Question

A `7 L` container holds `38` mol and `15` mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to five molecules of gas A and the reaction changes the temperature from `140^oK` to `210^oK`. How much does the pressure change?

Answer 1

The final pressure is 0.93 times whatever the initial pressure was.

Explanation:

`5A + 3B → A_5B_3` With an initial ratio of 38:15 gas B is the limiting reagent. Only 15 * 3/5 = 9.0 moles of A will be combined with 15 moles of B.

Thus, the final composition in the container will be 29 moles of A and 9 moles of `A_5B_3` for a total of 38 moles. The original number of moles was 53.

Assuming ideal gas behavior where n = PV/RT , with a constant volume , the ratio of the temperature change in `.^oK` is the same as the ratio of the change in the total number of moles of gas in the container times the inverse ratio of the pressures.

`(n_1*(PV)/T)_1 = (n_2*(PV)/T)_2` ; `(n_1/n_2)*(T_1/T_2) = P_2/P_1`

`P_1/P_2 = (n_2/n_1)*(T_2/T_1)`

`P_1/P_2 = (38/53)*(210)/(140)`

`P_1/P_2 = 1.07 ; P_2 = P_1 * 0.93`
so the final pressure is 0.93 times whatever the initial pressure was.