Question

Determine the partial pressure and number of moles of each gas in a 14.25-L vessel at 30 degrees C containing a mixture of Xenon and neon gases only. The total pressure in the vessel is 6.70 atm and the mole fraction of xenon is 0.721 What is the partial pressure of Xenon? What is the partial pressure of neon? What is the number of moles of Xenon? What is the number of moles of neon?

Answer 1

partial pressure of Xenon = Total pressure x mole fraction of Xenon

= 6.70 atm x 0.721

= 4.83 atm

So pertial pressure of Neon = total pressure - partial pressure of Xenon

= 6.70 - 4.83

= 1.87 atm

We know that mole fraction of Xenon = number of moles of Xe / total number of moles = 0.721

n _Xe / ( n_Xe + n _Ne ) = 0.721

We know that ideal gas equation is PV = nRT

Where

T = Temperature = 30 oC = 30+273 = 303 K

P = pressure = 6.70 atm

n = No . of moles = ?

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

V= Volume of the vessel = 14.25 L

Plug the values we get n = PV / (RT) = 3.84 moles

So n = n_Xe + n _Ne = 3.84

n _Xe / ( n_Xe + n _Ne ) = 0.721

n _Xe / 3.84 = 0.721

n _Xe = 2.77 moles

n_Xe + n _Ne = 3.84

n _Ne = 3.84 - 2.77 = 1.07 atm