Then, using $p V=n R T$.

$\text { and }$

$n_{1} =\frac{p V}{R(300)}$

$n_{2} =\frac{5 p(4 V)}{R(400)}$

When value is opened gas flows from $C_{2}$ to $C_{1}$ till pressure in $C_{1}$ and $C_{2}$ is equal. Let after equalisation of pressure in both $C_{1}$ and $C_{2}$, its value is $p_{0}$,

Then, using $p V=n R^{\prime} T$.

$p_{0} V=n_{1}^{\prime} R(300)\left(\right.$ Container $\left.C_{1}\right)$

and $p_{0} 4 V=n_{2}{ }^{\prime} R(400)$ (Container $C_{2}$ )

So, $\quad n_{1}^{\prime}=\frac{p_{0} V}{R(300)}$

and $\quad n_{2}^{\prime}=\frac{p_{0}(4 V)}{R(400)}$

As no gas is leaked from containers,

$n_{1}+n_{2}=n_{1}{ }^{\prime}: n_{2}{ }^{\prime}$

$\Rightarrow \quad \frac{p V}{R(300)}+\frac{20 p V}{R(400)}$ = $\quad \frac{p_{0} V}{R(300)}+\frac{4 p_{0} V}{R(400)}$

So, $p\left(\frac{1}{300}+\right.\left.\frac{20}{400}\right)$ $=p_{0}\left(\frac{1}{300}+\frac{4}{400}\right)$

$\Rightarrow \quad 4 p=p_{0}$

$\text { Now, } \frac{n_{2}{ }^{\prime}}{n_{1}{ }^{\prime}}=\frac{\frac{4 p_{0} V}{R(400)}}{\frac{p_{0} V}{R(300)}}=3$

Finally number of moles of $C_{2}$ is thrice of $C_{1}$.