[Mayhem101]

1. The van der Waals constants for the gas He are a = 0.2452 barL^2mol^-2 and b = 0.0265 Lmol^-1. Estimate the diameter of a He atom assuming it is spherical (the volume of a sphere is V=(4/3)pir^3.

2. A 2.4 L box of He gas at 77K is found to have a unit area collision frequency Z_wall = 4.01 x 10^28 s^-1m^-2. A hole 0.2 mm in diameter is drilled in the box. How many grams of He will effuse through this hole in 1 hour (3600 s).

Is it just me, or do professors enjoy giving questions involving equations never given in the textbook or lectures?

[Borek]

Van Der Waals constant suggests use of the http://en.wikipedia.org/wiki/Van_Der_Waals_Equation

[sjb]

1. The van der Waals constants for the gas He are a = 0.2452 barL^2mol^-2 and b = 0.0265 Lmol^-1. Estimate the diameter of a He atom assuming it is spherical (the volume of a sphere is V=(4/3)pir^3.

2. A 2.4 L box of He gas at 77K is found to have a unit area collision frequency Z_wall = 4.01 x 10^28 s^-1m^-2. A hole 0.2 mm in diameter is drilled in the box. How many grams of He will effuse through this hole in 1 hour (3600 s).

Is it just me, or do professors enjoy giving questions involving equations never given in the textbook or lectures?

1. What do the a and b terms mean in the van der Waals correction to the ideal gas law?

[Mitch]

As Borek suggested start with:

[itex]\left(p + \frac{n^2 a}{V^2}\right)\left(V-nb\right) = nRT[/itex]

[Mayhem101]

This is a question from a previous year's test; the solution goes something like this:

V_He = (0.0265 L/mol) x (1 mol/6.22x10^23) = 4.401 x 10^-26 L/atom

(4/3)pir^3 = 4.401x10^-26 L = 4.401x10^-29 m^3
r^3 = 1.051x10^-29 m^3
r = 2.190 x 10^-10 m
d = 4.4x10^-10

Why did he multiply b by 1 mol/Avogadro's constant?

[Mayhem101]

Actually, I understand now. Thanks everyone for the help. How about #2?

[Jorriss]

Actually, I understand now. Thanks everyone for the help. How about #2?

It says effusion. How many equations do you know involving effusion?