Chapter 6 Properties of Gases: The Air We Breathe.

Chapter 6

Properties of Gases:

The Air We Breathe

30.36 in x 2.54 cm/in x 10 mm/cm= 771 mm (Hg)!!!

Is the barometric pressure in Butte greater than 1 atm?

The barometer in the P-Chem laboratory read:

628 mmHg

A Gas

• Has neither a definite volume nor shape.

• Uniformly fills any container.

• Mixes completely with any other gas

• Exerts pressure on its surroundings.

Composition of Earth’s AtmosphereCompound %(Volume) Mole Fractiona

Nitrogen 78.08 0.7808

Oxygen 20.95 0.2095

Argon 0.934 0.00934

Carbon dioxide 0.033 0.00033

Methane 2 x 10-4 2 x 10-6

Hydrogen 5 x 10-5 5 x 10-7

a. mole fraction = mol component/total mol in mixture.

Earth-like Atmosphere

A mercury barometer

The column height is proportional to the atmospheric pressure.

Atmospheric pressure results from the mass of the atmosphere and gravitational forces.

The pressure is the force per unit area.

P = F/A

1 atm = 760 mmHg

1 atm = 1.01325 E5 Pa

1 mmHg = 1 torr

Units for Expressing Pressure

Unit Value

Atmosphere 1 atm

Pascal (Pa) 1 atm = 1.01325 x 105 Pa

Kilopascal (kPa) 1 atm = 101.325 kPa

mmHg 1 atm = 760 mmHg

Torr 1 atm = 760 torr

Bar 1 atm = 1.01325 bar

mbar 1 atm = 1013.25 mbar

psi 1 atm = 14.7 psi

Pressure• is equal to force/unit area• SI units = Newton/meter2 = 1 Pascal (Pa)• 1 standard atmosphere = 101,325 Pa

(100,000 Pa = 1 bar)• 1 standard atmosphere = 1 atm =

760 mm Hg = 760 torr = 1013.25 hPa

= 14.695 psi

Meteorologists often report pressure in

millibar; 1 mbar =0.001bar =0.1 kPa = 1hPa

Variables Affecting Gases

• Pressure (P)

• Volume (V)

• Temperature (T)

• Number of Moles (n)

Elevation and Atmospheric Pressure

ManometerManometers are used to measure gas pressure in closed systems. For instance in a reaction vessel.

Boyle Law…pressure is inversely proportional to volume (at constant T and moles, n).

Boyle’s Law

• P 1/V (T and n fixed)

• P V = Constant

• P1V1 = P2V2

Problem. The pressure on a sample of an ideal gas was increased from 715 torr to 3.55 atm at constant temperature. If the initial volume of the gas was 485. mL, what would be the final volume?

Charles’s Law

• The volume of a gas is directly proportional to Kelvin temperature, and extrapolates to zero at zero Kelvin.

V T (P & n are constant)

V1 = V2

T1 T2

Problem 26. A 7.9 L sample of gas was cooled from 79°C to a temperature at which the volume of the gas was 4.3 L. Assuming the pressure remains constant, calculate the final temperature.

Avogadro’s Law

• For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures).

V n

V1 = V2

n1 n2

Amonton’s Law

• P T

• P/T = Constant

• P1 = P2

T1 T2

The temperature in the troposphere is also proportional to pressure.

Combined Gas Law

• Combining the gas laws the relationship P T(n/V) can be obtained.

• If n (number of moles) is held constant, then PV/T = constant.

P1V1

T1

= P2V2

T2

Ideal Gas Law

PV = nRT

R = universal gas constant = 0.08206 L atm K-1 mol-1

P = pressure in atm V = volume in liters n = moles T = temperature in Kelvin

Standard Temperature and Pressure (for gases)

• “STP” P = 1 atmosphere T = 0C The molar volume of an ideal gas is 22.42

liters at STP (put 1 mole, 1 atm, R, and 273 K in the ideal gas law and calculate V)

Note STP is different for other phases, e.g. solutions or enthalpies of formation.

Problem Calculate the pressure in atmospheres and pascals of a 1.2 Calculate the pressure in atmospheres and pascals of a 1.2

mol sample of methane gas in a 3.3 L container at 25°C.mol sample of methane gas in a 3.3 L container at 25°C.

Gas Laws and Stoichiometry

Oxygen respirators use canisters containing KO2 in the following reaction:

4KO2(s) + 2CO2(g) ↔ 2K2CO3(s) + 3O2(g)

What mass of KO2 is required to consume 8.90 L of CO2 at 22.0°C and 767 mm Hg?Answer: 52.7 g

TR

Pd

ddensityV

n

TR

P

V

n

TR

P

Μ

ΜΜ

The Ideal Gas Law can be rearranged to give a relationship between the density of a gas and its molar mass (M).

Multiplying both sides of the rearranged equation and solving for density gives this relationship.

Problem

An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.

Problem

A gas containing chlorine and oxygen has a density of 2.875 g/L at 756 mmHg and 11.0°C.

What is the most likely molecular formula for this compound?

Answer: ClO2

Dalton’s Law of Partial Pressures

• For a mixture of gases in a container

• PTotal = P1 + P2 + P3 + . . .

Mole Fraction & Partial Pressure

• Mole Fraction: the ratio of the number of Mole Fraction: the ratio of the number of moles of a given component in a mixture to moles of a given component in a mixture to the total number of moles in a mixture.the total number of moles in a mixture.

• Mole Fraction in terms of pressure (n = PV/RT)Mole Fraction in terms of pressure (n = PV/RT)

1 = n1 = n1

nTOTAL n1 + n2 + n3 + •••

1 = PP11(V/RT)(V/RT)

PP11(V/RT) + P(V/RT) + P22(V/RT) + P(V/RT) + P33(V/RT) + •••(V/RT) + •••

Continued

1 = P P1 1 == P P1 1

PP11 + P + P22 + P + P33 + ••• P + ••• PTOTAL TOTAL

1 = n1 = PP1 1

nTOTAL P PTOTALTOTAL

Mole Fraction Example At 250C, a 1.0 L flask contains 0.030 moles

of nitrogen, 150.0 mg of oxygen and

4 x 1021 molecules of ammonia.

A. What is the partial pressure of each gas?

B. What is the total pressure in the flask?

C. What is the mole fraction of each?

A sample of KClO3 is heated and decomposes to produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 754 torr. How many moles of oxygen formed?

Hint: The gas collected is a mixture so use Dalton’s Law Hint: The gas collected is a mixture so use Dalton’s Law to calculate the pressure of oxygen then the ideal gas law to calculate the pressure of oxygen then the ideal gas law to find the number of moles oxygen.to find the number of moles oxygen.

PT = PO2 + PH2O

Problem

Collecting a Gas Over Water

PH2O(25°C) = 23.8 torr

Problem 110

A mixture of 0.156 mol of C is reacted with 0.117 mol of O2 in a sealed, 10.0 L vessel at 500 K, producing a mixture of CO and CO2 gases. The total pressure is measure to be 0.640 atm.

What is the partial pressure of CO and CO2?

PCO = 0.349 atm

Kinetic Molecular Theory1. Gas molecules have tiny volumes

compared with their container’s volume.

2. Gas molecules move randomly and constantly.

3. The motion of these molecules is associated with their average kinetic energy that is proportional to the absolute temperature of the gas.

The temperature of an ideal monatomic gas is a measure related to the average kinetic energy of its atoms as they move. In this animation, the size of helium atoms relative to their spacing is shown to scale under 1950 atmospheres of pressure.

These room-temperature atoms have a certain, average speed (slowed down here two trillion fold).

Calculate the root mean square velocity of O2 at 60C (333 K).

[3·(8.314 kg m2 s-2 mol-1 K-1)·(333 K)/(0.032 kg mol-1)]1/2

= 509 m s-1

velocitysquaremeanrootRT

urms M3

The path of individual particles is a “Random Walk”

The Mean Free Path is the average distance traveled between collisions.

Kinetic Molecular Theory

4 Gas molecules continuously and elastically collide with one another and container walls.

5 Each molecule acts independently of all the other molecules in the sample. There are no forces of attraction (or repulsion) between molecules.

Kinetic Molecular Theory• K.E. = 1/2mu2

rms

Urms - the root-mean-squared speed of the molecules

•

At Constant Temp

Urms = 3RTM

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

• Effusion: describes the escape of gas through a tiny hole into a space of lower pressure.

• At a given temperature the urms velocity of a gas particle is dependent on temperature and mass of the particle. (from Kinetic Molecular Theory)

• The urms equation can be used to derive Graham’s Law of effusion (or diffusion).

)1(

)2(

2

1

gas

gas

gaseffusionofrate

gaseffusionofrate

MM

• What will be the relative rate of effusion of hydrogen gas as compared to oxygen?

98.301.2

0.32

2

2

2

2 H

O

O

H

r

r

M

M

• A sample of helium diffuses 4.58 times faster than an unknown gas.

• What is the likely identity of the unknown gas? (use Graham’s Law)

3.8)][krypton(89.38M

MM58.4

58.4M

M

r

r

unk

unkHe2

He

unk

unk

He

Problem

List the following gases, which are at List the following gases, which are at the same temperature, in the order of the same temperature, in the order of increasing rates of diffusion. increasing rates of diffusion.

OO22, He, & NO, He, & NO

Real GasesFor real gasses correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important).

Deviations from Ideal Behavior

For 1 mole of an ideal gas under standard conditions we can define a compressibility factor, Z = 1

deviations of Z from 1 reflect non-ideal behavior.

In general, deviations from ideal behavior become more significant the closer a gas is to a phase change, the lower the temperature or the larger the pressure.

The value of Z generally increases with pressure and decreases with temperature.

The value of Z generally increases with pressure and decreases with temperature.

Briefly…

•At high pressures molecules are colliding more often, and at low temperatures they are moving less rapidly.•This allows attractive forces between molecules to have a notable effect, making the volume of the real gas (Vreal) less than the volume of an ideal gas (Videal) which causes Z to drop below one•When pressures are lower or temperatures higher, the molecules are more free to move. In this case repulsive forces dominate, making Z > 1. The closer the gas is to its critical point or its boiling point, the more Z deviates from the ideal case.

Real Gases

[ ]P a V nb nRTobs2( / )+ x -( ) =n V

corrected pressurecorrected pressure corrected volumecorrected volume

PPidealideal VVidealideal

• van der Waals Equation

Substance a (L2 ۰atm/mol2) b (L2/mol)

He 0.0341 0.02337

H2 0.244 0.0266

CH4 2.25 0.0428

CO2 3.59 0.0427

SO2 6.71 0.05636

Table 8.4 Van der Waal Constants for Selected Gases

TRnbnV2V

a2nP

At high pressures, real gases do not behave ideally. Use the van der Waals equation and data in the text to calculate the pressure exerted by 17.55 mol H2 at 25°C in a 1.00 L container.

Repeat the calculation assuming that the gas behaves like an ideal gas. (check yourself, ans. 429 atm)

atm729P

L1.00mol

atmL0.244mol17.55

molL0.0266mol17.55L1.00

K298Kmol

atmL0.082057mol17.55

P

Van

nbVnRT

P

2

2

22

2

2

6.142. (a) Calculate the pressure exerted by 1.00 mol of CO2 in a 1.00 L vessel at 300 K, assuming that the gas behaves ideally. (b) Repeat the calculation by using the van der Waals equation.

6.142. (b) Repeat the calculation by using the van der Waals equation.

ChemTour: Ideal Gas Law

Click to launch animation

PC | Mac

In this ChemTour, students manipulate the variables of the ideal gas law to explore the relationship between the pressure, volume, and temperature of a gas. Includes worked examples and interactive Practice Exercises.

ChemTour: Dalton’s Law


PC | Mac

This ChemTour uses animation to explore Dalton’s law of partial pressures.

ChemTour: Molecular Speed


PC | Mac

This ChemTour explores kinetic molecular theory. Interactive graphs illustrate the concepts of kinetic energy and root-mean- square speed.

Diving and Gas Pressure

The hydrostatic pressure on a diver’s lungs is 1 atm at the surface and increases by 1 atm for each additional 10 m below the surface.

A) 10 m → 0 m B) 40 m → 20 m C) 60 m → 40 m

Which of the following ascents poses the gravest danger to a diver holding his or her breath?

Diving and Gas Pressure

Please consider the following arguments for each

answer and vote again:

A. The ascent from 10 m to 0 m is the most dangerous because the twofold decrease in pressure causes the greatest increase in lung volume.

B. The ascent from 40 m to 20 m is the most dangerous because it involves a larger change in pressure.

C. The ascent from 60 m to 40 m is the most dangerous because it occurs at the lowest depth.

Ideal Gas Law: PV versus P

Suppose 1 mole of an ideal gas is held at a constant temperature (T). Which of the following plots shows the correct relationship between the product of pressure and volume (PV) and the pressure (P)?

A) B) C)

Ideal Gas Law: PV versus P

Please consider the following arguments for each answer and vote again:

A. As the pressure increases, the product of the pressure and the volume also should increase.

B. As the pressure increases at a constant temperature, the volume should decrease so that the product PV remains constant.

C. Pressure and volume are inversely proportional, so as the pressure increases, the product PV should decrease.

Ideal Gas Law: Pressure Valve

An ideal gas at 10 atm and 50 ºC is placed in a rigid cylinder fitted with a pop-off valve set to open when the pressure reaches 20 atm. How high must the temperature be raised to open the pop-off valve?

A) < 100 ºC B) 100 ºC C) > 100 ºC

Ideal Gas Law: Pressure Valve



A. Any ideal gas that is heated will expand and open the valve.

B. If the temperature is increased by a factor of 2, the pressure also will increase by a factor of 2 to 20 atm.

C. A 50 ºC increase in the temperature will only produce a 15% increase in the pressure, which is not enough to open the valve.

Pressure of Water Vapor Product

A reaction of 0.50 atm of H2 and 0.50 atm of O2 occurs in a sealed vessel at a constant temperature to form H2O gas. If the reaction goes to completion, what will be the final pressure?

A) 0.5 atm B) 0.75 atm C) 1.0 atm

Pressure of Water Vapor Product



A. Because hydrogen is the limiting reactant, only 0.5 atm of H2O will be formed. So the final pressure will be 0.5 atm.

B. Although all 0.5 atm of H2 will be converted into 0.5 atm of H2O, only 0.25 atm of O2 will be consumed. Therefore, the final pressure will be 0.75 atm.

C. Because the total number of moles must be conserved, the total pressure also must be conserved. So the total pressure will remain at 1.0 atm.

Production of CO2 From CaCO3

Consider the reaction of calcium carbonate, CaCO3(s), with perchloric acid, HClO4(λ):

If there is initially 100 grams (1 mole) of CaCO3, how many grams of HClO4 must be added to produce ~10 L of CO2 at STP (273 K and 1 atm)?

A) 50 grams B) 100 grams C) 200 grams

CaCO3(s) + 2 HClO4(λ) → Ca(ClO4)2(s) + CO2(g) + H2O(λ).

Production of CO2 From CaCO3


A. 100 grams (1 mole) of CaCO3 is capable of producing ~20 L (~1 mole) of CO2, so only 50 grams (0.5 mole) of HClO4 should be added.

B. 100 grams of HClO4 will react completely with 50 grams of CaCO3, producing ~10 L of CO2 at STP.

C. Two HClO4 molecules are required for every one CaCO3 molecule, so 200 grams (2 moles) of HClO4 is needed.

Temperature and Molecular Speeds of Ideal Gases

The plot to the left shows the root-mean-velocity (vrms) of Br2 as a function of the square root of the temperature, assuming ideal behavior. Which of the following plots shows the correct relationship between vrms and for N2?

A) B) C)

T

Temperature and Molecular Speeds of Ideal Gases


It takes less energy (and hence a lower temperature) to increase the velocity of Br2 molecules, so the slope of the Br2 line should be steeper.

Since both Br2 and N2 are assumed to be ideal gases, their slopes should be the same. Only when you consider nonideal behavior will the slopes diverge.

N2 is a lighter molecule than Br2, and so at any given temperature its root-mean-square speed is always higher than that of Br2.

Distribution of Molecular Speeds

Shown to the left is a plot of the speed distributions for two gases, X and Y, at temperatures of 400 K and 800 K, respectively. Which of the following gases could be X and Y, respectively?

A) Br2 and Ar B) 2H2 and 1H2 C) N2 and C2H4

Distribution of Molecular Speeds


A. Br2 is four times heavier than Ar, so the root-mean-square speed (vrms) for Br2 should be half that of argon.

B. Deuterium gas (2H2) has twice the molecular mass of 1H2 and is at a temperature half that of 2H2. As a result, the vrms for 2H2 is a factor of 2 slower.

C. N2 and C2H4 have the same molecular mass. However, because the temperature of C2H4 is twice that of N2, the vrms is twice as fast.

Reaction of Metal with Cl2 Gas

One mole of metal, when added to 2 atm of chlorine gas (Cl2) in a 25-L vessel at 25 ºC, reacts to form a metal chloride, thereby decreasing the pressure to ~1 atm. Which of the following could be the unknown metal?

A) Na B) Ca C) Al

Reaction of Metal with Cl2 Gas


A. In a volume of 25 L at 25 ºC, 1 atm of a gas corresponds to ~1 mole. The only metal that can react in an equal molar quantity (1:1) with chlorine is Na, which can form NaCl.

B. At 25 ºC, 1 atm of Cl2 in a 25-L vessel is ~1 mole. Since 1 mole of Ca can react with 1 mole of Cl2 to form 1 mole of CaCl2, the unknown metal must be Ca.

C. Al will form AlCl3, using up three-fourths of the Cl2 and leaving an atmosphere of Cl atoms.

Expansion of Soda Can upon Heating

An unopened soda can expands when left outside on a hot day. Based on this observation, what can be concluded about the dissolution of CO2(g) in water?

A) It is exothermic. B) It is endothermic. C) It is isothermal.

Expansion of Soda Can upon Heating

Consider the following arguments for each answer and vote again:

A. The expansion is due to the decreased solubility of CO2(g) in water at higher temperatures, so the dissolution of CO2(g) is exothermic.

B. Energy is always required to dissolve a solute molecule in water, because to do so requires the breaking of hydrogen bonds within the water.

C. The dissolution of a gas into a liquid corresponds to an isothermal compression of the gas.

Injection of H2O into a Balloon of NH3

A single drop of water is injected into a balloon at 25 °C that contains ammonia, NH3(g), at 1.0 atm. What will happen to the volume of the balloon? Assume that the Henry’s constant for NH3(g) is ~0.02 atm/M.

A) It stays the same. B) It increases. C) It decreases.

Injection of H2O into a Balloon of NH3

Consider the following arguments for each answer and vote again:

A. A single drop of water takes up very little space and can hardly affect the volume of a 1-L balloon.

B. Once added, the drop of water will immediately vaporize to mix with the NH3(g), thus increasing the pressure and hence the volume.

C. The pressure inside the balloon will drop as part of the NH3(g) dissolves in the drop of water, thus decreasing the volume of the balloon.

Chapter 6 Properties of Gases: The Air We Breathe.

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Transcript of Chapter 6 Properties of Gases: The Air We Breathe.