A 5 L container holds 8 mol and 10 mol of gasses A and B, respectively. Every five of molecules of gas B bind to two molecule of gas A and the reaction changes the temperature from 360^oK to 270 ^oK. By how much does the pressure change?

1 Answer
Feb 2, 2018

The total pressure P inside the V = 5 L container is due to the pressures of both gasses A and B:

P = P_A + P_B.

According to the ideal gas law:

PV = nRT.

If we consider gasses A and B to be ideal, then this equation must apply to each one of them. For the total pressure, this means that:

P = (n_ART_A)/V_A + (n_BRT_B)/V_B.

If we also assume that both gasses can occupy all the container's volume and that they will always be at thermal equilibrium with each other, then:

P = (RT)/V(n_A + n_B).

This means that we can obtain the pressure's value if we know the above quantities. Since R = 8,2 J/(molK) is constant and we know the values for T and V, we must figure out what happens to the number of mols of each gas after the reaction takes place.

Since it takes 5 molecules of gas B to bind to 2 molecules of gas A, we can extend this proportion to their corresponding number of mol. Since initially we have n_(1A) = 8 mol and n_(1B) = 10 mol, then, after the reaction takes place, we will end up with n_(2A) = 4 mol, n_(2B) = 0 mol and n_C = 2 mol of the new gas, the product of the A and B reaction.

Then, our final pressure will be described by:

P_2 = (RT)/V(n_A + n_B + n_C).

(Remember that P_1 = (RT)/V(n_A + n_B).

Taking the values to perform the calculations:

P_1 = (0.082 (atm cancel(L))/(cancel(mol)cancel(K)) * 360 cancel(K))/(5 cancel(L)) * (8 + 10) cancel(mol);

P_1 = 106.3 atm.

After the reaction takes place:

P_2 = (0.082 (atm cancel(L))/(cancel(mol)cancel(K)) * 270 cancel(K))/(5 cancel(L)) * (4 + 0 + 2) cancel(mol);

P_2 = 26.6 atm.

Therefore, the change in pressure, DeltaP, is:

DeltaP = P_2 - P_1;

DeltaP = -79,7 atm.