mass of hydrogen $=14\; g$

Molar mass of $H_{2}=2\; g m mol ^{-1}$

No. of moles, $n_{H_{2}}=\frac{14}{2}=7 \;moles$

mass of oxygen $=96 \;gm$

Molar mass of $O _{2}=32 \;gmol ^{-1}$

No. of moles, $n_{0_{2}}=\frac{96}{32}=3 \;moles$

Total no. of moles in the vessel, $n=n_{H_{2}}+n_{O_2}$

$n=(7+3)=10 \text { moles }$

Let STP $P =1 atm =10^{5} Pa$

$T=273\; K$

We knows by ideal gas equation, $P V=n R T$

where $P=$ pressure of gas

$V=$ volume of gas

$n=$ amount of substance

$R=$ universal gas coustant

$T=$ absolute temperature

To find $V$, puts all values of $P, n, R, T$ :

$V=\frac{n R T}{P}=\frac{10 \times 8.3 \times 273}{10^{5}} \quad[R=8.3$

$V=0.23 m ^{2}$

chemical reaction is induced bypassing elictric spark in the vessel till one of the gas is consumed. When electric spark is passed. The reaction is

$2 H _{2}+ O _{2} \rightarrow 2 H _{2} O$

Each $1 gm$ of $H _{2}$ reacts with $8 gm ^{ m }$ of $O _{2}$

$\therefore 96 gm$ of oxygen $\rightarrow 12 g$ of $H _{2}$

After consumption:

orygen left $=0$

Hydrogen left $=14-12=2 gm$

No. of moles of $2\; gm$ of $H _{2} \Rightarrow n^{\prime}=\frac{2}{2}=1 \;mol$

$\therefore$ New ideal gas eq.: $P^{\prime} V=n^{\prime} R T$ $\dots (2)$

Dividing eq. $(1)$ and $(2)$:

$\frac{p^{\prime}}{p}=\frac{n^{\prime}}{n} \Rightarrow p^{\prime}=\frac{n^{\prime} p}{n}=\frac{1 \times 10^{5}}{10}=10^{4}$

$p^{\prime}=0.1 \; atm$

The presoure in the vessel is $0.1 \; atm$