Solution

For $$\underset{1\\(1-\alpha)}{nA}\ \rightleftharpoons \underset{0\\ \frac{\alpha}{n}}{{[A]}_n}$$ Total mole at equilibrium $$ = 1 - \alpha + \displaystyle\frac { \alpha }{ n }$$ $$ \therefore$$ Gas equation is : $$ \displaystyle\frac { PV }{ RT } = 1 - \alpha + \displaystyle\frac { \alpha }{ n }$$ or $$ \displaystyle\frac { P }{ RT } = \displaystyle\frac { 1 - \alpha + \displaystyle\frac { \alpha }{ n } }{ V }$$ ...(i) Given, $$\displaystyle\frac { PV }{ RT } = 1 - \displaystyle\frac { \left( n - 1 \right) { K }_{ c } }{ { V }^{ n - 1 } }$$ $$ \displaystyle\frac { P }{ RT } = \displaystyle\frac { 1 }{ V } - \displaystyle\frac { \left( n - 1 \right) { K }_{ c } }{ { V }^{ n } }$$ $$ = \displaystyle\frac { 1 }{ V } - \displaystyle\frac { \left( n - 1 \right) \times \displaystyle\frac { \alpha }{ n\cdot V } }{ { V }^{ n }\times { \left( \displaystyle\frac { 1 - \alpha }{ V } \right) }^{ n } }$$ $$ \left( \because { K }_{ c } = \displaystyle\frac { \displaystyle\frac { \alpha }{ nV } }{ { \left( \displaystyle\frac { 1 - \alpha }{ V } \right) }^{ n } } \right)$$ $$ = \left[ 1 - \displaystyle\frac { \left( n - 1 \right) \alpha }{ n{ \left( 1 - \alpha \right) }^{ n } } \right] \displaystyle\frac { 1 }{ V }$$ or $$ \displaystyle\frac { P }{ RT } = \displaystyle\frac { 1 }{ V } \left[ 1 - \displaystyle\frac { \left( n - 1 \right) \alpha }{ n } \right]$$ $$ \left[ \because { \left( 1 - \alpha \right) }^{ n } = 1 \right]$$ or $$ \displaystyle\frac { P }{ RT } = \displaystyle\frac { 1 }{ V } \left[ 1 - \alpha + \displaystyle\frac { \alpha }{ n } \right]$$ $$ \displaystyle\frac { P }{ RT } = \displaystyle\frac { 1 - \alpha + \displaystyle\frac { \alpha }{ n } }{ V } $$ ...(ii) by these equations it can be seen that, $$\displaystyle \frac{PV}{RT}=\left [ 1-\frac{\left ( n-1 \right )K_{c}}{V^{n-1}} \right ]$$