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?

1 Answer
Dec 26, 2016

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.