A 300 mL balloon containing the gases CO and CH4 has a total pressure of 450mm Hg. The partial pressure of CO is 140mm HG. What is the mole fraction of each gas

Question

Chemistry problem 

Answer 1

Recall Dalton's Law of Partial Pressures.  It states that the sum of the partial pressures is equal to the total pressure, and that the mole fraction of a gas times the total pressure is equal to its partial pressure.  Here, we know the total pressure is 450 mm Hg.  We also know the partial pressure of CO is 140 mm Hg.  This means the partial pressure of CH₄ must be 450 mm - 140 mm = 310 mm Hg

 

Mole fraction of CO = 140 mm/450 mm = 0.311

Mole fraction CH4 = 310 mm/450 mm = 0.689

 

Done another way:

Calculate moles of each gas using PV = nRT (assuming ideal gas behavior and constant temperature of 298K)

For CO, n = PV/R = (140 mm)(0.300 L)/(62.36 mm-L/Kmol)(298) = 0.00226 moles

For CH4, n = (310 mm)(0.300 L)/(62.36)(298) = 0.00500

Total moles = 0.00726

mole fraction CO = 0.00226/0.00726 = 0.311 √

mole fractio CH4 = 0.00500/0.00726 = 0.689 √

 

Answer 2

Here is yet another way, which does not require knowing the temperature:

 

Both gases (which I will call 1 and 2) occupy the same volume at the same temperature, so

 

(1)   P₁V = n₁RT

(2)   P₂V = n₂RT

(3)   (P₁ + P₂)V = (n₁ + n₂)RT

 

Divide equation (1) by equation (3) and get

 

n₁/(n₁ + n₂) = P₁/(P₁ + P₂) = 140/450 = 0.311

 

Divide equation (2) by equation (3) and get

 

n₂/(n₁ + n₂) = P₂/(P₁ + P₂) = (450 - 140)/450 = 0.689

 

This is equivalent to the first method employed in J.R.S.' answer.  Note you do not need to know the temperature.