Question

A 9.15-L container holds a mixture of two gases at 27 °C. The partial pressures of gas A and gas B, respectively, are 0.214 atm and 0.618 atm. If 0.210 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

The equation for an ideal gas is: PV = nRT

and Pt = Pp₁ + Pp₂ + ........Pp_n

Now, if the problem says that there are 2 gases initially and then, a third gas is added and there are no change in volume and temperature,the best way to know the total pressure is calculating the number of moles of 1 and 2 (we can do it per separate or in the innitial mix of A and B) and then, add the moles of C, and finally, calculating the pressure with the ideal gas equation. In this case, I will calculate numbers of moles of 1 and 2 directly into the mix so:

1. Calculate the total pressure of the innitial mixture:

Pt = Pp1 + Pp2 = 0.214 + 0.618 = 0.832 atm

2. Calculate number of moles:

n = PV / RT ----> Where R = 0.082 L atm / mol K and T = 27 + 273 = 300 K

n = 0.832 x 9.15 / 0.082 x 300 = 0.3095 moles

3. Add the moles of C:

n = 0.3095 + 0.210 = 0.5195 moles

4. Calculate the Total pressure:

P = 0.5195 x 0.082 x 300 / 9.15

P = 1.3967 atm