Question

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?

Answer 1

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`.