# A 6 L container holds 5  mol and 6  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 480^oK to 270^oK. How much does the pressure change?

##### 1 Answer
Jul 1, 2017

the pressure change from $P = 7312800 \frac{N}{m} ^ 2$ to $P = 1121850 \frac{N}{m} ^ 2$

#### Explanation:

To calculate the initial pressure you can use the general law of perfect gasses: PV = nRT .

$P = \frac{n R T}{V} = \frac{11 m o l 8 , 31 \frac{J}{K m o l} 480 K}{0 , 006 {m}^{3}} = 7312800 \frac{N}{m} ^ 2$
to know the final pressure, i suppose that there are still 3 moles of A, 3 moles of B and it was formed 1 mol of gas from the union of the moles reacted toward a total of 7 gass moles. The new pressure will be:

$P = \frac{n R T}{V} = \frac{7 m o l 8 , 31 \frac{J}{K m o l} 270 K}{0 , 006 {m}^{3}} = 2617650 \frac{N}{m} ^ 2$
if instead all the molecules (6moles) of B react with the maximum possible number of the moles of A (4 moles, as the ratio is 3:2) at the end you will have still a mole of A and it was formed 2 moles of ${A}_{2} {B}_{3}$ toward a total of 3 gass moles, the final pressure will be:
$P = \frac{n R T}{V} = \frac{3 m o l 8 , 31 \frac{J}{K m o l} 270 K}{0 , 006 {m}^{3}} = 1121850 \frac{N}{m} ^ 2$
I think that this second hypothesis is what is required