# A 5 L container holds 11  mol and 7  mol of gasses A and B, respectively. Every four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from 360^oK to 420 ^oK. By how much does the pressure change?

Nov 5, 2017

It decreases by 22.2%

#### Explanation:

This is really chemistry question, although it does overlap with the physics topic indicated.
Volume is directly related to molar amounts, temperature and pressure according to the Ideal Gas Law: $P \cdot V = \left(n \cdot R \cdot T\right)$
The change in molar amounts is derived from the balanced chemical equation.

11A + 7B → A_3B_4 + 8A + 3B (unreacted) So, we initially have 18 moles of gases and end up with 12 moles in the 5L volume.

The volume is constant in this case, and the gas constant is constant, so we only need the equation that shows the change in pressure with respect to molar quantities (n) and temperature (T) for a calculation of the ratio change.
${P}_{2} / {P}_{1} = \left({n}_{2} / {n}_{1}\right) \times \left({T}_{2} / {T}_{1}\right)$
${P}_{2} / {P}_{1} = \left(\frac{12}{18}\right) \times \left(\frac{420}{360}\right)$ ; ${P}_{2} / {P}_{1} = 0.778$
So, the pressure decreases by 1 – 0.778 = 0.222 or 22.2%