$\mathrm{H}_{2}$ gas is mixed with air at $25^{\circ} \mathrm{C}$ under a pressure of 1 atmosphere a

Question

$\mathrm{H}_{2}$ gas is mixed with air at $25^{\circ} \mathrm{C}$ under a pressure of 1 atmosphere and exploded in a closed vessel. The heat of the reaction, $\mathrm{H}_{2(g)}+\frac{1}{2} \mathrm{O}_{2(g)} ightarrow \mathrm{H}_{2} \mathrm{O}_{(v)}$ at constant volume, $\Delta U_{298 \mathrm{~K}}=-240.60 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $C_{V}$ values for $\mathrm{H}_{2} \mathrm{O}$ vapour and $\mathrm{N}_{2}$ in the temperature range 298 K and 3200 K are $39.06 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ and $26.40 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ respectively. The explosion temperature under adiabatic conditions is (Given : $n_{\mathrm{N}_{2}}=2$ )\

Professor PC Thomas Classes · Answer

(c) : If the process is carried out adiabatically and isochorically,
$
\begin{aligned}
& \Delta U=\Delta U_{	ext {heating }}+\Delta U_{298 \mathrm{~K}}=0 	ext { or } \Delta U_{	ext {heating }}=-\Delta U_{298 \mathrm{~K}} \
& =\int_{298 \mathrm{~K}}^{T_f} n \sum C_v d T=+240.60 \mathrm{~kJ} \mathrm{~mol}^{-1} \
& \sum n C_v=n C_{v\left(\mathrm{H}_2 \mathrm{O}_{(v)}ight)}+n C_{v\left(\mathrm{~N}_{2(g)}ight)} \
& \quad=(39.06+2 	imes 26.40)=91.86 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
$
by using the value of $\Sigma n C_\gamma$ in the above equation
$(91.86)\left(T_f-298ight)=240600 \mathrm{~J} \mathrm{~mol}^{-1}$
$
T_f-298=\frac{240600}{91.86}=2619 \mathrm{~K} \Rightarrow T_f=2619+298=2917 \mathrm{~K}
$