The equilibrium constants K_P1 and K_P2 for the reactions X↔2Y and Z↔P +Q, respectively are in the ratio of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of total pressure at these equilibria is

 Given: X   $\to$  2Y ------(1)
             Z  $\to$  P  +  Q---(2)
             reactions (1) and (2) having same degree of dissociation  
​             $\frac{K_{p1}}{K_{p2}}$= $\frac{1}{9}$
To find: Ratio of total pressures
Solve: Suppose $\alpha$  is the degrre of dissociation
                                      X   $\to$  2Y                        Z  $\to$  P  +  Q
​Initial concentration =    1           0                          1        0        0
Final concentration =  (1-$\alpha$)        $\alpha$                        (1-$\alpha$)    $\alpha$       $\alpha$  

Total number of moles at equilibrium: For reaction 1st= $1-\alpha +2\alpha =1+\alpha$
                                                             For reaction 2nd =$1-\alpha +\alpha +\alpha =1+\alpha$
Suppose P₁ and P₂ is the total pressure for reaction 1 and 2 respectively.
$$\begin{gathered} P_X= \frac{1-\alpha }{1+\alpha }\times P_1 P_Y=\frac{2\alpha }{1+\alpha }\times P_1 \\ {P}_Z=\frac{1-\alpha }{1+\alpha }\times P_2 P_P=\frac{\alpha }{1+\alpha }\times P_2 P_Q=\frac{\alpha }{1+\alpha }\times P_2 \end{gathered}$$

$$\begin{gathered} K_{P1}=\frac{(P_Y{)}^2}{P_X}=\frac{({\displaystyle \frac{2\alpha }{1+\alpha }}{)}^2\times P_1^2}{{\displaystyle \frac{\alpha }{1+\alpha }}}-----\left(3\right) \\ {K}_{P2}=\frac{(P_P\times P_Q)}{P_Z}=\frac{\frac{\alpha }{1+\alpha }{P}_2\times \frac{\alpha }{1+\alpha }{P}_2}{{\displaystyle \frac{1-\alpha }{1+\alpha }}\times P_2}----\left(4\right) \end{gathered}$$
Divide (3) by (4), we get

 $\frac{K_{p1}}{K_{p2}}$=$\frac{4P_1^2}{P_2^2}$=$\frac{1}{9}$(given)

$$\begin{gathered} \frac{P_1^2}{P_2^2}=\frac{1}{36} \\ \frac{P_1}{P_2}=\frac{1}{6} \end{gathered}$$
Therefore, the ratio of total pressures, P₁:P₂ = 1:6