Question

One mole of $N{H}_{4}Cl\left(s\right)$ is kept in an open container & then covered with a lid. The container is now heated to $600$K where all $N{H}_{4}Cl$(s) dissociates into $N{H}_3$ and $HCl\left(g\right)$. If volume of the container is $24.63$ litres, calculate what will be the final pressure of gases inside the container. Also find whether the lid would stay or bounce off if it can with stand a pressure difference of $5.5$ atm. Assume that outside air is at $300$K and $1$ atm pressure.

Answer 1

Given that $\underset{1mole}{N{H}_{4}Cl}⟶\underset{1mole}{N{H}_3}+\underset{1mole}{Cl}$

$PV=nRT$

$=\frac{2\times \frac{82.1}{1000}\times 600}{24.63}$

$=\frac{821\times 3}{25\times 24.63}=4$

Pressure= $4-1=3$ atm

So, the lid does not bounce.