what I want to ask is that the $dx$ in that formula is the displacement of piston or the displacement of the center of mass of the gas. also is there any situation where this clarity is useful.
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$\begingroup$ It depends entirely on the system you are describing. The $dx$ is just the differential length over which a force is applied to a body to do work. $\endgroup$– Matt HansonCommented May 18 at 4:24
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$\begingroup$ For the piston-cylinder situation, dx=dV/A. $\endgroup$– Chet MillerCommented May 18 at 10:39
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3 Answers
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The parameter $dx$ is the infinitesimal displacement where the force is applied. The key concept is that the system particles affected by the force are all changing energy in concert through the displacement—that's the distinguishing aspect of thermodynamic work.
Perhaps an object is being accelerated or raised, for example, respectively gaining kinetic or gravitational energy. Or a body deforming in place is gaining strain energy. Or gas molecules are receiving a momentum kick from an advancing boundary, gaining thermal energy. Or an interface is being extended, gaining surface energy. Switch the direction of energy transfer if either the force or the displacement (but not both) is reversed.
(Note that the minus sign in your relation means that work is being defined by the energy the system transfers to the external mechanism applying the force through the displacement. The associated First Law for this convention is $\Delta U=Q-W$, with change in system internal energy $\Delta U$, heat delivered to the system $Q$, and work done by the system $W$.)
answered May 18 at 5:13
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Great conceptual doubt. Firstly, lets get clear with what work exactly is.
Work done on a system is not force times displacement of center of mass. It is actually Force times the displacement of the point where the force is being applied(also known as point of application of force).
So if you take the gas as a system, the work done will be force times displacement of the point of application, that it the displacement of piston, not Center of mass.
Now, we cannot just say that W = F$\Delta x$ because we are assuming that the Force is constant throughout the process. It is the same thing like speed, you cannot say speed at a instant = $\frac{distance}{time}$. To find the speed at a instant, you look at a small displacement in a small time period. This gives you a much more accurate measure of speed at a given "instant".
Similarly, the Force isn't constant, but you can say it is constant for a very short time period, call is dt. "dx" just means a infinitely small change in x.
So you can say work done in dt time is: dW = Fdx. (F=external force on the piston at an instant, W=Work done by external agent)
We just used the formula of work, just for a very small displacement. Now even if the force keeps changing, we can find it. Usually we can derive a equation of Force in terms of x then integrate the equations and get the total work.
So the actual meaning of dx is just the small displacement of piston
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For a cylinder fitted with a piston, the pressure volume work, where the version of the first law is $\Delta U=Q-W$, is
$$dW=P(V)dV=\frac{F(x)}{A}Adx=F(x)dx$$
So $dx$ is the displacement of the piston.
The displacement of the center of mass (COM) will be less than the displacement of the piston, $dx$. See Figure below for an example.
