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This is a question from IIT-JEE - ADVANCE 2018

A closed tank has two compartments A and B, both filled with oxygen (assumed to be ideal gas). The partition separating the two compartments is fixed and is a perfect heat insulator (Figure 1). If the old partition is replaced by a new partition which can slide and conduct heat but does NOT allow the gas to leak across (Figure 2), the volume (in $\pu{m3}$) of the compartment A after the system attains equilibrium is ___? enter image description here

I've understood on how to solve the Question, we can equate the number of moles before and after on either side seperately (as gas transfer isn't allowed, so amount of gas remains same), take the volumes by considering one to increase and the other to decrease, and the pressure final to be the same for both to attain equilibrium.

My doubt is how do I find the final temperature on either side, are we allowed to take average ?

And also if the gases on both sides were different are we still allowed to take average, or then there's another way.

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  • $\begingroup$ The answer is "2.22" == "(20/9)" if anyone is wondering $\endgroup$
    – Paras Gupta
    Commented Sep 4, 2022 at 22:24
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    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. It looks like you are asking for a recipe to solve the problem. $\endgroup$
    – PAEP
    Commented Sep 4, 2022 at 22:42
  • $\begingroup$ It can be helpful if you calculate how many moles of oxygen are in the compartment A and in comparment B using the ideal gas laws. If I am not mistaken $n_A \neq n_B$. $\endgroup$
    – PAEP
    Commented Sep 4, 2022 at 22:45
  • $\begingroup$ No, you cannot just take the average. To make it more "intuitive" why not, think about a more extreme situation. Let's say compartment A had 1 mol of gas at 400 K and compartment B was very small and contained only 0,000001 mol of gas at 300 K. Do you expect the final temperature to be 350 K, or something very close to 400 K? You have to write a heat balance: $Q_A = -Q_B \implies m_A c \Delta T_A = -m_B c \Delta T_B$. $\endgroup$
    – Domen
    Commented Sep 4, 2022 at 22:50
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    $\begingroup$ It would be nice if I put 1 drop of water 60 °C into 100 L of water 20 °C and would get warm water 40 °C. Unfortunately, it does not happen. $\endgroup$
    – Poutnik
    Commented Sep 5, 2022 at 4:37

1 Answer 1

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As you have correctly said the final pressure must be the same to balance the forces on the movable partition. Also, note that the partition is a conductor, which means it wont allow any temperature difference across it - i.e. it will keep conducting heat and doing work until the temperature difference is 0.

Instead of finding the exact number of moles, just find a ratio of the number of moles on either side using n = PV/RT; which gives us a ratio of 5:4. Honestly speaking, you don't need to find the individual temperatures in the final state to attain your answer. Just by equating pressure in final state you get

n1RTfinal⁄(1+x) = n2RTfinal⁄(3-x) , solving which you get x = 11⁄9 , where x is the increase in volume of the left chamber

However, just pondering back to your original question of the temperatures. Since no energy loss occurs to the surroundings, the change in internal energy of the left gas = change in internal energy of right gas. You might think that some amount of work have been done by the piston, but actually that work is responsible for transferring energy between two chambers. So, n1Cv(400 - Tfinal) = n2Cv(Tfinal-300). The ratio of n1:n2 is sufficient here to guide you towards answer. Had the gases been different in nature - a monoatomic and a diatomic then you had to consider the ratio of n1Cv1:n2Cv2 for finding the temperature difference

Doing some algebraic manipulation, you can easily find the final temperature. What is interesting to note is - the temperature isn't a mean or an average as you say. Rather it is a weighted mean - something very similar to finding location of center of mass between two unequal masses. The final temperature is at temperature differences from initial temperature of chambers A and B in a ratio of 4:5; closer to A's 400K and farther from B's 500K frankly because of the quantitative dominance of A's molecules. Had the ratio of number of moles been same, the final temperature would have been an average of the two as you said. You can always bring physical analogy of masses, if two equal masses are placed, center of mass is the exact midpoint of the line joining them, which ain't true if the masses are unequal

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    $\begingroup$ +1 for discussing the possibility of two gases having different heat capacities, and for saying that you don't need any temperature data to solve the problem, realizing the two gases will be at thermal equilibrium. Giving additional, unnecessary information in a problem makes it much more difficult for many students. $\endgroup$
    – Karsten ♦
    Commented Sep 5, 2022 at 11:57
  • $\begingroup$ Thanks for taking your time to point out the positive points of my answer. The feedback is really going to be helpful as I am a new user out here.. $\endgroup$
    – Sam
    Commented Sep 5, 2022 at 15:24
  • $\begingroup$ Looking forward to more! $\endgroup$
    – Karsten ♦
    Commented Sep 5, 2022 at 16:35
  • $\begingroup$ @Sam Thanks for the answer, your solution does make sense, the only doubt that I have is, how does the temperature stay the same even if heat transfer takes place $\endgroup$
    – Paras Gupta
    Commented Sep 5, 2022 at 17:11
  • $\begingroup$ @Paras Gupta Temperature doesn't stay the same, read carefully. The heat exchange causes the temperature of one chamber to drop and another to rise, but temperatures of the two chambers must come to a fixed value in the final state - the reason of which I mentioned. Heat flow occurs because of temp difference and in final state heat flow will cease only if the temp difference = 0 $\endgroup$
    – Sam
    Commented Sep 5, 2022 at 17:19

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