Question

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?

Answer 1

the pressure change from `P= 7312800 N/m^2` to `P= 1121850 N/m^2`

Explanation:

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

`P= (nRT)/V = (11 mol 8,31 J/(K mol) 480 K)/(0,006 m^3) = 7312800 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= (nRT)/V = (7 mol 8,31 J/(K mol) 270 K)/(0,006 m^3) = 2617650 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_2B_3` toward a total of 3 gass moles, the final pressure will be:
`P= (nRT)/V = (3 mol 8,31 J/(K mol) 270 K)/(0,006 m^3) = 1121850 N/m^2`
I think that this second hypothesis is what is required