A 6L 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 480oK to 270oK. How much does the pressure change?

1 Answer
Jul 1, 2017

the pressure change from P=7312800Nm2 to P=1121850Nm2

Explanation:

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

P=nRTV=11mol8,31JKmol480K0,006m3=7312800Nm2
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=nRTV=7mol8,31JKmol270K0,006m3=2617650Nm2
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 A2B3 toward a total of 3 gass moles, the final pressure will be:
P=nRTV=3mol8,31JKmol270K0,006m3=1121850Nm2
I think that this second hypothesis is what is required