# How does the state variables $(P,V)$ remain constant for a system with adiabatic wall?

I have been very confused regarding how my physics textbook differentiates b/w an adiabatic wall and a diathermic wall.

The term ‘equilibrium’ in thermodynamics appears in a different context: we say the state of a system is an equilibrium state if the macroscopic variables that characterize the system do not change in time.

In general, whether or not a system is in a state of equilibrium depends on the surroundings and the nature of the wall that separates the system from the surroundings. Consider two gases $$A$$ and $$B$$ occupying two different containers. We know experimentally that pressure and volume of a given mass of gas can be chosen to be its two independent variables. Let the pressure and volume of the gases be $$(P_A, V_A)$$ and $$(P_B, V_B)$$ respectively. Suppose first that the two systems are put in proximity but are separated by an adiabatic wall – an insulating wall (can be movable) that does not allow flow of energy (heat) from one to another. The systems are insulated from the rest of the surroundings also by similar adiabatic walls. The situation is shown schematically in $$\text{Fig. 12.1(a)}$$. In this case, it is found that any possible pair of values $$(P_A, V_A)$$ will be in equilibrium with any possible pair of values $$(P_B, V_B )$$.

The full text on this topic in the textbook is here on the second page in section 12.1

What was my train of thought after reading this was as follows:-

The text tells that we can select the pressure and volume of the gas independently that leaves the temperature of the system to be determined which must depend on the independent variables $$P$$ & $$V$$. Now, if the system $$A$$ and $$B$$ have been assigned the independent state variables $$(P_A,V_A)$$ and $$(P_B,V_B)$$ respectively then lets say that their temperature initially are $$T_A$$ & $$T_B$$ respectively. So, through this what the text wants to convey is that that by assigning any two independent variables independently we assign the third statet variable too, well that was a given.

Now, let me state my concern regarding the text that bugs me:-

The text states that the adiabatic wall "can" be movable. Now, since the adiaabatic wall doesn't allow transfer of heat from the system to the surroundings, so the adiabatic wall moves(recall that I assumed the wall to be movable) till the system $$A$$ & $$B$$ exert equal pressure on the adiabatic wall, hence changing there state variables to, say, $$(P,V_A',T_A)$$ and $$(P,V_B',T_B)$$ respectively as time passes following and isothermal process while achieving the equilibrium.

But, the text states that:-

In this case, it is found that any possible pair of values $$(P_A,V_A)$$ will be in equilibrium with any possible pair of values $$(P_B,V_B)$$.

But, I see that for any value of $$T_A\; \& \; T_B$$ which we get from the initial state variables $$(P_A,V_A)\; \&\; (P_B,V_B)$$ resp. we will arrive at some other values of state variable function for both of the system so how can the book's statement be correct.

Edit:- To conclude my post my main problem with the text is as follows:-

If two systems are separated by a movable adiabatic wall and their initial state functions are $$(P_A,V_A)$$ and $$(P_B,V_B)$$, then due to the wall being movable wouldn't their state functions would change with time to, say, $$(P,V_A')$$ and $$(P,V_B')$$. So, for no arbitrary state function of both the systems would the systems remain in equilibrium initially, but the book states otherwise.

• "In this case" ... what case? Without studying the original book myself I can't make much sense of what you are asking for. Best to summarize with a specific question at the end of your post if you want an answer. Jan 16, 2017 at 14:37
• @docscience On it, though I would like to ask are you implying that my post is too long? If so, then, well it will take me a lot of time to shorten it. And about the "this case" you would like to read the first quote and if you have some time and patience you could also read from the link I provide below the first quote but still I will try to shorten my post. Is that the reason for the short view count...but how do people decide that it is long post even before viewing O_o ?!? Jan 16, 2017 at 14:45
• Long is OK, but you need to summarize, focus on your main question at the end. It's hard to gather what your question is Jan 16, 2017 at 14:50
• @docscience Hmmm....yeah on rereading it from the readers point of view this post is indeed horrible I will have to do a whole lot of editing. Jan 16, 2017 at 14:52
• @docscience - After thinking a lot I could only come up with the edit that I have written to state my concern in the most precise yet complete(somehow) manner, does that suffice. Jan 16, 2017 at 16:53

If two systems are separated by a movable adiabatic wall and their initial state functions are ($$P_A$$,$$V_A$$) and ($$P_B$$,$$V_B$$), then due to the wall being movable wouldn't their state functions would change with time to, say, ($$P$$,$$V'_A$$) and ($$P$$,$$V'_B$$). So, for no arbitrary state function of both the systems would the systems remain in equilibrium initially, but the book states otherwise.
If and only if $$P_A$$ = $$P_B$$ intially, then the system will remain in equilibrium, with $$P_A$$ = $$P_B$$ = $$P$$. So you are right, it is not true that for any pair of values will the system remain in equilibrium. Only if their pressures are equal, will it happen.
It states that the wall that separates the two systems is adiabatic, and that it ''can be movable'' but doesn't tell us that it is. If it is adiabatic but not movable, it is technically true that the two boxes would be in "equilibrium"–as in their states would not change–at any $$(P_A,T_A)$$ and $$(P_B,T_B)$$, but this is because they do not communicate with each other and therefore they do not affect each other's state. What this describes is two non interacting systems, each in in internal equilibrium but not in equilibrium with each other. As you correctly point out, if the wall is movable, then pressures will equilibrate.