Question

Container A holds 732 mL of ideal gas at 2.30 atm. Container B holds 114 mL of ideal gas at 4.30 atm. If the gases are allowed to mix together, what is the resulting pressure?

Answer 1

Calculation of number of moles of ideal gas in container A :

We know that PV = nRT

Where

T = Temperature = 298 K

P = pressure = 2.30 atm

n = No . of moles = ?

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = 732 mL = 0.732 L

Plug the value we get n = (PV) / (RT)

                                  = 0.0688 moles

Calculation of number of moles of ideal gas in container B :

We know that PV = nRT

Where

T = Temperature = 298 K

P = pressure = 4.30 atm

n = No . of moles = ?

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = 114 mL = 0.114 L

Plug the value we get n = (PV) / (RT)

                                  = 0.020 moles

So total number of moles , N = 0.020 + 0.0688 = 0.0888 moles

Total volume after mixing is , V = 114+732 = 846 mL = 0.846 L

So Total pressure after mixing P = NRT / V

                                                 = ( 0.0888x0.0821x298) / 0.846

                                                 = 2.57 atm

Therefore the resulting pressure is 2.57 atm