Question

The system shown in the figure is in equilibrium, where $A$ and $B$ are isomeric liquids and form an ideal solution at $TK$. Standard vapour pressures of $A$ and $B$ are ${P^0}_A$ and ${P^0}_B$, respectively, at $T$ K. We collect the vapour of $A$ and $B$ in two containers of volume $V$, first container is maintained at 2 $T$ K and second container is maintained at $3T/2$. At the temperature greater than $T$ K, both $A$ and $B$ exist in only gaseous form.
We assume than collected gases behave ideally at 2 $T$ K and there may take place an isomerisation reaction in which $A$ gets converted into $B$ by first-order kinetics reaction given as:
$A\stackrel{k}{\to }B$, where $k$ is a rate constant.
In container $\left(II\right)$ at the given temperature $3T/2,A$ and $B$ are ideal in nature and non reacting in nature. A small pin hole is made into container. We can determine the initial rate of effusion of both gases in vacuum by the expression.
$r=K.\frac{P}{\sqrt{M_0}}$
where $P=pressuredifferencebetweensystemandsurrounding$
$K=positiveconstant$
$M_0=molecularweightofthegas$

Vapours of $A$ and $B$ are passed into a container of volume 8.21 L, maintained at $2T$ K, where $T=50K$ and after 5 min, moles of $B=8/3$. The pressure developed into the container after two half lives is

• A. 3 atm
• B. 4 atm
• C. 5 atm
• D. 0.5 atm

Answer 1

The correct option is B 4 atm

$\frac{n_A}{n_B}=\frac{P_A}{P_B}=2;TotalV.P.=2atm$;
$PV=nRT\Rightarrow 2\times 8.21=(n_A+n_B)\times 0.0821\times 50$
$\Rightarrow n_A+n_B=4$
Also,
$n_A/n_B=2$
by solving these two equations,
$n_A=8/3,n_B=4/3$
$\frac{n_A}{n_B}=\frac{2}{1}$
$A$ $\stackrel{K}{\to }$ $B$
8/3 4/3
$t=5min$ $\frac{8}{3}-x$ $\frac{4}{3}+x=\frac{8}{3}\Rightarrow x=\frac{4}{3}$
$\equiv 4/3$ $\Rightarrow t=t_{1/2}=5min$
$\Rightarrow Att=10min$ $A$ $\stackrel{K}{\to }$ $B$
$x/3$ $10/3$
$\Rightarrow P=\frac{4\times 0.0821\times 100}{8.21}=4atm$