# A 3 L container holds 5  mol and 5  mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from 340^oK to 320^oK. How much does the pressure change?

Dec 26, 2016

We have to take two things into account:
the change in moles and the change in temperature.

#### Explanation:

(1) change caused by change in number of molecules:
Reaction equation: $2 A + 3 B \to {A}_{2} {B}_{3}$
So $5$ moles of B will react with $\frac{2}{3} \times 5 = 3 \frac{1}{3}$ moles of A.
$1 \frac{2}{3}$ moles of A will be left.

Since every 3 moles of B will form 1 mole of ${A}_{2} {B}_{3}$:
$\frac{5}{3} = 1 \frac{2}{3}$ mole of ${A}_{2} {B}_{3}$ will be formed

Total moles before reaction: $5 + 5 = 10$ moles
Total moles after reaction: $1 \frac{2}{3} + 1 \frac{2}{3} = 3 \frac{1}{3}$ moles
(mixture of ${A}_{2} {B}_{3}$ and left-over $A$)

Pressure change because of amount of matter:
$3 \frac{1}{3} \div 5 = \frac{10}{3} \div 5 = \times \frac{2}{3}$

(2) Change caused by temperature change:
Temperature goes down from $34 o K$ to $320 K$

Pressure change: $\frac{320}{340} = \times \frac{16}{17}$

Total change:
Pressure is decreased by a factor of $\frac{2}{3} \times \frac{16}{17} = \frac{32}{51} \approx \times 0.63$

Or: pressure decreases to 63% of original (or by 37%)

Note:
Of course you could have used the general gas-formula

$p = \frac{n R T}{V}$ to calculate both pressures.