# A 4 L container holds 4  mol and 4  mol of gasses A and B, respectively. Groups of four of molecules of gas B bind to three molecule of gas A and the reaction changes the temperature from 280^oK to 320^oK. By how much does the pressure change?

Apr 19, 2017

$10 a t m$

#### Explanation:

First, you need to turn the information into a chemical equation:
$3 A \left(g\right) + 4 B \left(g\right) \to {A}_{3} {B}_{4} \left(g\right)$

Then you find the moles of product formed and moles of reactant left. You should get 1 mol of ${A}_{3} {B}_{4} \left(g\right)$ and 1 mol $A \left(g\right)$ left.

After that you just input the information into the formula ${P}_{T o t a l} = \frac{{n}_{T o t a l} R T}{V}$

${P}_{T o t a l} = \frac{2 m o l s \cdot 0.08206 \text{L atm } m o {l}^{-} 1 {K}^{-} 1 \cdot 320 K}{4 L}$
${P}_{T o t a l} = 13.1296 a t m$

Since you only have one significant figure, the answer would be $10 a t m$.