# 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?

Oct 15, 2017

-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 ${A}_{4} {B}_{5}$ 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. $P = n R \frac{T}{V} = \frac{\left(8 + 5\right) m o l \times 0 , 082 \frac{L \times A t m}{m o l \times K} \times 320 K}{5 L} = 68 A t m$
at the end you will have $P = n R \frac{T}{V} = \frac{\left(4 + n A B\right) m o l \times 0 , 082 \frac{L \times A t m}{m o l \times K} \times 450 K}{5 L} = 29 , 5 A t m$
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