A 2.0 L flask, initially containing one mol of each CO and $H_{2}O,$ was sealed and heated to 700 K, where the following equilibrium was established $CO\left(g\right)+H_{2}O\left(g\right)\stackrel{ }{\rightleftarrows }C{O}_2\left(g\right)+H_2\left(g\right);K=9$ Now the flask was connected to another flask containing some pure $C{O}_2\left(g\right)$ at same temperature and pressure, by means of a narrow tube of negligible volume. When the Equilibrium was restored, moles of $CO$ was found to be double of its mole at first equilibrium. Determine volume of $C{O}_2\left(g\right)$ flask

$CO\left(g\right)+H_{2}O\left(g\right)\underset{ }{\overset{ }{\rightleftharpoons }}C{O}_2\left(g\right)+H_2\left(g\right);$

$t=0$                  1 mole       1 mole         O            O

t = eq.state             $1-x$          $1-x$      $x$           $x$

$9=\frac{x.x}{(1-x)(1-x)}\Rightarrow \frac{x}{1-x}=3\Rightarrow x=3-3x\Rightarrow x=0.75$

$CO\left(g\right)+H_{2}O\left(g\right)\underset{ }{\overset{ }{\rightleftharpoons }}C{O}_2\left(g\right)+H_2\left(g\right)$

t = old eq. state                 0.25         0.25        0.75           0.75
t = y mole $C{O}_2$ added    0.25         0.25        0.75+y      0.75
t = at new eq. state           0.5          0.5         0.5+y        0.5
$9=\frac{(0.5+y)\left(0.5\right)}{0.5\times 0.5}\Rightarrow y=4\text{\hspace{0.17em}mole}$

Initially 2 mole gas present in 2 ltr Vessel. So at same T,P; 4 moles of $C{O}_2$ will be present in 4 ltr Vessel