Chapter 10: Properties of Gases: The Air We Breathepostonp/ch222/pdf/Ch10_s18.pdf · 10.9 Mixtures...

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Chapter 10: Properties of Gases:

The Air We Breathe

http://ozonewatch.gsfc.nasa.gov

Sept, 2006 Sept, 2016

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Chapter Outline

10.1 The Properties of Gases

10.2 The Kinetic Molecular Theory of Gases*

10.3 Atmospheric Pressure

10.4 Relating P,T, and V: The Gas Laws

10.5 The Combined Gas Law (we will add moles to the equation)

10.6 Ideal Gases and the Ideal Gas Law

10.7 Densities of Gases

10.8 Gases in Chemical Reactions

10.9 Mixtures of Gases

10.10 Solubilities of Gases and Henry’s Law

10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly

10.12 Real Gases

*Effusion has been added to Section 10.11

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The Properties of Gases

Neither definite shape nor definite volume


Gases can be compressed.

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All gases are miscible with all other gases.

http://catalog.flatworldknowledge.com/bookhub/4309?e=averill_1.0-ch13_s01

Chapter Outline Chapter Outline












10.12 Real Gases


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1. Gas particles have tiny volumes compared with their container’s volume

Kinetic Molecular Theory of Gases

2. They don’t interact with other gas molecules, e.g. no intermolecular forces.


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3. They move randomly and constantly


4. Elastic collisions with walls of container and other gas molecules


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5. Have average kinetic energy that is proportional to absolute Kelvin temperature: KE = ½ mu2

rms


½ mAu2rms,A = ½ mBu2

rms,B

<KE>A = <KE>B since both at the same temperature

mAu2rms,A = mBu2

rms,B

u2rms,B u2

rms,B

mAu2rms,A = mB

u2rms,B

mA u2rms,A = mB

mA u2

rms,B mA

u2rms,A = mB

u2rms,B mA

𝑢𝑟𝑚𝑠,𝐴

𝑢𝑟𝑚𝑠,𝐵 =

𝑚𝐵

𝑚𝐴

𝑟𝑎𝑡𝑒 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛,𝐴

𝑟𝑎𝑡𝑒 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛,𝐵 =

𝑀𝑀𝐵

𝑀𝑀𝐴



𝑀𝑀𝐵

𝑀𝑀𝐴

MM =

molar

mass



𝑚𝐵

𝑚𝐴

Graham’s Law of Effusion:

We will return to this in Section 10.11

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rms velocity inversely proportional to the Molar Mass

= 32 g/mol

= 28

= 18

= 4

Chapter Outline












10.12 Real Gases


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Pressure = force/unit area

Molecules collide with the inside surface of the container.

The force of the collision is measured as pressure.

Pounds/in2 (psi)

Atmospheres (atm)

Pascals (N/m2)

Torr (mmHg)

14.7 psi

1 atm

101.325 X 103 Pa

760 mmHg

Pressure at Sea Level

Torricelli’s Barometer

vacuum

Column of

mercury

760 mm Hg

Atmospheric

pressure

The pressure of the

atmosphere on the surface

of the mercury in the dish

is balanced by the

downward pressure

exerted by the mercury in

the column.

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Elevation and Atmospheric

Pressure

0.83 atm

0.62 atm

0.35 atm

Sea level

Chapter Outline












10.12 Real Gases


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State Variables for a Gas

P = pressure V = volume

T = temperature n = number of

moles

Some math before continuing - y = mx + b

m

b

0

10

20

30

40

50

60

0 5 10 15 20 25

yaxis

xaxis

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Boyle’s Law: P and V

(n and T held constant) • P and V are inversely proportional -

• P then V

• If you plot P vs. 1/V the result

is a straight line

• P = mx + b = m(1/V) + b

• so P = m/V and PV = m

• P1V1 = m and P2V2 = m, so

• P1V1 = P2V2

Boyle’s Law and Respiration

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Applying Boyle’s Law Example

A bubble of oxygen at the bottom of a lake floats up to the surface. The pressure at the bottom of the lake is 4.75 atm and the volume is 5.65 mL. At the surface, the new pressure is 0.985 atm. Assuming that the temperature and number of moles remained constant, what is the final volume of the bubble?

P1 = 4.75 atm

V1 = 5.65 mL

P2 = 0.985 atm

V2 = ? mL

Explaining Boyle’s Law Using Kinetic

Molecular Theory

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Charles’s Law: V and T

(n and P held constant)

2

2

1

1

T

V

T

V

• T and V are directly proportional -

• T then V

• If you plot V vs. T the result is

a straight line

• V = mT + b = mT

• so V = mT and V/T = m

• V1/T1 = m and V2/T2 = m

Volume of a gas extrapolates to zero at

absolute zero (0 K = −273°C) so you

MUST use the Kelvin temperature.

Jacques Alexandre Charles (1796-1823)

The French chemist Charles was most famous in his lifetime for his experiments in

ballooning. The first such flights were made by the Montgollier brothers in June 1783,

using a large spherical balloon made of linen and paper and filled with hot air. In

August 1783, however, a different group. supervised by Jacques Charles, tried a

different approach. Exploiting his recent discoveries in the study of gases, Charles

decided to inflate the balloon with hydrogen gas. Because hydrogen would escape

easily from a paper bag, Charles made a bag of silk coaled with a rubber solution.

Inflating the bag to its final diameter took several days and required nearly 500

pounds of acid and 1000 pounds of iron to generate the hydrogen gas. A huge crowd

watched the ascent on August 27, 1783. The balloon stayed aloft for almost 45

minutes and travelled about 15 miles. When it landed in a village, however, the

people were so terrified they tore if to shreds.

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Sample Exercise 10.4:

Applying Charles’ Law

Several students at a northern New England campus are hosting a party celebrating the mid-January start of “spring” semester classes. They decide to decorate the front door of their apartment building with party balloons. The air in the inflated balloons is initially 70 oF. After an hour outside, the temperature of the balloons is -12 oF. Assuming no air leaks from the balloons and the pressure in them does not change significantly, how much does their volume change? Express your answer as a percentage of the initial volume.

Explaining Charles’ Law Using Kinetic

Molecular Theory

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Avogadro’s Law: V and n

(T and P held constant)

2

2

1

1

n

V

n

V

• V and n are directly proportional

• n then V

• If you plot V vs. n the result is a

straight line

• V = mn + b = mn

• so V = mn and V/n = m

• V1/n1 = m and V2/n2 = m

Avogadro’s Law Problem

(none given in text, p. 427)

V1 = 3.5 L

n1 = 0.140 mol

V2 = 10.1 L

n2 = ? mol

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Chapter Outline

Note: we are skipping ahead to Sec. 10.6, then

back to Sec. 10.5












10.12 Real Gases


Ideal Gas Equation

Charles’ law: V a T (at constant n and P)

Avogadro’s law: V a n (at constant P and T)

Boyle’s law: V a (at constant n and T) 1 P

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Standard Temperature and Pressure (STP).

PV = nRT

Experiments show that at STP,

1 mole of an ideal gas

occupies 22.414 L.


Applying the Ideal Gas Law

Bottles of compressed O2 carried by climbers ascending Mt. Everest are

designed to hold one kilogram of the gas. What volume of O2 can one

bottle deliver to a climber at an altitude where the temperature is -38 oC

and the atmospheric pressure is 0.35 atm? Assume that each bottle

contains 1.00 kg of O2.

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Chapter Outline

Note: we are now back to Sec. 10.5












10.12 Real Gases


Deriving the Combined Gas Law

P, V, T, and n

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Applying the Combined Gas Law

The pressure inside a weather balloon as it is released is 798 mmHg. If the volume and temperature of the balloon are 131 L and 20 oC, what is the volume of the balloon when it reaches an altitude where its internal pressure is 235 mmHg and T = -52 oC?

Chapter Outline












10.12 Real Gases


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Densities and Molecular Weights of

Gases Using PV = nRT


Calculating the Density of a Gas

According to the U.S National Weather Service, the air

temperature in Phoenix, AZ reached 25.56 oC on January 1,

2012, when the atmospheric pressure was 1.0106 atm. What is

the density of the air? Assume the average molar mass of air is

28.8 g/mol, which is the weighted average of the molar masses

of the various gases in dry air.

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Example: Calculating the Molecular

Weight from PV = nRT

1.018 g of Freon-113 gas is trapped in a 145 mL container at

760 mmHg and 50.0°C. What is the molar mass of Freon-113?

Chapter Outline












10.12 Real Gases


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Gas Laws & Stoichiometry

Example: Combining Stoichiometry and

the Ideal Gas Law

Chlorine gas can be prepared in the laboratory by the reaction of

manganese dioxide with hydrochloric acid:

MnO2(s) + 4 HCl(aq) MnCl2(aq) + 2 H2O(l) + Cl2(g)

How many grams of MnO2 should be added to excess HCl to obtain 275

mL of chlorine gas at 5.0°C and 650 mmHg?

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Chapter Outline












10.12 Real Gases


Dalton’s Law of Partial Pressures

For a mixture of gases in a container:

• Ptotal = P1 + P2 + P3 + . . .

Total pressure depends only on total number moles

of gas, not on their identities

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Mole Fraction & Partial Pressure

Mole Fraction:

• Ratio of the # of moles of a given component

in a mixture to the total # of moles in a

mixture:

Mole Fraction in Terms of Pressure:

• When V and T are constant, P n

1 11

total 1 2 3

n nx

n n n n ...

Since P n

And:

Then…

1 11

total total

n Px

n P

1 1 totalP x P

11

total

Px

P

Mole Fraction & Partial Pressure

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Sample Exercise 10.11: Calculating

Mole Fractions and Partial Pressures

Scuba divers who dive to depths below 50 meters may breathe a gas mixture called Trimix during the deepest parts of their dives. One formulation of the mixture, called Trimix 10/70, is 10% oxygen, 70% helium, and 20% nitrogen by volume. What is the mole fraction of each gas in this mixture, and what is the partial pressure of oxygen in the lungs of a diver at a depth of 60 meters (where the ambient pressure is 7.0 atm)?

Collecting a Gas over Water

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

Gases collected:

O2(g) and H2O(g)

2 2total O H OP P P

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During the decomposition of KClO3, 92.0 mL of gas is collected by the displacement of water at 25.0 oC. If atmospheric pressure is 756 mmHg, what mass of O2 is collected?

Sample Exercise 10.12: Calculating the Quantity

of a Gas Collected by Water Displacement

Chapter Outline












10.12 Real Gases


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Solubility of Gases

Solubility of gases

depends on T and P

Solubility ↑ as Pressure ↑

Solubility ↓ as Temperature ↑

Henry’s Law

Henry’s Law:

• The higher the partial

pressure of the gas

above a liquid, the

more soluble

• Cgas Pgas

• Cgas kH Pgas

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Gas Solubility Using Henry’s Law

Lake Titicaca is located high in the Andes Mountains between Peru and Bolivia. Its surface is 3811 m above sea level, where the average atmospheric pressure is 0.636 atm. During the summer, the average temperature of the water’s surface rarely exceeds 15 oC. What is the solubility of oxygen in Lake Titicaca at that temperature? Express your answer in molarity and in mg/L.

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Chapter Outline












10.12 Real Gases


Graham’s Law of Effusion, p. 415

Effusion is the process

where a gas escapes

through a small pore in the

container wall into a region

of lower pressure.

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Graham’s Law of Effusion



𝑀𝑀𝐵

𝑀𝑀𝐴



𝑀𝑀𝐵

𝑀𝑀𝐴

MM = molar mass


Relative Rates of Effusion

An odorous gas emitted by a hot spring was found to effuse at 0.342 times

the rate at which helium effuses. What is the molar mass of the emitted gas?



𝑀𝑀𝐵

𝑀𝑀𝐴

A = X (MM = ?)

B = He (MM = 4.003 g/mol)

𝑟𝑎𝑡𝑒 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑋

𝑟𝑎𝑡𝑒 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛,𝐻𝑒 =

𝑀𝑊𝐻𝑒

𝑀𝑊𝑋

0.342 =4.003 𝑔/𝑚𝑜𝑙

𝑀𝑀𝑋

(0.342)2=4.003 𝑔/𝑚𝑜𝑙

𝑀𝑀𝑋 and so MMX = 34.2 g/mol or H2S

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𝑈𝑟𝑚𝑠,𝐴

𝑈𝑟𝑚𝑠,𝐵=

𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝐴

𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝐵=

𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝐴

𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝐵=

𝑀𝐵

𝑀𝐴

Diffusion - Same Formula as

Graham’s Law of Effusion:

Problem 10.121

A flask of ammonia is connected to a flask of an unknown acid HX by a 1.00 m

glass tube. As the two gases diffuse down the tube, a white ring of NH4X forms

68.5 cm from the ammonia flask. Identify element X.

Gas Diffusion: Molecules Moving Rapidly

𝑢𝑟𝑚𝑠 =3𝑅𝑇

𝑀

R = 8.314 J/mol K

Maxwell’s Equation

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Practice Exercise 10:14

Calculating Root-Mean-Square Speeds

Calculate the root-mean-squared speed of nitrogen molecules at 25 oC.

Chapter 10: Properties of Gases: The Air We Breathepostonp/ch222/pdf/Ch10_s18.pdf · 10.9 Mixtures...

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