Two vessels of equal volumes are connected to each other by a valve of negligible volume. One vessel containing $0.1$ mole $\ce {N_2}(g)$ and $0.05$ mole $\ce {I_2}(s)$ at temperature $T_1$. The other vessel is completely evacuated. The vessel containing gases is heated to temperature $T_2$ whereas the other vessel is kept at $\dfrac{T_2}{3}(\lt T_1)$. Calculate the mass of $\ce {N_2}$ in container at temperature $\dfrac{T_2}{3}$ when equilibrium is attained.

My attempt: Assuming the final pressures to be $P$ in both the container at equilibrium and using the respective ideal gas equations for both the containers, $$PV=(0.1-x +0.05-y)RT_2$$ $$PV=(x+y)R \frac{T_2}{3}$$ Where $x$ and $y$ are the final moles in the right container. Using this we get $$x+y=\frac{0.45}{4}$$ But now I have no clue for finding another equation for $x$ and $y$. In these type of questions, I always get stuck for the $2^{\text{nd}}$ equation.
  1. Another equation for $x$ and $y$

  2. Is it valid to assume final temperatures of both the containers same as initial?

  3. And why they have given the information about $T_1$, means how can we use it?
    Thank You

  • 2
    $\begingroup$ (1) There is none. (2) Yes. (3) You use $T_1$ to deduce the phase state of iodine in the second vessel. $\endgroup$ Sep 9, 2021 at 15:20
  • $\begingroup$ @IvanNeretin,then how will we solve for x and Y? $\endgroup$
    – UNAN
    Sep 9, 2021 at 16:29
  • 1
    $\begingroup$ There is no $y$, just $x$. $\endgroup$ Sep 9, 2021 at 16:33

1 Answer 1


Let's assume the problem can be solved using only nitrogen data. The amount of iodine is independent of the amount of nitrogen. With a little bit of luck, the final temperature in the cold container is so low (= $T_2/3$) that the iodine is entirely condensed.

In the final state, there is $n_1$ moles in the hot container, and $n_2$ moles nitrogen in the cold container. The pressures are respectively $p_1$ and $p_2$ in both containers. The temperatures are : $T_2$ and $T_2/3$. Equalizing the pressures, one gets : $p_1 = n_1RT_2/V = p_2 = n_2RT_2/3V$, which gives : $n_1 = n_2/3$.

Now the total number of moles is $n = n_1 + n_2 = 4 n_2/3 = 0.1 $. As a consequence, the total amount of nitrogen in the cold volume is : $n_2 = 0.1·3/4 = 0.075$ mole.

  • 1
    $\begingroup$ The condensed state of iodine is not left to luck, but stated explicitly. $\endgroup$ Sep 9, 2021 at 16:48

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