Question

A `3 L` container holds `5` mol and `5` mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from `340^oK` to `320^oK`. How much does the pressure change?

Answer 1

We have to take two things into account:
the change in moles and the change in temperature.

Explanation:

(1) change caused by change in number of molecules:
Reaction equation: `2A+3B->A_2B_3`
So `5` moles of B will react with `2/3xx5=3 1/3` moles of A.
`1 2/3` moles of A will be left.

Since every 3 moles of B will form 1 mole of `A_2B_3`:
`5/3=1 2/3` mole of `A_2B_3` will be formed

Total moles before reaction: `5+5=10` moles
Total moles after reaction: `1 2/3 + 1 2/3=3 1/3` moles
(mixture of `A_2B_3` and left-over `A`)

Pressure change because of amount of matter:
`3 1/3div5=10/3div5=xx2/3`

(2) Change caused by temperature change:
Temperature goes down from `34oK` to `320K`

Pressure change: `320/340=xx16/17`

Total change:
Pressure is decreased by a factor of `2/3xx16/17=32/51~~xx0.63`

Or: pressure decreases to `63%` of original (or by `37%`)

Note:
Of course you could have used the general gas-formula

`p=(nRT)/V` to calculate both pressures.