Graham’s Law

KE = ½mv2

Speed of diffusion/effusionSpeed of diffusion/effusion– Kinetic energy is determined by the temperature of

the gas.

– At the same temp & KE, heavier molecules move more slowly.

• Larger m smaller v

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Graham’s Law

2

21

222

2112

21 v m

1v mv m

v m

1

1

22

2

21

mm

v

v

1

2

2

1

mm

vv

Consider two gases at same temp.Gas 1: KE1 = ½ m1 v1

2

Gas 2: KE2 = ½ m2 v22

Since temp. is same, then… KE1 = KE2

½ m1 v12 = ½ m2 v2

2

m1 v12 = m2 v2

2

Divide both sides by m1 v22…

Take square root of both sides to get Graham’s Law:

2

2

22

2

2

2

2

Cl

COCOCl

Cl

CO

CO

Cl

1

2

2

1

m

m v v

m

m

v

v

mm

vv

m/s 320 g 71g 44

m/s 410 v2Cl

On average, carbon dioxide travels at 410 m/s at 25oC.

Find the average speed of chlorine at 25oC.

**Hint: Put whatever you’re looking for in the numerator.

2

2

2

2

F

unk

2

unk

F

F

unk

unk

F

1

2

2

1

m

m

v

v

m

m

v

v

mm

vv

Kr amu 82.9 394582

amu 38 v

v m m

2

unk

FFunk

2

2

At a certain temperature fluorine gas travels at 582 m/s

and a noble gas travels at 394 m/s. What is the noble gasnoble gas?

unkCH v 1.58 v4

444

4

CH

unk2

CH

unk

unk

unk

CH

unk

unk

CH

m

m (1.58)

m

m

v

v 1.58

m

m

v

v

Ar amu 39.9 amu) (16 (1.58) m (1.58) m 2CH

2unk 4

CH4 moves 1.58 times faster than which noble

gas?Governing relation:

So HCl dist. = 1.000 m/s (0.487 s) = 0.487 m

HClNHNH

HCl

HCl

NH v 1.465 v 1.465 17

36.5

m

m

v

v3

3

3

m/s 1.000 vHCl m/s 1.465 v

3NH

s 0.487 t t 1.465 t 1.000 dist. NH dist. HCl 1.20 3

HCl NH3

1.20 m

DISTANCE = RATE x TIME

HCl and NH3 are released at same time from opposite ends

of 1.20 m horizontal tube. Where do gases meet?

Velocities are relative; pick easy #s:

Graham’s Law

2

21

222

2112

21 v m

1v mv m

v m

1

1

22

2

21

mm

v

v

1

2

2

1

mm

vv

Consider two gases at same temp.Gas 1: KE1 = ½ m1 v1

2

Gas 2: KE2 = ½ m2 v22

Since temp. is same, then… KE1 = KE2

½ m1 v12 = ½ m2 v2

2

m1 v12 = m2 v2

2

Divide both sides by m1 v22…

Take square root of both sides to get Graham’s Law:

“mouse in the house”

Gas Diffusion and Effusion

Graham's law governs effusion and diffusion of gas molecules.

Thomas Graham(1805 - 1869)

Rate of effusion is inversely proportional to its molar mass.

Rate of effusion is inversely proportional to its molar mass.

Aof massB of mass

B of Rate Aof Rate

Graham’s Law

Graham’s LawGraham’s Law– Rate of diffusion of a gas is inversely related to the

square root of its molar mass.

– The equation shows the ratio of Gas A’s speed to Gas B’s speed.

A

B

B

A

m

m

v

v

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Graham’s Law• The rate of diffusion/effusion is

proportional to the mass of the molecules

– The rate is inversely proportional to the square root of the molar mass of the gas

v 1

m

250 g

80 g

Large molecules move slower than small molecules

Step 1) Write given information

GAS 1 = helium

M1 = 4.0 g

v1 = x

GAS 2 = chlorine

M2 = 71.0 g

v2 = x

He Cl2

Step 2) Equation

1

2

2

1

mm

vv

Step 3) Substitute into equation and solve

v1

v2

=71.0 g

4.0 g

4.21

1

Find the relative rate of diffusion of helium and chlorine gas

He diffuses 4.21 times faster than Cl2

Cl35.453

17

He4.0026

2

If fluorine gas diffuses at a rate of 363 m/s at a certain temperature, what is the rate of diffusion of neon gas at the same temperature?

Step 1) Write given information

GAS 1 = fluorine

M1 = 38.0 g

v1 = 363 m/s

GAS 2 = Neon

M2 = 20.18 g

v2 = x

F2 Ne

Step 2) Equation

1

2

2

1

mm

vv

Step 3) Substitute into equation and solve

363 m/s

v2

=20.18 g

38.0 g498 m/s

Rate of diffusion of Ne = 498 m/s

Ne20.1797

10

F18.9984

9

Find the molar mass of a gas that diffuses about 4.45 times faster than argon gas.

What gas is this?

Hydrogen gas: H2

Step 1) Write given information

GAS 1 = unknown

M1 = x g

v1 = 4.45

GAS 2 = Argon

M2 = 39.95 g

v2 = 1

? Ar

Step 2) Equation

1

2

2

1

mm

vv

Step 3) Substitute into equation and solve

4.45

1=

39.95 g

x g2.02 g/mol

H1.00794

1

Ar39.948

18

Where should the NH3 and the HCl meet in the tube if it is approximately 70 cm long?

41.6 cm from NH3

28.4 cm from HCl

Ammonium hydroxide (NH4OH) is ammonia (NH3) dissolved in water (H2O)

NH3(g) + H2O(l) NH4OH(aq)

Stopper

1 cm diameter

Cotton plug

Cotton plug

Stopper

Clamps

70-cm glass tube

Graham’s Law of Diffusion

HCl NH3

100 cm 100 cm

Choice 1: Both gases move at the same speed and meet in the middle.

NH4Cl(s)

Diffusion

HCl NH3

81.1 cm 118.9 cm

NH4Cl(s)

Choice 2: Lighter gas moves faster; meet closer to heavier gas.

Calculation of Diffusion Rate

NH3V1 = XM1 = 17 amu

HCl V2 = XM2 = 36.5 amu

Substitute values into equation

V1 moves 1.465x for each 1x move of V2

NH3 HCl

1.465 x + 1x = 2.465

200 cm / 2.465 = 81.1 cm for x

1 2

2 1

v m

v m

1

2

36.5

17

v

v

1

2

1.465v

xv

Calculation of Diffusion Rate

V1 m2

V2 m1

= NH3V1 = XM1 = 17 amu

HCl V2 = XM2 = 36.5 amu

Substitute values into equation

V1 36.5V2 17

=

V1 V2

= 1.465

V1 moves 1.465x for each 1x move of v2

NH3 HCl

1.465 x + 1x = 2.465

200 cm / 2.465 = 81.1 cm for x

Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

Determine the relative rate of diffusion for krypton and bromine.

1.381

Kr diffuses 1.381 times faster than Br2.

Kr

Br

Br

Kr

m

m

v

v2

2

A

B

B

A

m

m

v

v

g/mol83.80

g/mol159.80

Graham’s Law

The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vA/vB”.

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Kr83.80

36

Br79.904

35

A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions?

A

B

B

A

m

m

v

v

2

2

2

2

H

O

O

H

m

m

v

v

g/mol 2.02

g/mol32.00

m/s 12.3

vH 2

Graham’s Law

3.980m/s 12.3

vH 2

m/s49.0 vH 2

Put the gas with the unknown

speed as “Gas A”.

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O15.9994

8

H1.00794

1

An unknown gas diffuses 4.0 times faster than O2. Find its molar mass.

Am

g/mol32.00 16

A

B

B

A

m

m

v

v

A

O

O

A

m

m

v

v2

2

Am

g/mol32.00 4.0

16

g/mol32.00 mA

2

Am

g/mol32.00 4.0

g/mol2.0

Graham’s Law

The first gas is “Gas A” and the second gas is “Gas B”. The ratio “vA/vB” is 4.0.

Square both sides to get rid of the square

root sign.

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O15.9994

8

H1.0

1H2 = 2 g/mol