A 9 L container holds 3  mol and 2  mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from 375^oK to 425 ^oK. How much does the pressure change by?

Feb 5, 2018

It decreases by 32%

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.

3A + 2B → AB_2 + 2A (unreacted)
So, we initially have 5 moles of gases and end up with 3 moles in the 9L 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.
${P}_{2} / {P}_{1} = \left({n}_{2} / {n}_{1}\right) \times \left({T}_{2} / {T}_{1}\right)$

${P}_{2} / {P}_{1} = \left(\frac{3}{5}\right) \times \left(\frac{425}{375}\right)$ ; ${P}_{2} / {P}_{1} = 0.68$
So, the pressure decreases by 1 – 0.68 = 0.32 or 32%