Question

A vessel of volume $V=5.0$ litre contains $1.4 \,g$ of nitrogen at temperature, $T=1800 \,K$. Find the pressure (in $N / m ^2$ ) of the gas if $30 \%$ of its molecules are dissociated into atoms at this temperature. ___ $\left(1.94 \times 10^5\right)$

Answer 1

Mass of molecular nitrogen
$=\frac{70}{100} \times 1.4=0.98\, g$
Mass of atomic nitrogen
$=\frac{30}{100} \times 1.4=0.42\, g$
Number of moles of molecular nitrogen,
$n_1=\frac{0.98}{28}=0.035$
Number of moles of atomic nitrogen
$n_2=\frac{0.42}{14}=0.03$
The pressure of the gas = pressure exerted by molecular nitrogen $+$ pressure exerted by atomic nitrogen
i.e., $ P=P_1+P_2$
$=\frac{n_1 R T}{V}+\frac{n_2 R T}{V}=\frac{\left(n_1+n_2\right) R T}{V}$
$=\frac{(0.035+0.03) \times 8.31 \times 1800}{5 \times 10^{-3}}$
$=1.94 \times 10^5 \,N / m ^2$