Question

$N_2{O}_5$ decomposes to $N_2{O}_4$ and $O_2$ as :
$N_2{O}_5\to N_2{O}_4+\frac{1}{2}{O}_2$
The pressure $P_1$ at any stage is related to $p_0$ and $x$, the fraction of dissociation, as:

• A. $p_0$
• B. $(1+\frac{1}{2}x)p_0$
• C. $1-\frac{3}{2}{p}_0$
• D. $\frac{3}{2}x{p}_0$

Answer 1

Answer: (B)

$(1+\frac{1}{2}x)p_0$

$N_2{O}_5⟶N_2{O}_4+\frac{1}{2}{O}_2$

$P_0$ - -

$P_0-P$ $P$ $P/2$

Total pressure=$(P_0-P/2)=P_1$ (given)

Since $x=\frac{P}{P_0}$ $\therefore P=x{P}_0$ $\therefore P_1=(P_0+\frac{x{P}_0}{2})$