The degree of dissociation of $PCl_5$ $(\alpha)$ obeying the equilibrium $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is related to the pressure at equilibrium $(P)$ by:

Consider the following reaction in a sealed vessel at equilibrium with concentrations of $N_2 = 3.0 \times 10^{-3} \ M$,$O_2 = 4.2 \times 10^{-3} \ M$,and $NO = 2.8 \times 10^{-3} \ M$.
$2 \ NO_{(g)} \rightleftharpoons N_{2_{(g)}} + O_{2_{(g)}}$
If $0.1 \ mol \ L^{-1}$ of $NO_{(g)}$ is taken in a closed vessel,what will be the degree of dissociation $(\alpha)$ of $NO_{(g)}$ at equilibrium?

For a reaction at equilibrium $A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$,the relation between dissociation constant $(K)$,degree of dissociation $(\alpha)$ and equilibrium pressure $(p)$ is given by?

$4 \ mol$ of $PCl_5$ is heated in a closed vessel at a constant temperature. If the degree of dissociation for $PCl_5$ is $0.5$,what will be the total number of moles at equilibrium?

What is the vapor density of a mixture containing $20\%\, NO_2$ by volume and $N_2O_4$?

For the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,the degree of dissociation is inversely proportional to the square root of the system pressure. Assuming constant temperature,if the volume of the system increases to $16$ times its initial volume,the degree of dissociation becomes ....... times.