If $PCl_5$ undergoes $80\%$ dissociation at $250 \, ^\circ C$,what is its vapour density at this temperature?

For the reaction: $NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}; K_p$. The degree of dissociation $(\alpha)$ of $NH_3$ is related to total equilibrium pressure $(P^o)$ as:

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.

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:

In a $100 \ L$ vessel,$3 \ moles$ of nitrogen and $3 \ moles$ of $PCl_5$ are taken and heated to $500 \ K$. The equilibrium pressure is $3.28 \ atm$. The percentage degree of dissociation of $PCl_5$ is : (Assume ideal behaviour for all gases).

For the reaction $A \rightleftharpoons \frac{1}{2} B + C$,the degree of dissociation $\alpha$ in terms of vapour density $D_t$ (theoretical) and $D_o$ (observed) is given by: