Solution 23: Dissociation of Diatomic Gas
(a) Change in Specific Heat:
Consider 1 mole of diatomic gas ($A_2$) dissociating into atoms ($2A$). $$ A_2 \rightleftharpoons 2A $$ Let $\alpha$ be the degree of dissociation. Mixture: $(1-\alpha)$ moles of $A_2$ (diatomic, $C_v = \frac{5}{2}R$) and $2\alpha$ moles of $A$ (monoatomic, $C_v = \frac{3}{2}R$). $$ C_{mix} = (1-\alpha)\frac{5}{2}R + (2\alpha)\frac{3}{2}R $$ $$ C_{mix} = \frac{5}{2}R – \frac{5}{2}\alpha R + 3\alpha R = \left( \frac{5}{2} + \frac{1}{2}\alpha \right)R $$ Since $\alpha > 0$, the specific heat capacity increases.
Consider 1 mole of diatomic gas ($A_2$) dissociating into atoms ($2A$). $$ A_2 \rightleftharpoons 2A $$ Let $\alpha$ be the degree of dissociation. Mixture: $(1-\alpha)$ moles of $A_2$ (diatomic, $C_v = \frac{5}{2}R$) and $2\alpha$ moles of $A$ (monoatomic, $C_v = \frac{3}{2}R$). $$ C_{mix} = (1-\alpha)\frac{5}{2}R + (2\alpha)\frac{3}{2}R $$ $$ C_{mix} = \frac{5}{2}R – \frac{5}{2}\alpha R + 3\alpha R = \left( \frac{5}{2} + \frac{1}{2}\alpha \right)R $$ Since $\alpha > 0$, the specific heat capacity increases.
(b) Fraction Dissociated:
Given specific heat changes by 8%. $$ \frac{\Delta C}{C_{\text{initial}}} = \frac{0.5\alpha R}{2.5 R} = 0.08 $$ $$ \frac{\alpha}{5} = \frac{8}{100} $$ $$ \alpha = \frac{40}{100} = \frac{2}{5} $$ Answer: The fraction dissociated is 2/5.
Given specific heat changes by 8%. $$ \frac{\Delta C}{C_{\text{initial}}} = \frac{0.5\alpha R}{2.5 R} = 0.08 $$ $$ \frac{\alpha}{5} = \frac{8}{100} $$ $$ \alpha = \frac{40}{100} = \frac{2}{5} $$ Answer: The fraction dissociated is 2/5.