Question

A `5 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

A decrease of `~~2700\ "Pa"`

Explanation:

We know that `PV=nRT`, where

• `P` is pressure
• `V` is volume
• `n` is number of moles
• `R` is the gas constant, and equal to `8.31\ "J"/("mol"\ "K")`
• `T` is temperature

Rearrange the formula to get `P=(nRT)/V`

Initially, the pressure `P=((4+4)*8.31*280)/5=3722.88\ "Pa"`.

With the reaction, every `4\ "mol"` of gas B bonds to `3\ "mol"` of gas A. Since there are `4\ "mol"` initially for both chemicals, there will be `1\ "mol"` of gas A left and `1\ "mol"` of the compound of A and B left, or `2\ "mol"` of gas in total.

(You can see more clearly if you write as an equation `3A+4B->A_3B_4`. Seven moles of gas become `1` mole after the reaction.)

The temperature rises to `320\ "K"`, while the volume stays constant.

Thus, the new pressure is `P=(2*8.31*320)/5=1063.68\ "Pa"`.

Thus, there is a decrease in pressure of `3722.88-1063.68=2659.2~~2700\ "Pa"`.