Solution 24: Adiabatic Dissociation
Conservation of Energy:
In an adiabatic container with constant volume, total energy is conserved. Bond energy is consumed from internal thermal energy. $$ U_{\text{initial}} = U_{\text{final}} + E_{\text{dissociation}} $$ $$ N \left( \frac{5}{2}kT_1 \right) = \left[ N(1-\alpha)\frac{5}{2}kT_2 + N(2\alpha)\frac{3}{2}kT_2 \right] + N\alpha\varepsilon $$ Dividing by $N$: $$ \frac{5}{2}kT_1 = \frac{5}{2}kT_2 + \frac{1}{2}kT_2\alpha + \alpha\varepsilon $$ $$ \frac{5}{2}k(T_1 – T_2) = \alpha \left( \varepsilon + \frac{kT_2}{2} \right) $$ $$ \alpha = \frac{5k(T_1 – T_2)}{2\varepsilon + kT_2} $$
In an adiabatic container with constant volume, total energy is conserved. Bond energy is consumed from internal thermal energy. $$ U_{\text{initial}} = U_{\text{final}} + E_{\text{dissociation}} $$ $$ N \left( \frac{5}{2}kT_1 \right) = \left[ N(1-\alpha)\frac{5}{2}kT_2 + N(2\alpha)\frac{3}{2}kT_2 \right] + N\alpha\varepsilon $$ Dividing by $N$: $$ \frac{5}{2}kT_1 = \frac{5}{2}kT_2 + \frac{1}{2}kT_2\alpha + \alpha\varepsilon $$ $$ \frac{5}{2}k(T_1 – T_2) = \alpha \left( \varepsilon + \frac{kT_2}{2} \right) $$ $$ \alpha = \frac{5k(T_1 – T_2)}{2\varepsilon + kT_2} $$
Pressure Ratio:
Using $PV = N_{total}kT$. $$ N_{total} = N(1-\alpha) + 2N\alpha = N(1+\alpha) $$ $$ \frac{P_2}{P_1} = \frac{N(1+\alpha)T_2}{N T_1} = (1+\alpha)\frac{T_2}{T_1} $$ $$ \frac{P_2}{P_1} = \left( 1 + \frac{5k(T_1 – T_2)}{2\varepsilon + kT_2} \right) \frac{T_2}{T_1} $$
Using $PV = N_{total}kT$. $$ N_{total} = N(1-\alpha) + 2N\alpha = N(1+\alpha) $$ $$ \frac{P_2}{P_1} = \frac{N(1+\alpha)T_2}{N T_1} = (1+\alpha)\frac{T_2}{T_1} $$ $$ \frac{P_2}{P_1} = \left( 1 + \frac{5k(T_1 – T_2)}{2\varepsilon + kT_2} \right) \frac{T_2}{T_1} $$