Question

Two vessels of equal volume are connected to each other by a value of negligible volume. One of the containers has $2.8g$ of $N_{2}12.7g$ of $I_2$ at a temperature $T_1.$ The other container is completely evacuated. The container that has $N_2$ and $I_2$ is heated to temperature $T_2$ while the evacuated container is heated $T_2/3.$ The value is now opened. Calculate the mass of $N_2$ in container (B) after a very long time $I_2$ sublimes at $T_2$. (report your answer in nearest integer form in grams)

Answer 1

Answer: 2

Number of moles of $N_2$ in vessel
$(A)$ at $T_1=\frac{2.8}{2.8}=0.1$
Number of moles of $I_2$ in
Vessel (A) at $T_1=\frac{12.7}{254}=0.05$

Vessel $(A)$ is heated to $T_2$ the evacuated vessel $(B)$ is heated to $T_2/3.$
On opening stop-cock $I_2$ sublimes in $(A)$ and its vapours condense to solid $I_2$ in $(B)$. L
et the number of moles of $N_2$ moved from $(A)$ to $(B)$ at equilibrium be $x$.
Pressure of $N_2$ in $(A)=$ Pressure of $N_2$ in $(B)$
$=\frac{(0.1-x)R{T}_2}{V}=\frac{xR{T}_2}{3V}$
One solving, $x=0.075$
$\therefore$ Mass of $N_2$ is vessel $(A)=0.7g$
Mass of $N_2$ in vessel $(B)=2.1g$