Question

Two vessels connected by a valve of negligible volume. One container $\left(I\right)$ has $2.8\text{g}$ of $N_2$ at termperature $T_1\left(\text{K}\right)$. The other container $\left(II\right)$ is completely evacuated. The container $\left(I\right)$ is heated to $T_2\left(\text{K}\right)$ while container $\left(II\right)$ is maintained at $\frac{T_2}{3}\left(\text{K}\right)$. volume of vessel $\left(I\right)$ is half that of vessel $\left(II\right)$. If the valve is opened then what is the weight ratio of $N_2$ in both vessel $\left(W_I/W_{II}\right)$?

• A. $1:2$
• B. $1:3$
• C. $1:6$
• D. $3:1$

Answer 1

The correct option is C $1:6$


Let $x$ mole of $N_2$ present in vessel $II$ and $P$ is final pressure of $N_2$
$P\left(2V\right)=xR\left(\frac{T_2}{3}\right)...\left(1\right)$ $P\left(V\right)=(0.1-x)R{T}_2...\left(2\right)$
Dividing the equation $1$ by equation $2$, we get
$2=\frac{x}{3(0.1-x)}$
$\Rightarrow x=0.6/7\text{mole};$
$\frac{0.6}{7}\times 28\Rightarrow 2.4\text{g}{N}_2$
$II$ has $2.4\text{g}{N}_2$ and $I$ has $0.4\text{g}of{N}_2$
$\frac{W_I}{W_{II}}=\frac{0.4}{2.4}\Rightarrow 1:6$