$1mole$ of $H_2$ gas is contained in a box of volume $V=1.00m^3$ at $T=300K$. The gas is heated to a temperature of $T=3000K$ and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)

The correct option is D $20$ times the pressure initially.
As we know, ideal gas equation is given by
$PV=nRT$

From the given equation, we get

$\frac{P}{nT}=$ constant

(Since the volume of the box remains same no matter which gas is present)

Therefore,
$\frac{P_1}{n_1{T}_1}=\frac{P_2}{n_2{T}_2}$

Now,
$\frac{P_2}{P_1}=\frac{n_2{T}_2}{n_1{T}_1}$

Given:

$T_2=3000K$ $T_1=300K$

Since $H_2$ splits into hydrogen atoms, therefore number of moles becomes twice the original.

$\therefore \frac{P_2}{P_1}=\frac{2n_1}{n_1}\times \frac{3000}{300}$

$\frac{P_2}{P_1}=20$

$P_2=20P_1$

$\text{Final Answer}:$ (𝑑).