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(A) In an adiabatic process,the heat released by the reaction is absorbed by the products to raise their temperature.Let $x$ moles of $H_2$ be taken.T

Vedclass · Accepted Answer

$3798$

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(A) In an adiabatic process,the heat released by the reaction is absorbed by the products to raise their temperature.Let $x$ moles of $H_2$ be taken.The combustion reaction is: $H_{2(g)} + \frac{1}{2} O_{2(g)} 	o H_2O_{(g)}$.Since air contains $20\% O_2$ and $80\% N_2$,for $0.5x$ moles of $O_2$,the moles of $N_2$ present are $0.5x 	imes (80/20) = 2x$ moles.The heat released by the combustion is $q = x 	imes |\Delta H^o_f| = x 	imes 56000\ cal$.The heat absorbed by the products ($H_2O$ and $N_2$) to reach temperature $T$ is $q = n_{H_2O} C_{v(H_2O)} \Delta T + n_{N_2} C_{v(N_2)} \Delta T$.Since $C_v = C_p - R$ and $R \approx 2\ cal\ deg^{-1}\ mol^{-1}$,$C_{v(H_2O)} = 8.0 - 2.0 = 6.0\ cal\ deg^{-1}\ mol^{-1}$ and $C_{v(N_2)} = 7.0 - 2.0 = 5.0\ cal\ deg^{-1}\ mol^{-1}$.$56000x = (x 	imes 6.0 + 2x 	imes 5.0) 	imes (T - 298)$.$56000 = (6 + 10) 	imes (T - 298)$.$56000 = 16 	imes (T - 298)$.$3500 = T - 298$.$T = 3798\ K$.

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