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Suppose there is an ideal, monatomic gas contained in a cylinder with a moveable piston and you bring that system (system = gas only) through some process. If that process changes the volume (or temperature) of the gas, the pressure remains constant while the temperature (or volume) changes.

Why does the use of a heavy, moveable piston ensure that any gas process will be isobaric?

I know that it is the case, but do not understand why. Is their a qualitative, microscopic explanation (modeling the gas as particles colliding with the walls of the container and with the piston) that can make sense of this?

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  • $\begingroup$ It's not necessarily isobaric throughout the entire process although, in the end, the amount of work done (the change in potential energy of the piston) will be the same as if it were isobaric. $\endgroup$ – Chet Miller Dec 7 '18 at 13:02
  • $\begingroup$ @ChesterMiller Does isobaric always refer to the external pressure exerted on the gas? Although the work done is always based on the external pressure, can we not differentiate the external pressure, which is isobaric, and the pressure of the gas, which may or may not be isobaric (pressure gradients within) depending on whether or not the process is quasi-static processes? $\endgroup$ – Bob D Dec 7 '18 at 21:51
  • $\begingroup$ @ Bob D For an irreversible expansion, the gas does not satisfy the ideal gas law because viscous stresses (which depend on the rate of volume change) also contribute to the force per unit area exerted by the gas on the inside face of the piston. So we can't use the ideal gas law to establish the gas force per unit area at the inside piston face where the displacement occurs and the actual work is being done. However, for a massless frictionless piston, it will always be equal to the external pressure, which is constant. (Continued) $\endgroup$ – Chet Miller Dec 7 '18 at 22:56
  • $\begingroup$ @Bob D If the piston has mass, the force the gas exerts must also accelerate the piston in any irreversible process. However, eventually, the motion of the piston will be damped out by viscous stresses in the gas. But, during part of the expansion, the gas actually does more work than just elevating the piston and pushing back the atmosphere. But, once the piston motion has been damped out, the work that the gas has done is just that required to raise the piston and push back the atmosphere. $\endgroup$ – Chet Miller Dec 7 '18 at 23:00
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I think the Kinetic Theory of Gases is a good intro.

Pressure is due to random motion of constituent particles. Moving walls "strikes" the particles increasing their energy. Changing the temperature adds or removes energy as well. Particles bounce off of the boundaris, changing their momentum to be equal and opposite. Pressure is related to force, force is change in momentum over time. The change in direction is our change in momentum. The relevant time scale is how long it takes a particle to leave one wall, bounce off the opposite wall, and then return. So in the expression $F=\Delta p/\Delta t$, $\Delta p$ is not directly effected by the change in length, but $\Delta t$ is, and this changes the Force and therefor the pressure.

The random motion of the particles can change because the moving walls do work on them. It can also change because of temperature. If the changes to the motion due to work and temperature cancel out the right way, pressure can be preserved.

The Ideal Gas Law gives some idea of how that happens:

$$PV=NkT$$.

P is pressure, V is volume, N is number of particles, k is Boltzman's constant, and T is absolute temperature in Kelvin.

Taking differentials. $$VdP+PdV=dNkT+NkdT$$

We are not adding particles so $dN=0$. We are holding pressure constant, so $dP=0$. We end up with:

$$PdV=NkdT$$

So constant pressure requires a balancing act between changes in volume and changes in temperature is the kinetic theory of gases implies.

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You do not really need a microscopic explanation in this case (although Romero's answer is nice). Simply, any excess in pressure in your system can be accomodated by pushing the piston.

The pressure acting on your system at the beginning is the total force $Mg$ (weight of the piston of mass $M$) divided surface $S$ (area of the cylinder base) plus atmosferic pressure $p_0$ (note that here the weight of the piston is irrelevant, it only changes the total pressure but not the principle...!)

In the final state, the pressure will be... exactly the same! That the gas expanded does not change the pressure, as $p_0$ and $Mg$ do not change (whereas if the piston were fixed an internal pressure would build up against the un-movable piston). In our case, at equilibrium, the pressure of the gas will balance the weight of the piston (and $p_0$..) so the two pressures are equal. If then the pressure of the gas slightly increases, by expanding a little bit, because of $PV=nRT$ (see Romero's...), such an increase can be vented on a volume change which then brings the pressure back to normal (i.e. balanced with the piston). The same for a decrease of pressure.

What happens in between the initial and final state depends on the transformation. In a violent explosion the process would not be isobaric I guess. But if you evolve slow enough (quasistatically) you can assume at every instant of time the piston and the gas balance out.

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Why does the use of a heavy, moveable piston ensure that any gas process will be isobaric?

The piston does not have to be “heavy” (a relative term, anyway). It simply has to have mass. If it has mass, $m$, it exerts a downward force of $mg$. If that force is uniformly distributed over a surface A, the force per unit area will be $\frac{mg}{A}$. If this weight is placed on top of a gas, the force per unit area is called pressure. If the piston/cylinder is surrounded by the atmosphere, the total external pressure is $\frac{mg}{A}$ + 1 atm.

A process is designated isobaric if the externally applied pressure is constant. For the piston that external pressure does not change, whether the piston is sitting on top of the gas in equilibrium, is accelerating due to a pressure differential, or is reversibly compressing or expanding the gas due to very slowly transferring heat out of or into the gas, respectively, with the surroundings.

Although the external pressure is constant, the pressure of the gas may not be constant except for reversible processes. For irreversible processes, temperature and /or pressure differentials may exist in the gas. Consequently, the ideal gas law would not apply to irreversible processes.

A mechanics of materials analogy to the above is a column subjected to a downward axial load (force). If the force is uniformly distributed over the cross sectional area of the column, the downward force per unit area is called normal compressive stress, $σ_N$. This external stress is constant; provided any axial deformation does not change the area over which the force acts. This is analogous to the piston surfaces and cylinder walls not expanding or contracting during the compression or expansion of the gas, respectively.

Hope this helps.

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