What best describes the collision between ideal gas molecules
1 answer:
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Explanation:
The given data is as follows.
= 250 mL,
= 750 mL
=
= 35 + 273 K = 308 K
= 35 + 273 K = 308 K
= 0.55 atm,
= 1.5 atm
P = ? , V = 10.0 L
Since, temperature is constant.
So, = PV
Now, putting the given values into the above formula as follows.
= PV
=
P = 0.126 atm
As, 1 atm = 760 torr. So, = 95.76 torr.
Thus, we can conclude that the final pressure, in torr, of the mixture is 95.76 torr.
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.
r₁ / r₂ =
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.