Question

A `5 L` container holds `12` mol and `9` mol of gasses A and B, respectively. Every three of molecules of gas B bind to four molecule of gas A and the reaction changes the temperature from `360^oK` to `420 ^oK`. By how much does the pressure change?

Answer 1

the pressure change from 126 Bar to 12 Bar

Explanation:

the reaction is
`3 B + 4 A =B_3A_4 + Q`
so if at the beginning you have (12+9) = 21 mol of gases, after the reaction you have 3 mol of gas( we don't know if we have still gas?)
at the beginning with the gases'law, we have:
`P= nRT/V= 21 mol xx 8,31 J/(mol xx K) xx (360K)/(0,005m^3)=126 xx 10^5 J/m^3 = 126 xx 10^5 N/m^2 = 126 xx 10^5 Pascal = 126 Bar`
after the reaction:
`P= nRT/V= 3 mol xx 8,31 J/(mol xx K) xx (420K)/(0,005m^3)=12 xx 10^5 J/m^3 = 12 xx 10^5 N/m^2 = 12 xx 10^5 Pascal = 12 Bar`