Answer:
Partial pressure of He=486 torr, partial pressure of Xe= 14 torr
Explanation:
Using the equation, PV=nRT----------------------------------------(1)
Making n the subject of the formula;
n= PV/RT---------------------(2)
Where n= number of moles, v= volume, T= temperature, P= volume.
n= (500 torr/760 torr × 1 atm)× 1L ÷ 373K ×0.082 L atmK^-1. Mol^-1
n= 0.6579 atm.L/ 30.6233
n= 0.0215 mol.
Let the total mass of the gas= 2b(in g).
Mass of helium gas = b (in g) = mass of Xenon gas
Mole of helium gas= b(in g) / 4 gmol^-1
=b/4 mol
Mole of Xenon= b g/131.3 gmol^-1
= b/131.3 mol.
Solving for b, we have;
b/4+b/131.3 = 0.0215 mole
(131.3+4)b/525.2= 0.0215
Multiply both sides by 1/135.3.
b= 11.2918/135.3
b= 0.0835 g
Mole of He gas= 0.0835/4= 0.0209
Mole of Xe gas= 0.0215- 0.0209
= 0.0006 mol
Mole fraction of He = 0.0209/0.0215
= 0.972
Mole fraction of Xe= 0.0006/0.0215
= 0.028
Partial pressure of He gas= 0.972× 500 torr= 486 torr
Partial pressure of Xe gas= 0.028 ×500 torr = 14 torr
