States of matter grams law of diffusion

Question

States of matter's GRAMS LAWS OF DIFFUSION  asked in class but didn't get the best way to get the answer of this questionThe pressure of bulb dropped from 2000 to 1500 mm HG in 47 min when contained o2 leaked through a small hole . The mix of o2 and another gas of molecular weight 79 in molar ratio of 1:1 . At total pressure of 4000 mm of HG was introduced .find molar ratio of two gases remaining in bulb after 74 min 

Answer 1

Graham's Law of Diffusion/Effusion states that the rate of diffusion is inversely proportional to the square root of the molar mass.  In equation form, rate gas 1/rate gas 2 = √MW gas 2/√MW gas 1

For O2, molar mass = 32 and √32 = 5.66

Rate effusion of O2 = 500 mm Hg/47 min = 10.6 mm Hg/min

For gas X, molar mass = 79 and √79 = 8.89

From Graham's Law 10.6/x = 8.89/5.66

x = rate of effusion of gas x = 6.75 (note gas x effuses at a slower rate than O2, as expected from the molar mass)

 

For the combined mixture of gases:

For O2:  10.6 mm Hg/min x 74 min = 784.4 mm Hg pressure lost due to O2 effusion

For X:  6.75 mm Hg/min x 74 min = 499.5 mm Hg pressure lost due to X effusion

 

Beginning mole ratio = 1:1 and total pressure = 4000 mm Hg

Po O2 = 2000 mm Hg

Po X = 2000 mm Hg

Pf O2 = 2000 - 784 = 1216 mm 

Pf X = 2000 - 500 = 1500 mm

mole ratio gas x to O2 = 1500/1216 = 1.23 : 1