Can we calculate final temperature of two gases in thermal contact? [closed]








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This is a question from IIT-JEE - ADVANCE 2018



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> 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](https://i.stack.imgur.com/NyOucm.jpg)](https://i.stack.imgur.com/NyOucm.jpg)
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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.




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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