Microscopic Gas Tutorial

1) - estimate um on the T2 = 1000oC curve and then calculate M (use a ruler to get a good estimate of um)

- the estimated um is about 870 ms-1

- using all SI units leads to M = 2.792 X 10-2 kg mol-1 or 27.92 g mol-1

- estimate of um limits answer to 2 sig. figs (i.e. 28 g mol-1)

- if this is a diatomic molecule then each atom has an atomic mass of 14 g mol-1 and therefore the gas must be nitrogen, N2

2) m = M/NA = (2 X 14.0067)/(6.02214 X 1023) = 4.65174 X 10-23 g or 4.65174 X 10-26 kg

3) - estimate um on the T1 curve (use a ruler); um is approximately 400 ms-1

- using all SI units gives T = 270 K (or about 0C)

4) - similar to question 1... estimate um on the T3 curve (roughly 1500 ms-1) and then solve for M

- using all SI units gives M = 1.66 X 10-2 kg mol-1 or 16.6 g mol-1

- hydrocarbons contain only hydrogen and carbon

- a M less than 24 indicates this molecule can only contain one carbon atom

- it must be CH4

5) PV = nRT = NkT

- very important to watch that units are consistent throughout these calculations

a)

b)

- use the number density from part a) to get Zw = 4.02 X 1021 m-2s-1

c)

- use average speed from part b) and number density from part a) to get ZA = 8.74 X 103 s-1

d)

- simplest method is to use answers from parts a) and c) to get ZAA = 1.55 X 1023 m-3s-1 (other formula will give the same answer)

e)

- simplest method is to use average speed from part b) and answer from part c) to get lambda = 5.20 X 10-2m (other formula will also give the answer)

6) use the second formula for mean free path in question 5e) and substitute PV = NkT yields

- the mean free path is directly proportional to T and indirectly proportional to P, thus

- solve for the mean free path under the latter conditions gives an answer of 7.46 X 10-8 m

7) The collision frequency, Z, is given by Z = ZW X A where A is the surface area of the inside of the bulb. The surface area of a sphere is given by A = 4pir2 so we need to be able to calculate r, the radius of the bulb. The volume of the bulb is exactly 1 L so we can use

to find r = 6.20350 X 10-2m. Substitution back into the original equation, including ZW from 5b), yields a collision frequency of 1.94 X 1020 s-1.

8) rate of effusion = ZW X A, where A is the area of the orifice given by A = pir2 (remember to divide the diameter that was provided in the question by 2 to get the radius)

- substitution of ZW from part 5b) and the area of the orifice gives

rate of effusion = 1.97119 x 1012 s-1 = (# of molecules)/(time)

- we need to know what 1% of the molecules is

1% of molecules = 0.01 X total number of molecules = 0.01 X number density X volume

- we calculated the number density in part 5a), the volume was given as exactly 1 L, so 1% of the molecules is 3.53519 X 1014

- the time required for 1% of the molecules to effuse can now be calculated by rearranging the formula used above to calculate the rate of effusion

time = (# of molecules)/(rate of effusion)

therefore, for 1% of the molecules to effuse, time = 1.8 X 102 s

9) It follows from Graham's Law that