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 $N_2$ and $12.7g$ of $l_2$ 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 containers is heated to $T_2/3$. The value is now opened. Calculate the mass of $N_2$ in containers $\left(B\right)$ after a very long time $l_2$ sublimes at $T_2$. (report your answer is nearest integer form in grams)

• A. Mass of $N_2$ is vessal $A$ is $0.3g$
• B. Mass of $N_2$ is vessal $B$ is $4.1g$
• C. Mass of $N_2$ is vessal $A$ is $0.7g$
• D. Mass of $N_2$ is vessal $A$ is $2.1g$

Answer 1

The correct option is A Mass of $N_2$ is vessal $A$ is $0.3g$

At container $B$

No of moles of $N_2=x$

Volume of $A$ and $B$ is equal to $V$

Temperature $T=\frac{T_2}{3}$

Applying Ideal gas euation

$PV=xR\frac{T_2}{3}$ (eq 1)

At container $A$

No. of moles $=0.1-x$

Temeperature$T=\frac{T_2}{3}$

Applying Ideal gas equation

$PV=(0.1-x)R{T}_2$ (eq 2)

eq 1 is divided by eq 2

$\frac{PV}{PV}=\frac{xR{T}_2}{3}\times \frac{1}{(0.1-x)R{T}_2}$

$1=\frac{x}{3(0.1-x)}$

$x=\frac{0.3}{4}$

No of moles of $N_2$ gas at container B is $\frac{0.3}{4}$

Mass at container B

$M_B=\frac{0.3}{4}\times 28=2.1$g

Moles at $N_2$container $A$ $=0.1-\frac{0.3}{4}=\frac{0.1}{4}$

Mass of $N_2$ at $A$ $=\frac{0.1}{4}\times 28=0.7$g