Question

Container A holds 712 mL of ideal gas at 2.40 atm. Container B holds 164 mL of ideal gas at 4.20 atm. If the gases are allowed to mix together, what is the partial pressure of each gas in the total volume?

Answer 1

Assume T = 298 K

V1 = 712 mL = 0.712 L

P1 = 2.40 atm

n1 = ?

n1 = PV/(RT) = 2.4*0.712 / (0.082*298) = 0.06992 moles

for gas 2

n2 = PV/(RT) = 4.2*0.164 / (0.082*298) = 0.028187 moles

then

n3 = n1+n2 = 0.06992 +0.028187 = 0.098107

substitute :

Vtotal = 0.712+0.164 = 0.876 L

T = 298K, P = ?

P = nRT/V

P = (0.098107*0.082*298)/(0.876 ) = 2.736 atm

now...

Partial pressure:

gas 1:

P-gas 1 = 0.06992 /(0.098107) * 2.736 = 1.949923 atm

P-gas 2 = 0.028187 /(0.098107) * 2.736 = 0.78607 atm