DALTON’S LAW, AVOGADRO, IDEAL

DALTON’S LAW OF DALTON’S LAW OF PARTIAL PRESSURE, PARTIAL PRESSURE, AVOGADRO’S LAW,AVOGADRO’S LAW,

IDEAL GAS LAW IDEAL GAS LAW

MS. ANA D. HIRANGMS. ANA D. HIRANGSY 2008 -2009SY 2008 -2009

RECALL on GAS LAWSRECALL on GAS LAWS

PP11 == PP22

TT11 T T22


Rate of gas 1Rate of gas 1 == MW of gas 2MW of gas 2

Rate of gas 2 MW of gas Rate of gas 2 MW of gas 11


PP11VV11 = P = P22VV22


PP11VV11 == PP22VV22

TT11 T T22


VV11 == VV22

TT11 T T22

DALTON’S LAW OF DALTON’S LAW OF PARTIAL PRESSUREPARTIAL PRESSURE

In 1801, Dalton found out that the total In 1801, Dalton found out that the total pressure of a mixture of gases is equal to pressure of a mixture of gases is equal to the sum of the partial pressure exerted the sum of the partial pressure exerted by each gas. This is known as by each gas. This is known as Dalton’s Dalton’s law of partial pressure.law of partial pressure.

Each gasEach gas in the mixture exerts a pressure in the mixture exerts a pressure that is independent of the other gases that is independent of the other gases present. These pressures are called present. These pressures are called partial pressures.partial pressures.

DALTON’S LAW OF DALTON’S LAW OF PARTIAL PRESSUREPARTIAL PRESSURE

Thus if we have a Thus if we have a mixture of two gases,mixture of two gases,

OO2 2 = 0.1 atm= 0.1 atm

NN22 = 0.7 atm = 0.7 atm

PPTOTALTOTAL= 0.8 atm= 0.8 atmN2 O2 N2 + O2

Mathematically this can be stated as:

PTOTAL = P1 + P2 + P3 + …

DALTON’S LAW OF PARTIAL DALTON’S LAW OF PARTIAL PRESSUREPRESSURE

DALTON’S LAW OF PARTIAL DALTON’S LAW OF PARTIAL PRESSUREPRESSURE

Dalton’s Law of Partial Pressure can Dalton’s Law of Partial Pressure can be explained by 2 concepts from be explained by 2 concepts from the Kinetic Molecular Theory.the Kinetic Molecular Theory.

1.1. The pressure of a gas is caused by The pressure of a gas is caused by the collision of molecules with the the collision of molecules with the walls of the container.walls of the container.

2.2. Gas molecules act independently of Gas molecules act independently of each other. each other.

DALTON’S LAW OF PARTIAL DALTON’S LAW OF PARTIAL PRESSURE: APPLICATIONPRESSURE: APPLICATION

Gas exchange between living Gas exchange between living organisms and the environment organisms and the environment depends on the properties of gases, depends on the properties of gases, in particular, partial pressure and in particular, partial pressure and solubility.solubility.

RESPIRATION is one of the most RESPIRATION is one of the most important processes because we important processes because we need to breathe OXYGEN and need to breathe OXYGEN and breathe out CObreathe out CO2 2 in order to live. in order to live.


GASGASPartial Pressure (kPa)Partial Pressure (kPa)

INHALED AIRINHALED AIR EXHALED AIREXHALED AIR

NN2 (g)2 (g) 79.379.3 75.975.9

OO2 (g)2 (g) 21.321.3 15.515.5

COCO2 (g)2 (g) 0.0400.040 3.73.7

HH22OO(g)(g)** 0.670.67 6.26.2

* The quantity of water in air varies. The value used * The quantity of water in air varies. The value used in this table is based on a relatively low humidity.in this table is based on a relatively low humidity.

TABLE 1: PARTIAL PRESSURE CHANGES DURING RESPIRATION


Oxygen is the most important gas in Oxygen is the most important gas in the atmosphere, it makes up 21 % of the atmosphere, it makes up 21 % of the volume of dry air. the volume of dry air.

Partial pressure of a gas is more Partial pressure of a gas is more useful than percentage composition useful than percentage composition because it is the pressure of oxygen because it is the pressure of oxygen that determines how much oxygen is that determines how much oxygen is absorbed by the lungs of the person.absorbed by the lungs of the person.


21 % of the volume and pressure of 21 % of the volume and pressure of the atmosphere is due to OXYGEN. the atmosphere is due to OXYGEN. The partial pressure of oxygen can be The partial pressure of oxygen can be calculated by multiplying the percent calculated by multiplying the percent in decimal form by the total pressure.in decimal form by the total pressure.

PPOXYGENOXYGEN = 0.21 X 760 torr = 159.6 torr = 0.21 X 760 torr = 159.6 torr We function best breathing this partial We function best breathing this partial

pressure of oxygen.pressure of oxygen.


DID YOU KNOWDID YOU KNOW You may have about 3.5 x 10You may have about 3.5 x 1022

tiny tiny ALVEOLI (air sacs) in each lung. The ALVEOLI (air sacs) in each lung. The surface area of contact with surface area of contact with capillaries for absorbing oxygen is capillaries for absorbing oxygen is about 75 mabout 75 m22, about four (4) times as , about four (4) times as much area as an average classroom much area as an average classroom floor!floor!

X –SECTION OF AN ALVEOLIX –SECTION OF AN ALVEOLI


When we live at higher elevations, the When we live at higher elevations, the partial pressure of oxygen is lower partial pressure of oxygen is lower and our bodies adjust accordingly.and our bodies adjust accordingly.

On top of the highest mountain, e.g. On top of the highest mountain, e.g. MT. EVEREST, the total atmospheric MT. EVEREST, the total atmospheric pressure is 270 torr, so the partial pressure is 270 torr, so the partial pressure of oxygen is only 56.7 torr, pressure of oxygen is only 56.7 torr, or about 1/3 of normal. or about 1/3 of normal.

PPOXYGENOXYGEN = 0.21 X 270 torr = 56.7 torr = 0.21 X 270 torr = 56.7 torr


A human being cannot survive for A human being cannot survive for long at such a low pressure of long at such a low pressure of oxygen.oxygen.

At that altitude, even conditioned At that altitude, even conditioned climbers must use an oxygen tank climbers must use an oxygen tank and mask, which give an and mask, which give an increased partial pressure of increased partial pressure of oxygen to the lungs. oxygen to the lungs.

DALTON’S LAW OF PARTIAL DALTON’S LAW OF PARTIAL PRESSURE: TOTAL PRESSUREPRESSURE: TOTAL PRESSURE

SAMPLE EXERCISESSAMPLE EXERCISES

1.1. An equilibrium mixture contains HAn equilibrium mixture contains H22 at 560 torr, Nat 560 torr, N22 at 180 torr and O at 180 torr and O22 at at 250 torr pressure. What is the total 250 torr pressure. What is the total pressure of the gases, in mm Hg pressure of the gases, in mm Hg and atm, in the system?and atm, in the system?



2. 2. In a compressed air tank for scuba In a compressed air tank for scuba diving to a depth of 30 m, a diving to a depth of 30 m, a mixture with an oxygen partial mixture with an oxygen partial pressure of 28 atm and a nitrogen pressure of 28 atm and a nitrogen partial pressure of 110 atm is partial pressure of 110 atm is used. What is the total pressure in used. What is the total pressure in the tank?the tank?



3. 3. The total pressure of a gas mixture in a The total pressure of a gas mixture in a cylinder is 6.40 atm. Gas A exerts a cylinder is 6.40 atm. Gas A exerts a pressure three times that of gas B. pressure three times that of gas B. Gas C exerts a pressure twice that of Gas C exerts a pressure twice that of gas A. What will be the pressure of gas A. What will be the pressure of gas A, B, C if they occupy the cylinder gas A, B, C if they occupy the cylinder alone?alone?

DALTON’S LAW OF PARTIAL DALTON’S LAW OF PARTIAL PRESSURE: WATER DISPLACEMENT PRESSURE: WATER DISPLACEMENT

METHODMETHOD Another applicationAnother application of partial of partial

pressures is THE COLLECTION OF GAS pressures is THE COLLECTION OF GAS BY THE DISPLACEMENT OF WATER.BY THE DISPLACEMENT OF WATER.

Hydrogen and oxygen are often Hydrogen and oxygen are often generated in the lab and collected by generated in the lab and collected by bubbling the gases into a container bubbling the gases into a container filled with water. Both of this gases filled with water. Both of this gases have a relatively low solubility in have a relatively low solubility in water.water.

DALTON’S LAW OF PARTIAL DALTON’S LAW OF PARTIAL PRESSURE: WATER DISPLACEMENT PRESSURE: WATER DISPLACEMENT

METHODMETHOD

HYDROGEN GAS GENERATION


However, water evaporates However, water evaporates relatively easily and the gas relatively easily and the gas collected will be mixed with some collected will be mixed with some water vapor.water vapor.

Water vapor is a gas like any other Water vapor is a gas like any other gas and the pressure exerted by a gas and the pressure exerted by a gas above its liquid is called gas above its liquid is called VAPOR VAPOR PRESSURE.PRESSURE.


The vapor pressure of water at The vapor pressure of water at different temperature is well different temperature is well known. See Table 13.2 , p. 253 of known. See Table 13.2 , p. 253 of your textbook.your textbook.

Dalton’s law of partial pressures Dalton’s law of partial pressures and a table of known vapor and a table of known vapor pressures of water can be used to pressures of water can be used to determine the determine the pressure of dry gaspressure of dry gas that has been collected. that has been collected.

DALTON’S LAW OF PARTIAL PRESSURE: DALTON’S LAW OF PARTIAL PRESSURE: WATER DISPLACEMENT METHODWATER DISPLACEMENT METHOD

SAMPLE PROBLEM SAMPLE PROBLEM

In a laboratory, oxygen gas was In a laboratory, oxygen gas was collected by water displacement at collected by water displacement at an atmospheric pressure of 726 torr an atmospheric pressure of 726 torr and a temperature of 22and a temperature of 22°C. °C. Calculate the partial pressure of dry Calculate the partial pressure of dry oxygen.oxygen.


SOLUTION to Sample ProblemSOLUTION to Sample Problem

PPTOTAL TOTAL = = 726 torr726 torr

PPWATERWATER= 19.8 torr (22 = 19.8 torr (22 °C°C))

PPOXYGEN OXYGEN = ?= ?

PPTOTAL TOTAL = P= POXYGEN OXYGEN + P+ PWATERWATER

PPOXYGEN OXYGEN = P= PTOTAL TOTAL – P– PWATERWATER

= 726 torr – 19.8 torr= 726 torr – 19.8 torr

PPOXYGEN OXYGEN = 706.2 torr= 706.2 torr



1. 1. A sealed container of bottled A sealed container of bottled water sits on a store shelf at a water sits on a store shelf at a temperature of 23temperature of 23°C. What is the °C. What is the partial pressure of water vapor in partial pressure of water vapor in the air space inside the container?the air space inside the container?



2.2. Nitrogen gas is collected at 20°C Nitrogen gas is collected at 20°C and a total ambient pressure of and a total ambient pressure of 735.8 torr using the method of 735.8 torr using the method of water displacement. What is the water displacement. What is the partial pressure of dry nitrogen?partial pressure of dry nitrogen?



3.3. Hydrogen gas is collected over Hydrogen gas is collected over water at a total pressure of 714.4 water at a total pressure of 714.4 mm Hg. The volume of hydrogen mm Hg. The volume of hydrogen collected is 30 mL at 25°C. What collected is 30 mL at 25°C. What is the partial pressure of hydrogen is the partial pressure of hydrogen gas?gas?


SAMPLE PROBLEM SAMPLE PROBLEM

A A 500 mL500 mL sample of oxygen was sample of oxygen was collected over water at collected over water at 2323°C°C and and 760 torr760 torr pressure. What volume pressure. What volume will the dry oxygen occupy at will the dry oxygen occupy at 2323°C°C and and 760 torr760 torr? The vapor pressure ? The vapor pressure of water at of water at 2323°C°C is is 21.121.1 torr (from torr (from table 13.2, TBK)table 13.2, TBK)


SOLUTION to Sample Problem SOLUTION to Sample Problem

PPTOTAL TOTAL = P= POXYGENOXYGEN - P - PWATERWATER

PPOXYGENOXYGEN = 760 torr – 21.1 torr = 760 torr – 21.1 torr

PPOXYGEN OXYGEN = 738.9 torr= 738.9 torr

Given: VGiven: V11 = 500 mL = 500 mL VV22 = ? = ?

PP11 = 738.9 torr = 738.9 torr PP22 = 760 torr = 760 torr

TT11=23=23c + 273 = 300K c + 273 = 300K

TT22=23=23c + 273 = 300Kc + 273 = 300K


SOLUTION to Sample Problem SOLUTION to Sample Problem VV22 = = PP11VV11

PP22

= (= (500 mL) (739 torr)500 mL) (739 torr) = 486 mL dry = 486 mL dry OO22

(760 torr)(760 torr)


SAMPLE EXERCISESAMPLE EXERCISE

Oxygen gas occupies 500 mL at Oxygen gas occupies 500 mL at 20°C and 760 mm Hg. What 20°C and 760 mm Hg. What volume will it occupy if is collected volume will it occupy if is collected over water at 30°C and 750 mm over water at 30°C and 750 mm Hg?Hg?

AVOGADRO’S LAWAVOGADRO’S LAW

Joseph Louis Gay-Lussac of France Joseph Louis Gay-Lussac of France studied the studied the volume relationship of volume relationship of reacting gases.reacting gases.

In 1809, he published his results. He In 1809, he published his results. He summarized in a statement known as summarized in a statement known as GAY-LUSSAC’S LAW OF COMBINING GAY-LUSSAC’S LAW OF COMBINING VOLUMES OF GASES: When measured at VOLUMES OF GASES: When measured at the same temperature and pressure, the the same temperature and pressure, the ratios of the volumes of reacting gases ratios of the volumes of reacting gases are the small whole numbers.are the small whole numbers.


HH22 + Cl + Cl22 2 HCl 2 HCl

1 volume 1 volume 2 1 volume 1 volume 2 volumesvolumes

1 molecule 1 molecule 2 1 molecule 1 molecule 2 moleculesmolecules

1 mol1 mol 1 mol 2 mol 1 mol 2 mol


AVOGADRO’S LAW states that equal AVOGADRO’S LAW states that equal volumes of different gases at the same volumes of different gases at the same temperature and pressure contain the temperature and pressure contain the same number of molecules.same number of molecules.

If the amount of gas in a container is If the amount of gas in a container is increasedincreased, the volume is , the volume is increasedincreased..

If the amount of gas in a container is If the amount of gas in a container is decreaseddecreased, the volume is , the volume is decreaseddecreased..


As you increase As you increase the amount of the amount of gas (i.e. through gas (i.e. through inhalation) the inhalation) the volume of the volume of the balloon balloon increases increases likewise.likewise.


This law was a real breakthrough This law was a real breakthrough in understanding the nature of in understanding the nature of gases.gases.

1. It offered a rational explanation of 1. It offered a rational explanation of Gay-Lussac’s law of combining Gay-Lussac’s law of combining volumes of gases and indicated volumes of gases and indicated the diatomic nature of elemental the diatomic nature of elemental gases, such as hydrogen, chlorine gases, such as hydrogen, chlorine and oxygen.and oxygen.


HH22 + Cl + Cl22 2 HCl 2 HCl

1 volume 1 volume 2 1 volume 1 volume 2 volumesvolumes

1 molecule 1 molecule 2 1 molecule 1 molecule 2 moleculesmolecules

1 mol1 mol 1 mol 1 mol 1 mol 1 mol


2. 2. It provided a method for It provided a method for determining the molecular determining the molecular weights of gases of known weights of gases of known molecular weight.molecular weight.

3. It afforded a firm foundation for 3. It afforded a firm foundation for the development of the kinetic the development of the kinetic molecular theory.molecular theory.


The mathematical form of Avogadro’s law is: The mathematical form of Avogadro’s law is: VV ; ; VV11 = = VV22

n nn n11 n n22

SAMPLE PROBLEM 1SAMPLE PROBLEM 1A sample of gas with a volume of 9.20 L is A sample of gas with a volume of 9.20 L is known to contain 1.225 mol. If the amount known to contain 1.225 mol. If the amount of gas is increased to 2.85 mol, what new of gas is increased to 2.85 mol, what new volume will result if the pressure and volume will result if the pressure and temperature remain constant?temperature remain constant?


SOLUTIONSOLUTION

Given: VGiven: V11 = 9.20 L = 9.20 L VV22 = ? = ?

nn11 = 1.225 mol = 1.225 mol nn22 = 2.85 mol = 2.85 mol

Solution:Solution:

VV22 = = nn22VV11 = = (2.85 mol) (9.20 L)(2.85 mol) (9.20 L) = 21.4 L = 21.4 L

nn11 (1.225 mol) (1.225 mol)



1. 1. If 0.25 mol of argon gas occupies a If 0.25 mol of argon gas occupies a volume of 7.62 mL at a particular volume of 7.62 mL at a particular temperature and pressure, what temperature and pressure, what volume would 0.43 mol of argon volume would 0.43 mol of argon have under the same conditions?have under the same conditions?



2. 2. At a certain temperature and At a certain temperature and pressure, a balloon with 10.0 g of pressure, a balloon with 10.0 g of oxygen has a volume of 7.00 L. oxygen has a volume of 7.00 L. What is the volume after 5.00 g of What is the volume after 5.00 g of oxygen is added to the balloon?oxygen is added to the balloon?

Ideal Gas LawIdeal Gas LawThe ideal gas law was first written in 1834 The ideal gas law was first written in 1834

by EMIL CLAPEYRON. It is a by EMIL CLAPEYRON. It is a combination of all the gas laws.combination of all the gas laws.

PV = nRTPV = nRT

Where, P = pressure , V = volume, Where, P = pressure , V = volume, n = amount of gas expressed in mol, n = amount of gas expressed in mol, T = temperature,T = temperature, R = gas constant, 0.0821 R = gas constant, 0.0821 L L atm atm

mol mol K K

Ideal Gas LawIdeal Gas LawThe kinetic molecular theory The kinetic molecular theory

assumes that the particles of an assumes that the particles of an IDEAL GAS have negligible IDEAL GAS have negligible volume and no attraction exists volume and no attraction exists between molecules.between molecules.

The temperature, pressure and The temperature, pressure and volume of an ideal gas are related volume of an ideal gas are related to each other by the ideal gas to each other by the ideal gas equation.equation.

Ideal Gas LawIdeal Gas LawGAS EQUATIONGAS EQUATION

R = R = PVPV = = (1.0 atm) (22.4L)(1.0 atm) (22.4L)

nT (1.0 mol) (273K)nT (1.0 mol) (273K)

R = 0.0821 R = 0.0821 L L atm atm

mol mol K K

Ideal Gas LawIdeal Gas LawSAMPLE PROBLEM 1SAMPLE PROBLEM 1What pressure will be exerted by 0.400 mol of gas in a What pressure will be exerted by 0.400 mol of gas in a 5.00 L container at 17.05.00 L container at 17.0°C?°C?

Given:Given:n = 0.400 moln = 0.400 molV= 5.00 LV= 5.00 LT= 17.0 °C + 273 = 290 KT= 17.0 °C + 273 = 290 KSolution:Solution: L L atm atmP = P = nRTnRT = = (0.400 mol) (0.0821 mol (0.400 mol) (0.0821 mol K) (290 K) K) (290 K) = 1.9atm = 1.9atm V (5.00 L)V (5.00 L)

Ideal Gas LawIdeal Gas LawSAMPLE PROBLEM 2SAMPLE PROBLEM 2Calculate the molecular weight of butane gas, if 4.96 g Calculate the molecular weight of butane gas, if 4.96 g occupy 2.13 L at occupy 2.13 L at 20.0 °C and 1 atm pressure.20.0 °C and 1 atm pressure.

GivenGiven::g = 4.96 gg = 4.96 gV= 2.13 LV= 2.13 LT= 20.0 °C + 273 = 293 KT= 20.0 °C + 273 = 293 KP= 1 atmP= 1 atmSolutionSolution:: L L atm atmMW = gMW = gRTRT = = (4.96 g) (0.0821 mol (4.96 g) (0.0821 mol K) (293 K) K) (293 K) = 56.0 = 56.0 g/molg/mol PV (1 atm) (2.13 L)PV (1 atm) (2.13 L)

Ideal Gas LawIdeal Gas LawSAMPLE PROBLEM 3SAMPLE PROBLEM 3What is the density of oxygen gas at STP? What is the density of oxygen gas at STP?

GivenGiven::MW = 32 g/molMW = 32 g/molT= 273 KT= 273 KP= 1 atmP= 1 atmSolutionSolution:: D = D = MWPMWP = = (32 g/mol) (1 atm) (32 g/mol) (1 atm) = 1.43 g/L = 1.43 g/L RT RT L L atm atm (0.0821 mol (0.0821 mol K) K) (273 K) (273 K)

Ideal Gas LawIdeal Gas LawSAMPLE PROBLEM 4SAMPLE PROBLEM 4

What is the density of oxygen gas at 20°C and 750 mm Hg? What is the density of oxygen gas at 20°C and 750 mm Hg?

GivenGiven::

MW = 32 g/molMW = 32 g/mol

T= 20T= 20°C°C + 273 K= 293 K + 273 K= 293 K

P= 750 mm Hg x P= 750 mm Hg x 1 atm 1 atm = 0.987 atm = 0.987 atm

760 mm Hg760 mm Hg

SolutionSolution::

D = D = MWPMWP = = (32 g/mol) (0.987 atm) (32 g/mol) (0.987 atm) = 1.31 g/L = 1.31 g/L

RT RT L L atm atm

(0.0821 mol (0.0821 mol K) K) (293 K) (293 K)

Ideal Gas LawIdeal Gas LawSAMPLE EXERCISESSAMPLE EXERCISES

1.1. What volume will 1.27 mol of What volume will 1.27 mol of helium gas occupy at STP?helium gas occupy at STP?

2.2. Calculate the volume of 0.360 Calculate the volume of 0.360 mole Hmole H22 at 52 at 52 °C and 1.5 atm °C and 1.5 atm pressure.pressure.


3. How many moles of gas are 3. How many moles of gas are contained in a 50.0-L cylinder at a contained in a 50.0-L cylinder at a pressure of 100.0 atm and a pressure of 100.0 atm and a

temperature of 35.0 °C?temperature of 35.0 °C? 4. Calculate the molecular weight of a 4. Calculate the molecular weight of a

gas having a density of 2.47 g/L at gas having a density of 2.47 g/L at 26 26 °C and 1 atm pressure?°C and 1 atm pressure?


5. What is the density of nitrogen gas 5. What is the density of nitrogen gas at STP?at STP?

SHORT QUIZSHORT QUIZSOLVE THE FOLLOWING PROBLEMSSOLVE THE FOLLOWING PROBLEMS

1.1. What is the total pressure in What is the total pressure in atmospheres of a gas mixture containing atmospheres of a gas mixture containing argon gas at 0.28 atm, helium gas at argon gas at 0.28 atm, helium gas at 280 mm Hg, and nitrogen gas at 390 280 mm Hg, and nitrogen gas at 390 torr?torr?

2.2. A sample of 8.00 moles of argon has a A sample of 8.00 moles of argon has a volume of 20.0 L. A small leak causes volume of 20.0 L. A small leak causes half of the molecules to escape. What is half of the molecules to escape. What is the new volume of the gas?the new volume of the gas?

SHORT QUIZSHORT QUIZSOLVE THE FOLLOWING PROBLEMSSOLVE THE FOLLOWING PROBLEMS

3. How many moles of O3. How many moles of O22 are present are present in 44.8 L of Oin 44.8 L of O22 at STP? at STP?

DALTON’S LAW, AVOGADRO, IDEAL

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