Partial pressure of gas A is 1.31 atm and that of gas B is 0.44 atm.
The partial pressure of a gas in a mixture can be calculated as
Pi = Xi x P
Where Pi is the partial pressure; Xi is mole fraction and P is the total pressure of the mixture.
Therefore we have Pa = Xa x P and Pb = Xb x P
Let us find Xa and Xb
Χa = mol a/ total moles = 2.50/(2.50+0.85) = 2.50/3.35 = 0.746
Xb = mol b/total moles = 0.85/(2.50+0.85) = 0.85/3.35 = 0.254
Total pressure P is given as 1.75 atm
Pa = Xa x P = 0.746 x 1.75 = 1.31atm
Partial pressure of gas A is 1.31 atm
Pb = Xb x P = 0.254 x 1.75 = 0.44atm
Partial pressure of gas B is 0.44 atm.
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According to
Graham's Law ," the rates of effusion or diffusion of two gases are inversely proportional to the square root of their molecular masses at given pressure and temperature".
r₁ / r₂ =

---- (1)
r₁ = Rate of effusion of He
r₂ = Rate of Effusion of O₃
M₁ = Molecular Mass of He = 4 g/mol
M₂ = Molecular Mass of O₃ = 48 g/mol
Putting values in eq. 1,
r₁ / r₂ =

r₁ / r₂ =

r₁ / r₂ =
3.46
Result: Therefore, Helium will effuse
3.46 times more faster than Ozone.