Transcript The Gas Laws
Diffusion vs. Effusion
Diffusion
- The tendency of the molecules of a given substance to move from regions of higher concentration to regions of lower concentration
Examples:
A scent spreading throughout a room or people entering a theme park
Effusion
- The process by which gas particles under pressure pass through a tiny hole
Examples:
Air slowly leaking out of a tire or helium leaking out of a balloon
Effusion
Particles in regions of high concentration spread out into regions of low concentration, filling the space available to them.
To use Graham’s Law, both gases must be at same temperature. diffusion : particle movement from high to low concentration
NET
MOVEMENT effusion : diffusion of gas particles through an opening For gases, rates of diffusion & effusion obey Graham’s law: more massive = slow; less massive = fast
Graham’s Law
Speed 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
KE = ½mv
2
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Derivation of Graham’s Law
• The average kinetic energy of gas molecules depends on the temperature:
KE
1 2
mv
2 where
m
is the mass and
v
is the speed
KE
1
KE
2 1 2 1 2
m
1
v
1 2
m
2
v
2 2
Graham’s Law
Consider two gases at same temp.
Gas 1: Gas 2: KE KE 1 2 = ½ m 1 = ½ m 2 v 1 2 v 2 2 Since temp. is same, then… ½ m 1 m 1 KE 1 v 1 2 v 1 2 = KE 2 = ½ m = m 2 2 v 2 2 v 2 2 Divide both sides by m 1 v 2 2 … 1 m 1 v 2 2 m 1 v 1 2 m 2 v 2 2 1 m 1 v 2 2 v 1 2 v 2 2 m 2 m 1 Take square root of both sides to get Graham’s Law: v 1 v 2 m 2 m 1
Graham’s Law
Graham’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 .
v A v B
m B m A
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250 g
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
80 g Large molecules move slower than small molecules
He
4.0026
2 Find the relative rate of diffusion of helium and chlorine gas 17 Cl
35.453
Step 1) Write given information GAS 1 = helium M 1
v
1 = 4.0 g = x Step 2) Equation v 1 v 2 m 2 m 1 He Step 3) Substitute into equation and solve v v 1 2 GAS 2 = chlorine M 2
v
2 = 71.0 g = x = 71.0 g 4.0 g Cl 2 4.21
1
He diffuses 4.21 times faster than Cl 2
9 F
18.9984
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?
10 Ne
20.1797
Step 1) Write given information GAS 1 = fluorine F 2 M 1
v
1 = 38.0 g = 363 m/s GAS 2 = Neon M 2
v
2 = 20.18 g = x Ne Step 2) Equation Step 3) Substitute into equation and solve v 1 v 2 m 2 m 1 363 m/s v 2 = 20.18 g 38.0 g
Rate of diffusion of Ne = 498 m/s
498 m/s
Find the molar mass of a gas that diffuses 18 Ar
39.948
about 4.45 times faster than argon gas.
Step 1) Write given information GAS 1 = unknown ?
M 1
v
1 = x g = 4.45
GAS 2 = Argon M 2
v
2 = 39.95 g = 1 Ar Step 2) Equation Step 3) Substitute into equation and solve v 1 v 2 m 2 m 1 What gas is this?
4.45
1 = 39.95 g x g
Hydrogen gas: H 2
2.02 g/mol 1 H
1.00794
HCl and NH 3 are released at same time from opposite ends of 200 cm horizontal tube. Where do the two gases meet?
Stopper Clamps 1 cm diameter Stopper Cotton plug 70-cm glass tube Cotton plug Ammonium hydroxide (NH 4 OH) is ammonia (NH 3 ) dissolved in water (H 2 O) NH 3 (g) + H 2 O (l) NH 4 OH (aq)
HCl
Graham’s Law of Diffusion
NH 4 Cl(s) NH 3 100 cm 100 cm
Choice 1: Both gases move at the same speed and meet in the middle.
HCl
Diffusion
NH 4 Cl(s) 81.1 cm 118.9 cm
Choice 2: Lighter gas moves faster; meet closer to heavier gas.
NH 3
Calculation of Diffusion Rate
v
1
v
2
m
2
m
1 NH 3 V 1 M 1 = X = 17 amu HCl V 2 M 2 = X = 36.5 amu Substitute values into equation
v
1
v
2 36.5
17 V 1 (NH 3 ) moves 1.465x for each 1x move of V 2 (HCl) DISTANCE = RATE x TIME Velocities are relative; pick easy #s: 3 200 cm = (1.000
cm s s HCl s v NH 3 t = 81.1 s cm s So HCl dist. … 1.000 cm/s (81.1 s) = 81.1 cm
Calculation of Diffusion Rate
v
1
v
2
m
2
m
1 NH 3 V 1 M 1 = X = 17 amu HCl V 2 M 2 = X = 36.5 amu Substitute values into equation
v
1
v
2 36.5
17
v
1
v
2 1.465
x
V 1 moves 1.465x for each 1x move of V 2 NH 3 HCl 1.465 x + 1x = 2.465
200 cm / 2.465 = 81.1 cm for x
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Br
79.904
35
Graham’s Law
Kr
83.80
36 Determine the relative rate of diffusion for krypton and bromine.
The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “v A /v B ”.
v
A
v
B
m
B
m
A
v
Kr
v
Br 2
m m
Br 2 Kr
159.80 g/mol 83.80 g/mol
1.381
Kr diffuses 1.381 times faster than Br 2 .
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1 H
1.00794
Graham’s Law
8 O
15.9994
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?
v A v B
m B m A v H 2 v O 2
m O 2 m H 2
v
H
2
12.3
m/s
32.00 g/mol 2.02
g/mol
Put the gas with the unknown speed as “Gas A”.
v
H
2
12.3
m/s
3.980
v
H
2
49.0 m/s
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1.0
H 1
v A v B v A v O 2
H 2 = 2 g/mol
Graham’s Law
An unknown gas diffuses 4.0 times faster than O 2 . Find its molar mass.
8 O
15.9994
The first gas is “Gas A” and the second gas is “Gas B”. The ratio “v A /v B ” is 4.0.
2
m B m A
Square both sides to get rid of the square root sign.
m O 2
16
32.00 g/mol m
A m A
m
A
32.00 g/mol
2.0 g/mol 16
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