Rates of effusion/diffusion problems? (will give best answer)?

Could you show as much work as possible because i have no clue how to do this. Thanks!

1) If a car is running and produces both CO and CO2, compare the rate of diffusion for the two particles.

2) Compare the rates of effusion of sulfur dioxide and oxygen gas if they are at the same temperature and pressure.

Thanks for your time!

Answer:
Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of the molecular mass of the gas:

R = k/sqrt(M)

where R is the diffusion rate, M is the molecular mass, and k is a constant of proportionality.

If we have two gases with masses M1 and M2, their rates will be therefore be related by:

R1/R2 = sqrt(M2/M1)

In this case, CO has a molecular mass of 12+16 = 28 gm/mol and CO2 has a molecular mass of 12+2*16 = 44 gm/mol, so

R_CO/R_CO2 = sqrt(M_CO2/M_CO) = sqrt(44/28)
R_CO/R_CO2 = 1.254

The CO will diffuse 1.254 times as fast as the CO2

Graham's law of effusion is similar, and states that the rates of effusion of two gases at the same temperature and pressure are related by:

E1/E2 = sqrt(M2/M1)

where E1 and E2 are the rates of effusion of the gases through a small aperture.

O2 gas has a molecular mass of 2*16 = 32 gm/mol, and SO2 has a molecular mass of 32+16*2 = 64 gm/mol, so:

E_O2/E_SO2 = sqrt(64/32)
E_O2/E_SO2 = 1.414

The O2 will effuse 1.414 times faster than the SO2.
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