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TOPIC 5 - EFFUSION

Examples of Solved Problems

1. Find the molecular formula of hydrocarbon gas that effuses 0.8 times as fast as O2

through a small hole, the temperatures and pressures being equal.

Answer:

From Graham’s law of effusion :

1

2

2

48.20

32)8.0/1(

,32)8.0/(

2

−=

=

=

=

gmolM

M

MN

N

M

M

OeffusionRate

HCgaseffusionRate

HC

HC

HCw

w

HC

O

The molecular formula : CH4

2. A Knudsen cell was used to determine the vapor pressure of germanium at 1000 oC. During an interval of 7200 s the mass loss through a hole of radius 0.50 mm

amounted to 43 µg. What is the vapor pressure of germanium at 1000 oC? Assume

the gas to be monatomic.(Ge: M=72.64).

Answer:

( )

2/1

13

111

23

8

2/1

2/1

2/1

106.72

12733145.82

7200)105.0(

103.4

22

)2(

×

××

×××

×=

∆

∆=

∆

∆=

=∆

∆==

−−

−−−

−

−

molkg

KmolKsmkg

sm

kg

M

RT

tA

wRTM

tmNA

wP

RTM

PN

tmA

wZwJ

oAo

A

o

N

π

π

ππ

π

= 7.3 ×10-3

Pa

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3. The best laboratory vacuum pump can generate a vacuum of about 1 nTorr. At 25 oC and assuming that air consists of N2 molecules with a collision diameter of 395

pm, calculate the mean free path of the gas.

Answer:

Change the pressure to suitable unit:

217215

9 1033.11

1

1

1001325.1

760

11011 −−−

−−− ×=×

××××== skgm

Pa

smkg

atm

Pa

Torr

atmTorrnTorrP

The unit of diameter, from pm change to meter, so d = 395×10-12

m

Derive the formula using the simple formula first , then do some substitutions:

APNd

RT

dvd

v

z

v222

11 2

1

2 πρπρπλ ==

><

><=

><=

)10022.6)(1033.1()10395(2

298314.8123217212

1122

−−−−−

−−−−

××××

×=

molskgmm

KKmolsmkg

πλ

= 4.46×104 m

Exercise 5a

1. It takes gas A 3.2 times as long to effuse through an orifice as the same amount of

carbon dioxide. What is the molar mass of gas A? (450.6 g mol-1

)

2. Find the molecular formula of hydrocarbon gas that effuses 0.8 times as fast as

CO through a small hole, the temperatures and pressures being equal. (C3H8)

3. The vapor pressure of solid beryllium (9.01 g mol-1

) was measured using a

Knudsen cell. The effusion hole was 0.318 cm in diameter, and found a mass loss

of 9.54 mg in 60.1 min at temperature of 1457 K. What is the vapor pressure?

(0.966 Pa)

4. A Knudsen cell containing crystalline benzoic acid (122 g mol-1

) is carefully

weighed and placed in an evacuated chamber thermostated at 80 oC for 45 min.

The circular hole through which effusion occurs in 0.60 mm in diameter.

Calculate the sublimation pressure of benzoic acid at 80 oC in Pa from the fact

that the weight loss is 55.7 mg.(42.0 Pa)

5. A substance of M = 250 g mol-1

has a vapor pressure of 10-5

Pa at 26 oC. What

mass of the substance will effuse from a Knudsen cell in 3 h through a hole 0.2

cm in diameter. (1.36×10-8

g )

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6. Cesium (132.9 g mol-1

) was introduced into a container and heated to 500 oC.

When a hole of diameter 0.50 mm was opened in the container for 100 s, a mass

loss of 385 mg was measured. Calculate the vapor pressure of liquid cesium at

500 oC.( 11 kPa or 83 torr)

7. How long would it take 1.0 g of cesium (132.9 gmol-1

) atoms to effuse out of the

oven under the same conditions in Q7? (260 s)

8. The vapor pressure of naphthalene (M = 128.16 g mol-1

) is 17.7 Pa at 30 oC.

Calculate the weight loss in a period of 2 h of a Knudsen cell filled with

naphthalene and having a round hole 0.250 mm in radius. (0.071 mg)

9. For a liquid in equilibrium with its vapor, the rate of evaporation of liquid

molecules equals the rate of condensation of gas molecules. For Octoil,

C6H4(COOC8H17)2,(M=390.6 g mol-1

) the vapor pressure is 0.010 torr at 393 K.

Calculate the number of Octoil molecules that evaporate into vacuum from a 1.0

cm2 surface of the liquid at 393 K in 1.0 s; also calculate the mass that evaporates.

(8.9×1017

molecues , 0.58 mg)