Gas Law Review

No definite shape◦ It fills its container

Compressible◦ With increases in pressure

Low Density◦ Molecules are far apart◦ Intermolecular forces are ignored

Diffusion◦ Molecules can randomly spread out

Properties of Gases

Theoretical description of gases◦ Disregard the volume of the molecule itself

◦ Disregard any attractive force between molecules

Real gases stray from ideal gases at:◦ LOW Temperatures

◦ HIGH Pressures

Ideal Gas

At STP (Standard temperature and pressure), all gases have a volume of 22.4 L.

Standard Temperature:◦ 0oC◦ 273 K

Standard Pressure:◦ 101.3 kPa◦ 1 atm◦ 760 torr (or mmHg)

The volume of a quantity of gas, at constant pressure, varies directly with the Kelvin temperature.

Temperature MUST be in Kelvin! (oC + 273)

Charles Law

2

2

1

1

T

V

T

V @ constant

Pressure

A gas is collected at 58oC and has a volume of 225mL. What volume will it occupy at standard temperature, if pressure remains constant?

133127358 TKCo 2

2

1

1

T

V

T

V

Since the temperature decreases, pressure must decrease!

T1= 331 KV1 = 225 mLT2 = 273 KV2 = ?

KK

mL

273

?

331

225

186 mL = 0.186 L

As pressure increases, volume decreases.

Temperature MUST be constant.

Boyles Law

2211 VPVP

A sample of O2 gas at 0.947atm has a volume of 150mL. What would its volume be at 0.987atm if the temperature stay constant?

2211 VPVP P1 = 0.947 atmV1 = 150 mLP2 = 0.987 atmV2 = ? ?987.0150947.0 atmmLatm

144mL = 0.144L

Gay-Lussac’s Law

2

2

1

1

T

P

T

P Volume is constant

* Remember, Temperature MUST be in Kelvin

Combined Gas Law

2

22

1

11

T

VP

T

VP

An aerosol can has a pressure of 103 kPa at 25oC. It is thrown into a fire and its temperature increases to 928oC. What will its pressure be?

2

2

1

1

V

P

V

P

P1 = 103 kPaP2 = ?T1 = 25 oC + 273 = 298 KT2 = 928 oC + 273 = 1201 K

K

P

K

kPa

1201298

103 2 P2 = 415 kPa

The volume of a gas-filled balloon is 30.0 L at 313 K and 153 kPa. What would the volume be at STP?

2

22

1

11

T

VP

T

VP

V1 = 30.0LT1 = 313 KP1 = 153 kPaV2 = ?T2 = 273 KP2 = 101.3 kPa

K

VkPa

K

LkPa

273

3.101

313

0.30153 2

2371.07.14 V

V2 = 39.6 L

Ideal Gas Law

nRTPV P = pressure

V = volume (in Liters!)

n = moles

T = temperatre (in Kelvin!)

R = gas constant

Gas constants

Kmol

kPaL

31.8

Kmol

atmL

082.0

Kmol

torrL

4.62

What volume would be occupied by 1.00 moles of gas at 0oC at 1 atm pressure?

Find Volume

nRTPV

P = 1 atmV = ?n = 1 molR = 0.082 (because pressure is in atm)T = 0oC + 273 = 273 K

KKmol

atmLmolVatm 273082.011

V = 22.4L

Equal volumes of gases at the same temperature and pressure contain equal numbers of particles.

Molar Volume- for a gas, the volume that one mole occupies at STP◦ Temp = 0oC or 273K◦ Pressure = 1 atm or 101.3 kPa or 760 torr (mmHg)

Avogadro’s Principle

1 mole = 22.4 Liters

Determine the volume of a container that holds 2.4 mol of gas at STP.

Practice

mol

Lmol

1

4.224.2 54 L

If 100L of hydrogen gas react at STP, how many grams of hydrogen chloride can form?

1 H2 + 1 Cl2 2 HCl

molHCl

gHCl

molH

molHCl

L

molHL

1

5.36

1

2

4.22

1100

2

2

326 g HCl

The sum of the individual gas pressures equals the overall pressure of the mixture of gases.

Dalton’s Law of Partial Pressure

totalPPPP .......321

If a container has 166 torr H2, 109 torr CO2 and 176 torr of O2, what is the total pressure of the mixture?

116 torr + 109 torr + 176 torr = 401 torr

Our atmosphere is made of 21% O2, 78% N2 and 1% other gases. At sea level (standard pressure), what is the partial pressure of oxygen?

760 torr x 0.21 = 159 torr O2

The spontaneous spreading of particles

The rate of diffusion depends on the velocities and masses of the molecules

Effusion – the process by which a gas escapes from a small hole in a container

Lighter gases ALWAYS diffuse/effuse faster than heavier molecules

Diffusion

The relative rates at which two gases, at the same temperature and pressure, will diffuse, vary inversely as the square root of the molecular mass of the gases.

Grahams Law of Diffusion

1

2

2

1

m

m

v

v **Always

consider gas 1 the lighter gas

Compute the relative rates of diffusion of helium and argon.◦ go to the periodic table for molar mass of He and

Ar

1

2

2

1

m

m

v

v Mass He = 4 g/mol

Mass Ar = 40 g/mol

16.3104

40

2

1 v

v

1

16.316.3

2

1 v

v So, helium diffuses 3x’s faster than argon.