Question

A `2 L` container holds `16` mol and `24` mol of gasses A and B, respectively. Groups of five of molecules of gas B bind to three molecules of gas A and the reaction changes the temperature from `340^oK` to `420^oK`. How much does the pressure change?

Answer 1

For an ideal gas, we know, `(PV)/(nT) = "constant"`

For our case, the volume remains constant. Hence

`P_1/(n_1T_1) = P_2/(n_2T_2) => P_1/P_2 = (n_1T_1)/(n_2T_2)`

Given, `T_1 = 340 K`, and `T_2 = 420 K`

`n_1 = 16+24 = 40`

For `n_2`, note that, `16/24>3/5`. Hence gas B gets used up in the reaction, and in the reaction, `24/5` moles of gas `A_3B_5` is produced and `(16- 24/5xx3) = 8/5` moles of gas A remains. Hence,

`n_2 = 24/5+8/5 = 32/5`

Hence, `P_1/P_2 = (n_1T_1)/(n_2T_2) = (40xx340)/(24/5xx420) ~~6.74`