A1.00-l gas sample at 100.°c and 500. torr contains 52.0% helium and 48.0% xenon by mass. what are the partial pressures of the individual gases?

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

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.