Question

A 7.50 -L container holds a mixture of two gases at 33 °C. The partial pressures of gas A and gas B, respectively, are 0.214 atm and 0.571 atm. If 0.220 mol of a third gas are added with no change in volume or temperature, what will the total pressure become?

Answer 1

Volume of the Container, V = 7.50 L

Temperature, T =

Gas constant, R = 0.082

Before addition of a third gas:

Partial pressure of gas A,

Partial pressure of gas B,

Assuming that the gases are ideal, the partial pressure of a gas is nothing but its contribution to the total pressure.

Hence, using the ideal gas law, we can calculate the number of moles of A and B in our container.

hence, the number of moles of gas A in the mixture is 0.064 mol.

Similarly for gas B,

Now we are adding 0.220 mol of a third gas(Lets call it C).

Hence, total number of moles of all three gases(A+B+C) is

The total pressure only depends on the total number of moles of gas at a particulat temperature and pressure assuming all the gases are non-interacting ideal gases.

Hence, using the ideal gas equation, total pressure P is

Hence, the total pressure after addition of a third gas is 1.510 atm.