Question

A `5 L` container holds `8` mol and `5` mol of gasses A and B, respectively. Every five of molecules of gas B bind to four molecule of gas A and the reaction changes the temperature from `320^oK` to `450 ^oK`. By how much does the pressure change?

Answer 1

-39,5 Atm

Explanation:

If every 5 of molecules of gas B bind to 4 molecules of gas A and You have at first
5 mol of B and 8 of A, at the end it remains 4 mol of A and one compound `A_4B_5` of which we don't know if it is condensed or gas, neither what is its formula so i suppose it is solid.

Al the beginning the Pressure is given by ideal gases law. `P = nRT/V = ((8+5)mol xx 0,082 (L xx Atm) /(mol xx K) xx 320K)/ (5L) = 68 Atm`
at the end you will have `P = nRT/V = ((4 + nAB)mol xx 0,082 (L xx Atm) /(mol xx K) xx 450K)/ (5L) = 29,5 Atm`
where with nAB i have indicate the possible number of gas moles formed by the reaction I considered null. the pressure change decreasing from 69 to 29,5 atm that is - 39,5 atm.

Pay attention!. With a pressure so high you cannot use the ideal gases law, but the real gases law and for that you must know what are your gases