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This way wouldn't the pressure applied by the air on the gas always be equal to the pressure applied by the gas on the air. If this is the case wouldn't the gas always be in equilibrium,even if the air pressure were to decrease,thus there wouldn't be any expansion.

What is going on here?What is the reason for this contradiction?

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  • $\begingroup$ If air pressure decreases then gas would expand until pressure of gas reduces to air pressure. $\endgroup$ – Shreyansh Pathak Mar 18 at 4:54
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You are right if you take a container initially filled with gas at 1 atm. Then when you open the container to the earth atmosphere, you get an equilibrium situation (although not exact).

There are other thermodynamic factors that has to be considered - Temperature and Entropy.

Equilibrium is defined to be when none of these quantities change with time, more precisely when the average fluctuations are zero.

Henceforth, its pretty easy to see why two different systems having equal pressure ONLY need to be in "equilibrium".

To answer your question, any two system need not be in isobaric equilibrium always. Consider your case, here you have two systems,

  1. Atmosphere
  2. The container

Now, you usually don't have control over the atmosphere, but you will usually have full control over the container, in the sense that you can fiddle around with the various thermodynamic quantities like Temperature and Pressure. Suppose you fill the container with idea gas, then it satisfies the relation $PV=NRT$, where you have two independent variables. So to get pressure, P > 1 atm, you could in principle change the volume of container(take a different container) and its temperature so that you get P > 1 atm.

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  • $\begingroup$ Say I increase P from 1atm to 2atm. The air is applying a pressure of 1 atm on the gas as an action which will have a reaction of a pressure of 1atm being applied on the air. The gas is applying 2 atm of pressure on the air as an action which will have the reaction of the air applying 2atm of pressure on the gas. So won't the overall effect be that the air and the gas are applying a common pressure of 3 atm on each other. In such a case how will expansion take place, because if this reasoning is correct then there is no way that either of the pressures can be more the other. $\endgroup$ – Aditya Bharadwaj Mar 18 at 7:47
  • $\begingroup$ So i think what you are essentially asking is "according to netwons third law action and reaction are equal and oppo", so wouldn't they cancel out, only that here we have two actions and their corresponding reactions? Is that the question you have in mind? $\endgroup$ – TheImperfectCrazy Mar 18 at 7:53
  • $\begingroup$ Yes I think thats the case... just draw the free body diagram of the two system and mark the action and reactions on each system, then find the net force on each system, that should clarify this issue. You will find that 2 atm of gas is opposed by only 1 atm so the gas expands due to an access 1 atm at its disposal. $\endgroup$ – TheImperfectCrazy Mar 18 at 7:58
  • $\begingroup$ If up is +ve and down is -ve, we get +2atm(action) pressure plus +1atm(reaction) pressure on the air, adding up to +3atm. For the gas -1atm(action) pressure plus -2atm(reaction) pressure on the gas, addding up to -3atm. So the pressure of the atmosphere on the gas is 3atm and the pressure of the gas on the atmosphere is also 3atm. $\endgroup$ – Aditya Bharadwaj Mar 18 at 8:12
  • $\begingroup$ ohh no no... I should have clarified something sorry... So there systems we are dealing with are not rigid systems meaning that they no longer have rigid boundary as soon as you open the container. Here what you do is consider a flexible diaphragm between the two systems, now draw the free body diagram of this diaphragm. Now you will find that there is an excess pressure of 1 atm on the diaphragm due to the gas in container. This is essentially the principle of measuring pressure as used in barometers. $\endgroup$ – TheImperfectCrazy Mar 18 at 9:24
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In equilibrium, the gas pressure would be equal to the atmospheric pressure.

But, if atmospheric pressure decreases, then the gas now has more pressure than the air around it , and thus it will expand. As, it expands , the pressure of the gas decreases due to the gas law. And when it has expanded to a point, where the gas pressure has reduced enough to become equal to the air pressure, it reaches a new equilibrium

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Per Newton's third law, the pressure at the boundary between the gas and the air is always the same regardless of whether the gas and air are in equilibrium. However, only if the gas is in equilibrium with the air will the pressure of the gas away from the boundary be the same. It the gas is not in equilibrium with the air, there will be pressure gradients beyond the boundary resulting in expansion or compression of the gas. It may be easier to visualize this if the boundary consisted of a massless piston.

Hope this helps.

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  • $\begingroup$ As per Newton's Third Law if the pressure at the boundary between the gas and the air is the same then won't the pressure of gas beyond the boundary also be the same, as per Newton's Third Law. What is the difference between the boundary and beyond the boundary? $\endgroup$ – Aditya Bharadwaj Mar 18 at 12:44
  • $\begingroup$ @AdityaBharadwaj Do you agree that if I apply heat to the bottom of the container of gas, the temperature and pressure of the gas near the bottom will be greater than the temperature and pressure at the top? That's a pressure and temperature gradient. The gas is not internally in equilibrium. If I remove the heat source the temperature and pressure throughout the gas will eventually be the same, i.e., the gradients disappear. $\endgroup$ – Bob D Mar 18 at 12:52
  • $\begingroup$ Ok, so that is what you were trying to say. My question wasn't that there will be no creation of pressure gradients. It was more like: even though there is a pressure difference by newton's third law wouldn't the pressure of A on B always be equal to the pressure of B on A and if there is a situation like this how would it be possible for gases to expand. $\endgroup$ – Aditya Bharadwaj Mar 18 at 13:18
  • $\begingroup$ @AdityaBharadwaj At the interface (e.g. a massless piston) the pressure of the air equals down equals the upward pressure of the gas per N3. But, to determine the effect of that pressure on the air and/or the gas you need to apply N2 individually to each. Try and visualize a local volume of gas adjacent to the piston. The pressure acting down at the top on this volume is the air pressure. But the pressure acting up on the bottom of this volume is greater owing to the pressure gradient created by heating. Result, a net upward force on the volume and overall expansion of the gas in the cylinder. $\endgroup$ – Bob D Mar 18 at 18:46
  • $\begingroup$ I can illustrate this with drawings if you can't visualize it. $\endgroup$ – Bob D Mar 18 at 18:48
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Recall kinetic molecular theory:

Gas is made up of a bunch of individual spherical particles that love moving straight until something bumps into it. When you have a bunch of these flying particles around, more bumping occurs, and we begin to define the force or pressure caused by the gas as being related to the number of molecule bumping/collisions. Increase the number of particles, the frequency of collision increase, therefore the pressure increases. If we decrease air pressure, the air molecules would be less constricted and the frequency of collisions on the other gas will decrease.

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