Question

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?

Answer 1

`10 atm`

Explanation:

First, you need to turn the information into a chemical equation:
`3A(g) + 4B(g) -> A_3B_4(g)`

Then you find the moles of product formed and moles of reactant left. You should get 1 mol of `A_3B_4(g)` and 1 mol `A(g)` left.

After that you just input the information into the formula `P_(Total) = (n_(Total)RT)/V`

`P_(Total) = (2mols * 0.08206 "L atm " mol^-1 K^-1 * 320 K)/(4L)`
`P_(Total) = 13.1296atm`

Since you only have one significant figure, the answer would be `10 atm`.