# 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?

Jul 6, 2018

the pressure change from 126 Bar to 12 Bar

#### Explanation:

the reaction is
$3 B + 4 A = {B}_{3} {A}_{4} + Q$
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:
$P = n R \frac{T}{V} = 21 m o l \times 8 , 31 \frac{J}{m o l \times K} \times \frac{360 K}{0 , 005 {m}^{3}} = 126 \times {10}^{5} \frac{J}{m} ^ 3 = 126 \times {10}^{5} \frac{N}{m} ^ 2 = 126 \times {10}^{5} P a s c a l = 126 B a r$
after the reaction:
$P = n R \frac{T}{V} = 3 m o l \times 8 , 31 \frac{J}{m o l \times K} \times \frac{420 K}{0 , 005 {m}^{3}} = 12 \times {10}^{5} \frac{J}{m} ^ 3 = 12 \times {10}^{5} \frac{N}{m} ^ 2 = 12 \times {10}^{5} P a s c a l = 12 B a r$