GasesGases

Chapter 10Chapter 10

HH22

Paris 1783

Gas Bag

NN22

Why gases are studied Why gases are studied separatelyseparately

Many common compounds exist as gasesMany common compounds exist as gases

Gases transport matter and energy Gases transport matter and energy across the globe (i.e., weather)across the globe (i.e., weather)

Compared with those of liquids and Compared with those of liquids and solids, the behavior of gases is easiest solids, the behavior of gases is easiest to to model.model.

• Gases assume the volume and shape of their containers.

• Gases are the most compressible state of matter.

• Gases will mix evenly and completely when confined to the same container. (No solubility rules!)

• Gases have much lower densities than liquids and solids.

• Density of a gas given in g/L (vice g/mL for liquids)

Physical Characteristics of Gases

NO2

Units of Pressure

1 pascal (Pa) = 1 N/m2

1 atm = 101,325 Pa = 101.325 kPa

1 atm = 760 mm Hg = 760 torr

Pressure = ForceArea

(force = mass x acceleration)

Height of column ∝ 1/density

Fig. 10.2mass

760 mm Hg

Fig. 10.3

For P ≈ 1 atm

Using a Manometer to Measure Gas Pressure

Pressure-Volume RelationshipPressure-Volume Relationship: Boyle’s Law: Boyle’s Law

P1V1 = P2V2

Temperature-VolumeTemperature-Volume Relationship Relationship: Charles’s Law : Charles’s Law

V T∝

Volume-Amount RelationshipVolume-Amount Relationship: Avogadro’s Law: Avogadro’s Law

V n∝

The Ideal Gas LawThe Ideal Gas Law:: PV = nRT

2

2

1

1

T

V

T

V

V1

P

As P (h) increases V decreases

Fig. 10.6

Add Hg

Add HgAdd Hg

P · V = constant

P1 · V1 = P2 · V2

Boyle’s Law

V1

P

Constant temperatureConstant amount of gas

Fig 10.7

A sample of argon gas occupies a volume of 500. mL at a pressure of 626 mm Hg. What is the volume of the gas (in mm Hg) if the pressure is reduced at constant temperature to 355 mm Hg?

P1 · V1 = P2 · V2

P1 = 626 mm Hg

V1 = 500. mL

P2 = 356 mm Hg

V2 = ?

V2 = P1 · V1

P2

(626 mm Hg) (500. mL)355 mm Hg

= = 882 mL

As T increases V increases

Variation of gas volume with temperatureat constant pressure: Charles’ Law

V ∝ TV = constant · T

T (K) = t (°C) + 273.15

Temperature must be in Kelvin

Constant pressureConstant amount of gas2

2

1

1

T

V

T

V

Fig 10.8

A sample of hydrogen gas occupies 13.20 L at 22.8 °C. At what temperature will the gas occupy half that volume if the pressure remains constant?

V1 = 13.20 L

T1 = 296.0 K

V2 = 1.60 L

T2 = ?

T2 = V2 · T1

V1

6.60 L · 296.0 K13.20 L

= = 148 K

T1 = 22.8 (°C) + 273.15 (K) = 296.0 K

2

2

1

1

T

V

T

V

148 K - 273.15 = -125 °C

Avogadro’s Law

V ∝ number of moles (n)

V = constant · n

Constant pressureConstant temperature2

2

1

1

n

V

n

V

Fig 10.10

Fig 10.11

Avogadro’s Hypothesis

Molar Volume

Propane burns in oxygen to form carbon dioxide and water vapor. How many volumes of carbon dioxide are obtained from one volume of propane at the same temperature and pressure?

C3H8 + 5O2 3CO2 + 4H2O

1 mole C3H8 3 mole CO2

At constant T and P

1 volume C3H8 3 volumes CO2

PV = nRT

The conditions 0 °C and 1 atm are called

standard temperature and pressure (STP).

K) 5mol)(273.1 (1.000L) 4atm)(22.41 (1.000

nTPV

R

Experiments show that at STP,1 mole of an ideal gasoccupies 22.414 L:

KmolatmL

08206.0R

Fig 10.13 Comparison of Molar Volumes at STP

• One mole of an ideal gas occupies 22.41 L STP

• One mole of various real gases at STP occupy:

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

PV = nRT

V = nRTP

T = 0 °C = 273.15 K

P = 1 atm

n = (49.8 g) (1 mol HCl)(36.45 g HCl)

= 1.37 mol

V = 1 atm

(1.37 mol)(0.08216 ) (273.15 K)L•atmmol•K

V = 30.6 L

PV = nRT useful when P, V, n, and T do not change

Modify equation when P, V, and/or T change:

RTVP

1

11

• Initial state (1) of gas:

• Final state (2) of gas:

RTVP

2

22

2

22

1

11

TVP

TVP

Eqn [10.8]

Combined Gas Law

Density (d) Calculations

d = mV

=P(MM)

RT

m is the mass of the gas in g

MM is the molar mass of the gas

Molar Mass (MM) of a Gaseous Substance

dRTP

MM = d is the density of the gas in g/L

What happens to the density of a gas if:

(a) it is heated at constant volume?(b) It is compressed at constant temperature?(b) Additional gas is added at constant volume?

Volumes of Gases in Chemical Reations

What is the volume of CO2 produced at 37.0 °C and 1.00 atm when 5.60 g of glucose are used up in the reaction:

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)

g C6H12O6 mol C6H12O6 mol CO2 V CO2

5.60 g C6H12O6

1 mol C6H12O6

180 g C6H12O6

x6 mol CO2

1 mol C6H12O6

x = 0.1867 mol CO2

V = nRT

P

(0.1867 mol) (0.0821 ) (310.15 K)L•atmmol•K

1.00 atm= = 4.75 L

Gas Mixtures and Partial Pressures

V and T are

constant

P1 P2 Ptotal = P1 + P2

Dalton’s Law of Partial Pressures

Consider a case in which two gases, A and B, are in a container of volume V.

PA = nART

V

PB = nBRT

V

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB XA = nA

nA + nB

XB = nB

nA + nB

PA = XA PT PB = XB PT

Pi = Xi PT mole fraction (Xi) = ni

nT

Dalton’s Law of Partial Pressures

2KClO3 (s) 2KCl (s) + 3O2 (g)

PT = PO + PH O2 2

Fig. 10.16 Collecting a water-insoluble gas over water

Bottle full of oxygen gas and water vapor

2KClO3 (s) 2KCl (s) + 3O2 (g)

A 0.811 g sample of KClO3 is partially decomposed to produceoxygen gas over water. The volume of gas collected is 0.250 Lat 26 °C and 765 torr total pressure. How many grams of O2 are collected?

Sample Exercise 10.12 p 413

PT = PO + PH O2 2 PO = - PH O2 2

PT

PV = nRT

22

22

3 O g 0.32O mol 1

O g 32.0)O mol10 x (9.92

23 O mol10 x 9.92

K) (299Kmol

atmL 0.0821

0L)torr)(0.25 760 / atm torr)(1 (740n

Pressure of water vapor

vs temperature

App B p 1111

Kinetic Molecular Theory of Gases

1. A gas is composed of widely-separated molecules. The molecules can be considered to be points; that is, they possess mass but have negligible volume.

2. Gas molecules are in constant random motion.

3. Collisions among molecules are perfectly elastic.

4. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins.

KE ∝ T

Fig 10.18 The Effect of Temperature on Molecular Speeds

urms ≡root-mean-square

speed

Hot molecules are fast, cold molecules are slow.

The distribution of speedsof three different gases

at the same temperature

urms = 3RT (MM)

R = 8.314 J/(mol K)

Fig 10.19 The Effect of Molecular Mass on Molecular Speeds

Heavy molecules are slow, light molecules are fast.

An “Ideal Gas”An “Ideal Gas”

Assumptions:

• Gas molecules do not exert any force (attractive or repulsive) on each other

• i.e., collisions are perfectly elastic

• Volume of molecules themselves is negligible compared to volume of container

• i.e., the molecules are considered to be points

• An ideal gas “obeys” PV = nRT

• i.e., calculated value ≈ experimental value

Deviations from Ideal Behavior

Assumptions made in the kinetic-molecular model:

negligible volume of gas molecules themselves

no attractive forces between gas molecules

These breakdown at high pressure and/or low temperature.