Instructor:

Dr. Orlando E. Raola

Santa Rosa Junior College

Chapter 5:Properties of Gases

Chemistry 1A

General Chemistry

Gases and the Kinetic Molecular Theory

5.1 An Overview of the Physical States of Matter

5.2 Gas Pressure and Its Measurement5.3 The Gas Laws and Their Experimental

Foundations5.4 Applications of the Ideal Gas Law

1. Gas volume changes greatly with pressure.

2. Gas volume changes greatly with temperature.

3. Gases have relatively low viscosity.

4. Most gases have relatively low densities under normal conditions.

5. Gases are miscible.

The distinction of gases from liquids and solids

The three states of matter

A mercury barometer

Units of pressure

Sample Problem 5.1 Converting Units of Pressure

PROBLEM: A geochemist heats a limestone (CaCO3) sample and collects the CO2 released in an evacuated flask attached to a closed-end manometer. After the system comes to room temperature, Dh = 291.4 mm Hg. Express the CO2 pressure in torr, atmosphere, and kilopascal.

SOLUTION:

PLAN: Construct conversion factors to find the other units of pressure.

291.4 mmHg 1torr

1 mmHg

= 291.4 torr

291.4 torr 1 atm

760 torr= 0.3834 atm

0.3834 atm 101.325 kPa

1 atm= 38.85 kPa

The relationship between the volume and pressure of a gas.

Boyle’s Law

The relationship between the volume and temperature of a

gas.

Charles’s Law

Boyle’s Law n and T are fixedV a1

P

Charles’s Law V a T P and n are fixed

V

T= constant V = constant x T

Amontons’s Law P a T V and n are fixed

P

T= constant P = constant x T

combined gas law V aT

PV = constant x

T

P

PV

T= constant

V x P = constant V = constant / P

Avogadro’s Law

For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas

V = a n

a = proportionality constantV = volume of the gas (m3)n = chemical amount of gas (mol)

Ideal Gas Law

An equation of state for a gas.

“state” is the condition of the gas at a given time.

PV = nRTR = 8.314 J∙mol-1·K-1

(in common units 0.082 atm∙L∙mol-1·K-1)

Standard molar volume.

THE IDEAL GAS LAW

PV = nRT

IDEAL GAS LAW

nRT

PPV = nRT or V =

Boyle’s Law

V =constant

PV = V =

Charles’s Law

constant X T

Avogadro’s Law

constant X n

fixed n and T fixed n and P fixed P and T

R is the universal gas constant

Sample Problem 5.2 Applying the Volume-Pressure Relationship

PROBLEM: Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8 cm3 at 1.12 atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64 atm. Assuming constant temperature, what is the new volume of air (in L)?

PLAN: SOLUTION:

V1 in cm3

V1 in mL

V1 in L

V2 in L

unit conversion

gas law calculation

P1 = 1.12 atm P2 = 2.64 atm

V1 = 24.8 cm3 V2 = unknown

n and T are constant

24.8 cm3 1 mL

1 cm3

L

103 mL= 0.0248 L

P1V1

n1T1

P2V2

n2T2

=P1V1 = P2V2

P1V1

P2

V2 = = 0.0248 L1.12 atm

2.46 atm= 0.0105 L

1cm3=1mL

103 mL=1L

xP1/P2

Sample Problem 5.3 Applying the Temperature-Pressure Relationship

PROBLEM: A steel tank used for fuel delivery is fitted with a safety valve that opens when the internal pressure exceeds 1.00x103 torr. It is filled with methane at 230C and 0.991 atm and placed in boiling water at exactly 1000C. Will the safety valve open?

PLAN: SOLUTION:

P1(atm) T1 and T2(0C)

P1(torr) T1 and T2(K)

P1 = 0.991atm P2 = unknown

T1 = 230C T2 = 1000C

P2(torr)

1atm=760torr

x T2/T1

K=0C+273.15

P1V1

n1T1

P2V2

n2T2

=P1

T1

P2

T2

=

0.991 atm1 atm

760 torr = 753 torr

P2 = P1 T2

T1 = 753 torr

373K

296K= 949 torr

Sample Problem 5.4 Applying the Volume-Amount Relationship

PROBLEM: A scale model of a blimp rises when it is filled with helium to a volume of 55 dm3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm3. How many more grams of He must be added to make it rise? Assume constant T and P.

PLAN:

SOLUTION:

We are given initial n1 and V1 as well as the final V2. We have to find n2 and convert it from moles to grams.

n1(mol) of He

n2(mol) of He

mol to be added

g to be added

x V2/V1

x M

subtract n1

n1 = 1.10 mol n2 = unknown

V1 = 26.2 dm3 V2 = 55.0 dm3

P and T are constant

P1V1

n1T1

P2V2

n2T2

=

V1

n1

V2

n2

= n2 = n1 V2

V1

n2 = 1.10 mol55.0 dm3

26.2 dm3= 2.31 mol

4.003 g He

mol He= 9.24 g He

Sample Problem 5.5 Solving for an Unknown Gas Variable at Fixed Conditions

PROBLEM: A steel tank has a volume of 438 L and is filled with 0.885 kg of O2. Calculate the pressure of O2 at 210C.

PLAN:

SOLUTION:

V, T and mass are given. From mass, find amount (n), and use the ideal gas law to find P.

V 438L

1m3

103 L0.438m3

T 21C 273.15 294.15K

m= 0.885 kg

n = m

M

0.885 kg1000g

1kg

32.00gmol-127.7mol

P nRT

V

P nRT

V

27.7mol 8.314 kgm2 s 2 294.15K

0.438m31.55 105 Pa

,m

Since n and PV nRT thenM

mRTPV

MmRT

MPV

mand because d

VdRT

finally MP

The density of a gas is directly proportional to its molar mass.

The density of a gas is inversely proportional to the temperature.

Relationship between density and molar mass for gases

Sample Problem 5.7 Calculating Gas Density

PROBLEM: To apply a green chemistry approach, a chemical engineer uses waste CO2 from a manufacturing process, instead of chlorofluorocarbons, as a “blowing agent” in the production of polystyrene containers. Find the density (in g/L) of CO2 and the number of molecules (a) at STP (00C and 1 atm) and (b) at room conditions (20.0C and 1.00 atm).

PLAN:

SOLUTION:

Density is mass/unit volume; substitute for volume in the ideal gas equation. Since the identity of the gas is known, we can find the molar mass. Convert mass/L to molecules/L with Avogadro’s number.

d = mass/volume PV = nRT V = nRT/P d =RT

M x P

d =44.01 g/mol x 1atm

atm*L

mol*K0.0821 x 273.15K

= 1.96 g/L

1.96 g

L

mol CO2

44.01 g CO2

6.022x1023 molecules

mol= 2.68x1022 molecules CO2/L

(a)

Sample Problem 5.6 Calculating Gas Density

continued

(b) = 1.83 g/Ld =44.01 g/mol x 1 atm

x 293Katm*L

mol*K0.0821

1.83g

L

mol CO2

44.01g CO2

6.022x1023 molecules

mol= 2.50x1022 molecules CO2/L

Determining the molar mass of an unknown

volatile liquid.

based on the method of J.B.A. Dumas (1800-1884)

Sample Problem 5.8 Finding the Molar Mass of a Volatile Liquid

PROBLEM: An organic chemist isolates a colorless liquid from a petroleum sample. She uses the Dumas method and obtains the following data:

PLAN:

SOLUTION:

Use unit conversions, mass of gas, and density-M relationship.

Volume of flask = 213 mL

Mass of flask + gas = 78.416 g

T = 100.00C

Mass of flask = 77.834 g

P = 754 torr

Calculate the molar mass of the liquid.

m = (78.416 - 77.834) g = 0.582 g

M = m RT

VP=

0.582 gatm*L

mol*K0.0821 373Kxx

0.213 L x 0.992 atm

= 84.4 g/mol

,m

Since n and PV nRT thenM

mRTPV

MmRT

MPV

mand because d

VdRT

finally MP

The density of a gas is directly proportional to its molar mass.

The density of a gas is inversely proportional to the temperature.

Relationship between density and molar mass for gases

Sample Problem 5.7 Calculating Gas Density

PROBLEM: To apply a green chemistry approach, a chemical engineer uses waste CO2 from a manufacturing process, instead of chlorofluorocarbons, as a “blowing agent” in the production of polystyrene containers. Find the density (in g/L) of CO2 and the number of molecules (a) at STP (273.15 K and 1 bar) and (b) at room conditions (20.0C and 1.00 atm).

PLAN:

SOLUTION:

Density is mass/unit volume; substitute for volume in the ideal gas equation. Since the identity of the gas is known, we can find the molar mass. Convert mass/L to molecules/L with Avogadro’s number.

d = mass/volume PV = nRT V = nRT/P d =RT

M x P

1.94 g

L

mol CO2

44.01 g CO2

6.022x1023 molecules

mol= 2.65x1022 molecules CO2/L

(a)

-1

-1

-1 -1

44.01 g·mol 1bard= =1.94 g·L

0.08314bar·L·mol ·K 275.15 K

Sample Problem 5.6 Calculating Gas Density

continued

(b) = 1.83 g/Ld =44.01 g/mol x 1 atm

x 293Katm*L

mol*K0.0821

1.83g

L

mol CO2

44.01g CO2

6.022x1023 molecules

mol= 2.50x1022 molecules CO2/L

Determining the molar mass of an unknown

volatile liquid.

based on the method of J.B.A. Dumas (1800-1884)

Sample Problem 5.8 Finding the Molar Mass of a Volatile Liquid

PROBLEM: An organic chemist isolates a colorless liquid from a petroleum sample. She uses the Dumas method and obtains the following data:

PLAN:

SOLUTION:

Use unit conversions, mass of gas, and density-M relationship.

Volume of flask = 213 mL

Mass of flask + gas = 78.416 g

T = 100.00C

Mass of flask = 77.834 g

P = 754 torr

Calculate the molar mass of the liquid.

m = (78.416 - 77.834) g = 0.582 g

M = m RT

VP=

0.582 gatm*L

mol*K0.0821 373Kxx

0.213 L x 0.992 atm

= 84.4 g/mol

Sample Problem 5.5 Solving for an Unknown Gas Variable at Fixed Conditions

PROBLEM: A steel tank has a volume of 438 L and is filled with 0.885 kg of O2. Calculate the pressure of O2 at 210C.

PLAN:

SOLUTION:

V, T and mass are given. From mass, find amount (n), and use the ideal gas law to find P.

V 438L

1m3

103 L0.438m3

T 21C 273.15 294.15K

m= 0.885 kg

n = m

M

0.885 kg1000g

1kg

32.00gmol-127.7mol

P nRT

V

P nRT

V

27.7mol 8.314 kgm2 s 2 294.15K

0.438m31.55 105 Pa

Dalton’s Law of Partial Pressures

Ptotal = P1 + P2 + P3 + ...

P1= c1 x Ptotal where c1 is the mole fraction

c1 = n1

n1 + n2 + n3 +...=

n1

ntotal

• Gases mix homogeneously in any proportions.

• Each gas in a mixture behaves as if it were the only gas present.

Mixtures of gases

Mole fraction and partial pressure

...B B

BA B C

n nx

n n n n

For each component we define the mole fraction xB

and because of Dalton’s law, we can calculate the partial pressure of each component as

B B totp x P

Sample Problem 5.9 Applying Dalton’s Law of Partial Pressures

PROBLEM: In a study of O2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mol% N2, 17 mol% 16O2, and 4.0 mol% 18O2. (The isotope 18O will be measured to determine the O2 uptake.) The pressure of the mixture is 0.75atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18O2 in the mixture.

PLAN:

SOLUTION:

Find the c and P from Ptotal and mol% 18O2. 18O218O2

mol% 18O2

c 18O2

partial pressure P18O2

divide by 100

multiply by Ptotal

c18O2

=4.0 mol% 18O2

100= 0.040

= 0.030 atmP = c x Ptotal = 0.040 x 0.75 atm 18O2

18O2

Table 5.3 Vapor Pressure of Water (P ) at Different TH2O

T(0C) P (torr) T(0C) P (torr)

05

10111213141516182022242628

3035404550556065707580859095

100

31.842.255.371.992.5

118.0149.4187.5233.7289.1355.1433.6525.8633.9760.0

4.66.59.29.8

10.511.212.012.813.615.517.519.822.425.228.3

Collecting a water-insoluble gaseous reaction product and determining its pressure.

Sample Problem 5.10 Calculating the Amount of Gas Collected Over Water

PLAN:

SOLUTION:

The difference in pressures will give us the P for the C2H2. The ideal gas law will allow us to find n. Converting n to grams requires the molar mass, M.

PROBLEM: Acetylene (C2H2), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC2) reaction with water:

CaC2(s) + 2H2O(l) C2H2(g) + Ca(OH)2(aq)

For a sample of acetylene that is collected over water, the total gas pressure (adjusted to barometric pressure) is 98.32 kPa and the volume is 523 mL. At the temperature of the gas (230C), the vapor pressure of water is 2.798 kPa. How many grams of acetylene are collected?

Ptotal PC2H2

nC2H2

gC2H2

PH2O n =

PV

RT

x M

PC2H2

= (98.32-2.798) kPa =95.52 kPa

Sample Problem 5.9 Calculating the Amount of Gas Collected Over Water

continued

0.0203mol26.04 g C2H2

mol C2H2

= 0.529 g C2H2

nC2H2

9.552 104 Pa 5.23 10 4 m3

8.314Jmol 1 K 1 296K 0.0203 mol

P,V,T

of gas A

amount (mol)

of gas A

amount (mol)

of gas B

P,V,T

of gas B

ideal gas law

ideal gas law

molar ratio from balanced equation

Summary of the stoichiometric relationships among the amount (mol,n) of gaseous reactant or product and the gas

variables pressure (P), volume (V), and temperature (T).

Sample Problem 5.11 Using Gas Variables to Find Amount of Reactants and Products

PROBLEM: Dispersed copper in absorbent beds is used to react with oxygen impurities in the ethylene used for producing polyethylene. The beds are regenerated when hot H2 reduces the metal oxide, forming the pure metal and H2O. On a laboratory scale, what volume of H2 at 765 torr and 2250C is needed to reduce 35.5 g of copper(II) oxide?

SOLUTION:

PLAN: Since this problem requires stoichiometry and the gas laws, we have to write a balanced equation, use the moles of Cu to calculate mols and then volume of H2 gas.

mass (g) of Cu

mol of Cu

mol of H2

L of H2

divide by M

molar ratio

use known P and T to find V

CuO(s) + H2(g) Cu(s) + H2O(g)

35.5 g Cumol Cu

63.55 g Cu

1 mol H2

1 mol Cu= 0.559 mol H2

0.559 mol H2 x 498Katm*L

mol*K0.0821 x

1.01 atm

= 22.6 L

Sample Problem 5.12 Using the Ideal Gas Law in a Limiting-Reactant Problem

PROBLEM: The alkali metals (Group 1) react with the halogens (Group 17) to form ionic metal halides. What mass of potassium chloride forms when 5.25 L of chlorine gas at 0.950 atm and 293K reacts with 17.0 g of potassium?

SOLUTION:

PLAN: After writing the balanced equation, we use the ideal gas law to find the number of moles of reactants, the limiting reactant and moles of product.

2K(s) + Cl2(g) 2KCl(s) V = 5.25 L

T = 293K n = unknown

P = 0.950 atm

n = PV

RT Cl2 x 5.25L

= 0.950 atm

atm*L

mol*K0.0821 x 293K

= 0.207 mol

17.0g39.10 g K

mol K= 0.435 mol K

0.207 mol Cl22 mol KCl

1 mol Cl2= 0.414 mol

KCl formed

0.435 mol K2 mol KCl

2 mol K= 0.435 mol

KCl formedCl2 is the limiting reactant.

0.414 mol KCl74.55 g KCl

mol KCl= 30.9 g KCl

Problem

Gaseous iodine pentafluoride can be prepared by the reaction between solid iodine and gaseous fluorine. A 5.00-L flask containing 10.0 g of I2 is charged with 10.0 g of F2 and the reaction proceeds until one of the reactants is completely consumed. After the reaction is complete, the temperature in the flask is 125 ºC.

a) What is the partial pressure of IF5 in the flask?

b) What is the mole fraction of IF5 in the flask?

Postulates of the Kinetic-Molecular Theory

Because the volume of an individual gas particle is so small compared to the volume of its container, the gas particles are considered to have mass, but no volume.

Gas particles are in constant, random, straight-line motion except when they collide with each other or with the container walls.

Collisions are elastic therefore the total kinetic energy(Ek) of the particles is constant.

Postulate 1: Particle Volume

Postulate 2: Particle Motion

Postulate 3: Particle Collisions

Distribution of molecular speeds as a function of temperature

Distribution of molecular speeds as a function of temperature

Ek T

Ek constant T

Molecular description of Boyle’s Law

Molecular description of Dalton’s law of partial pressures

Molecular description of Charles’s law

Molecular description of Avogadro’s law

Relationship between molecular speed and mass

E

k

1

2mu2 u

rms

3RT

M

The meaning of temperature

E

k

3

2

R

NA

T

Absolute temperature is a measure of the average kinetic energy of the molecular random motion

Effusion and difussionEffusion: describes the passage of gas through a small orifice into an evacuated chamber.

Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.

Effusion:

Diffusion:

2

1

MRateof effusionfor gas1=

Rateof effusionfor gas2 M

2

1

MDistance traveled by gas1=

Distance traveled by gas2 M

Effusion and KMT

distribution of molecular speeds

mean free path

collision frequency

Diffusion through space

Sample Problem 5.13 Applying Graham’s Law of Effusion

PROBLEM: Calculate the ratio of the effusion rates of helium and methane (CH4).

SOLUTION:

PLAN: The effusion rate is inversely proportional to the square root of the molar mass for each gas. Find the molar mass of both gases and find the inverse square root of their masses.

M of CH4 = 16.04g/mol M of He = 4.003g/mol

CH4

Herate

rate= √ 16.04

4.003= 2.002

Table 5.4 Molar Volume of Some Common Gases at STP (00C and 1 atm)

GasMolar Volume

(L/mol)Condensation Point

(0C)

HeH2

NeIdeal gasArN2

O2

COCl2NH3

22.43522.43222.42222.41422.39722.39622.39022.38822.18422.079

-268.9-252.8-246.1----185.9-195.8-183.0-191.5-34.0-33.4

The behavior of several real gases with increasing external pressure

Effect of molecular attractions on pressure

Effect of molecular volume on measured volume

van der Waals equation

0.0340.2111.352.324.190.2441.391.366.493.592.254.175.46

HeNeArKrXeH2

N2

O2

Cl2CO2

CH4

NH3

H2O

0.02370.01710.03220.03980.05110.02660.03910.03180.05620.04270.04280.03710.0305

Van der Waalsequation for n

moles of a real gas

Gas a (atm·L2·mol-2) b (L·mol-1)

2

nP+a V-nb =nRT

V