1. Initial moles of $\mathrm{N_2O_5}$

Given mass: $37.8\text{ g}$

Molar mass of $\mathrm{N_2O_5}$: $2\times 14+5\times 16=28+80=108\text{ g mol}^{-1}$

So initial moles are: $n_0=\frac{37.8}{108}=0.35\text{ mol}$

Since the vessel is $1\text{ L}$, initially only $\mathrm{N_2O_5}$ is present.

2. Set up equilibrium using extent of reaction

Reaction: $2\mathrm{N_2O_5(g)}\rightleftharpoons 2\mathrm{N_2O_4(g)}+\mathrm{O_2(g)}$

Let the extent of reaction be $\xi$.

Then at equilibrium:

• $\mathrm{N_2O_5}: \;0.35-2\xi$
• $\mathrm{N_2O_4}: \;2\xi$
• $\mathrm{O_2}: \;\xi$

Total moles at equilibrium: $n_{\text{tot}}=(0.35-2\xi)+2\xi+\xi=0.35+\xi$

3. Use total pressure to find total moles at equilibrium

For ideal gas: $P V=nRT$

Given:

• $P=18.65\text{ bar}$
• $V=1\text{ L}$
• $R=0.082\text{ bar L mol}^{-1}\text{K}^{-1}$
• $T=500\text{ K}$

So, $n_{\text{tot}}=\frac{PV}{RT}=\frac{18.65\times 1}{0.082\times 500}$ $n_{\text{tot}}=\frac{18.65}{41}=0.454878\approx 0.4549$

Thus, $0.35+\xi=0.454878$ $\xi=0.104878$

4. Equilibrium moles of each species

$n(\mathrm{N_2O_5})=0.35-2\xi=0.35-2(0.104878)=0.140244$

$n(\mathrm{N_2O_4})=2\xi=0.209756$

$n(\mathrm{O_2})=\xi=0.104878$

Check total: $0.140244+0.209756+0.104878=0.454878$

5. Calculate partial pressures

Since $p_i=\frac{n_i}{n_{\text{tot}}}P_{\text{tot}}$

we get:

$p_{\mathrm{N_2O_5}}=\frac{0.140244}{0.454878}\times 18.65\approx 5.75\text{ bar}$

$p_{\mathrm{N_2O_4}}=\frac{0.209756}{0.454878}\times 18.65\approx 8.60\text{ bar}$

$p_{\mathrm{O_2}}=\frac{0.104878}{0.454878}\times 18.65\approx 4.30\text{ bar}$

6. Expression for $K_p$

For $2\mathrm{N_2O_5(g)}\rightleftharpoons 2\mathrm{N_2O_4(g)}+\mathrm{O_2(g)}$

$K_p=\frac{\left(p_{\mathrm{N_2O_4}}\right)^2\left(p_{\mathrm{O_2}}\right)}{\left(p_{\mathrm{N_2O_5}}\right)^2}$

Substitute values: $K_p=\frac{(8.60)^2(4.30)}{(5.75)^2}$

$K_p\approx \frac{73.96\times 4.30}{33.06}$

$K_p\approx \frac{318.028}{33.06}\approx 9.62$

Thus, $K_p=9.62=962\times 10^{-2}$

Nearest integer multiplying $10^{-2}$ is: $\boxed{962}$

7. Comparison with stored answer

Stored correct answer = $962$

Our derived answer also = $962$.

So the stored answer is correct.