# 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?

Apr 18, 2018

A decrease of $\approx 2700 \setminus \text{Pa}$

#### Explanation:

We know that $P V = n R T$, 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 $P = \frac{n R T}{V}$

Initially, the pressure $P = \frac{\left(4 + 4\right) \cdot 8.31 \cdot 280}{5} = 3722.88 \setminus \text{Pa}$.

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

(You can see more clearly if you write as an equation $3 A + 4 B \to {A}_{3} {B}_{4}$. Seven moles of gas become $1$ mole after the reaction.)

The temperature rises to $320 \setminus \text{K}$, while the volume stays constant.

Thus, the new pressure is $P = \frac{2 \cdot 8.31 \cdot 320}{5} = 1063.68 \setminus \text{Pa}$.

Thus, there is a decrease in pressure of $3722.88 - 1063.68 = 2659.2 \approx 2700 \setminus \text{Pa}$.