# Calculation of the mass of N2 in the container

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.
Confusions:
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

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

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.