitational potential energy of the piston.

Now as for Newton's third law. It is not actually correct. There are many examples where the third law does not hold. Where there is acceleration occurring, be weary of trusting the third law. Don't get me wrong, in static situations, the third law is a great tool and should be used. But in dynamics is can usually be thrown away.

  • $\begingroup$ I beg to differ. The third law is always correct, even in situations where acceleration is occurring. When you do a force balance on a body, you only include the forces acting on that body, not the forces that that body exerts on other bodies. $\endgroup$ Commented Oct 29, 2016 at 11:49
  • $\begingroup$ First, E&M proves that the third law is incorrect. The problem with the third law is it looks at forces. And forces are not the actual physics. $\vec{F} = -\nabla V$. We use forces early is physics because it minimizes the math which if taught too early obfuscates the physics fundamentals. In that paradigm the resultant laws can only be trusted to a point. It this case we are speaking of the third law, but no physics law should ever be completely above suspicion. $\endgroup$
    – Michael
    Commented Oct 29, 2016 at 12:14
  • $\begingroup$ @John Rennie What are your thoughts on this? $\endgroup$ Commented Oct 29, 2016 at 13:06
  • $\begingroup$ @ChesterMiller I agree with you about N3L in this example. You need to have a look at this reference and the links within it physics.stackexchange.com/questions/43269/… and this one animations.physics.unsw.edu.au/jw/Newton.htm $\endgroup$
    – Farcher
    Commented Oct 29, 2016 at 23:33
  • $\begingroup$ Here is a nice paper on the violation of the third law. phys.org/news/2015-05-newton-law-broken.html $\endgroup$
    – Michael
    Commented Oct 30, 2016 at 1:24

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