chapter 10; gases

Chapter 10;

Gases

Elements that exist as gases at 250C and 1 atmosphere

• 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.

• Gases have much lower densities than liquids and solids.

Physical Characteristics of Gases

Gas LawsIn the first part of this chapter we will

examine the quantitative relationships, or empirical laws, governing gases.

• First, however, we need to understand the concept of pressure.

PressureForce exerted per unit area of surface by

molecules in motion.

– 1 atmosphere = 14.7 psi– 1 atmosphere = 760 mm Hg– 1 atmosphere = 101,325 Pascals– 1 Pascal = 1 kg/m.s2

P = Force/unit area

Units of Pressure

1 pascal (Pa) = 1 N/m2

1 atm = 760 mmHg = 760 torr

1 atm = 101,325 Pa

Barometer

Pressure = Force

Area

The Empirical Gas LawsBoyle’s Law: The volume of a sample of

gas at a given temperature varies inversely with the applied pressure. V 1/P (constant moles and

T) or

iiff VPVP

P 1/V

P x V = constant

P1 x V1 = P2 x V2

Boyle’s Law

Constant temperatureConstant amount of gas

A Problem to ConsiderA sample of chlorine gas has a volume of

1.8 L at 1.0 atm. If the pressure increases to 4.0 atm (at constant temperature), what would be the new volume?

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?P1 x V1 = P2 x V2

As T increases V increases

The Empirical Gas LawsCharles’s Law: The volume occupied by any

sample of gas at constant pressure is directly proportional to its absolute temperature.

V Tabs (constant moles and P)

or

i

i

f

f

TV

TV

Variation of gas volume with temperatureat constant pressure.

V T

V = constant x T

V1/T1 = V2/T2T (K) = t (0C) + 273.15

Charles’ Law

Temperature must bein Kelvin

A Problem to ConsiderA sample of methane gas that has a

volume of 3.8 L at 5.0°C is heated to 86.0°C at constant pressure. Calculate its new volume.

A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?

V1/T1 = V2/T2

The Empirical Gas LawsGay-Lussac’s Law: The pressure exerted

by a gas at constant volume is directly proportional to its absolute temperature. P Tabs (constant moles and V)

or

i

i

f

f

TP

TP

A Problem to Consider

An aerosol can has a pressure of 1.4 atm at 25°C. What pressure would it attain at 1200°C, assuming the volume remained constant?

i

i

f

f

TP

TP

using

)K298()K1473)(atm4.1(

TTP

fi

fiP

atm9.6Pf

The Empirical Gas LawsCombined Gas Law: In the event that all

three parameters, P, V, and T, are changing, their combined relationship is defined as follows:

f

ff

i

ii

TVP

TVP

A Problem to ConsiderA sample of carbon dioxide occupies 4.5 L

at 30°C and 650 mm Hg. What volume would it occupy at 800 mm Hg and 200°C?

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?

• The volume of one mole of gas is called the molar gas volume, Vm.

• Volumes of gases are often compared at standard temperature and pressure (STP), chosen to be 0 oC and 1 atm pressure.

The Empirical Gas LawsAvogadro’s Law: Equal volumes of any

two gases at the same temperature and pressure contain the same number of molecules.

– At STP, the molar volume, Vm, that is, the volume occupied by one mole of any gas, is

22.4 L/mol– So, the volume of a sample of gas is

directly proportional to the number of moles of gas, n.

The Empirical Gas LawsAvogadro’s Law

nV

A Problem to ConsiderA sample of fluorine gas has a volume of

5.80 L at 150.0 oC and 10.5 atm of pressure. How many moles of fluorine gas are present?First, use the combined empirical gas law

to determine the volume at STP.

)K423)(atm0.1()K273)(L80.5)(atm5.10(

TPTVP

Vistd

stdiiSTP

L3.39VSTP

A Problem to ConsiderSince Avogadro’s law states that at STP the

molar volume is 22.4 L/mol, then

L/mol 22.4V

gas of moles STP

L/mol 22.4L 39.3

gas of moles

mol 1.75 gas of moles

Avogadro’s LawV number of moles (n)

V = constant x n

V1/n1 = V2/n2

Constant temperatureConstant pressure

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

The Ideal Gas LawFrom the empirical gas laws, we See that

volume varies in proportion to pressure, absolute temperature, and moles.

Law sBoyle' 1/PV

Law sAvogadro' nV Law Charles' TV abs

– Combining the three proportionalities, we can obtain the following relationship.

The Ideal Gas LawThis implies that there must exist a

proportionality constant governing these relationships.

)( PnTabs R""V

where “R” is the proportionality constant referred to as the ideal gas constant.

The Ideal Gas LawThe numerical value of R can be derived

using Avogadro’s law, which states that one mole of any gas at STP will occupy 22.4 liters.

nTVP R

K) mol)(273 (1.00atm) L)(1.00 (22.4 R

KmolatmL 0.0821

The Ideal Gas LawThus, the ideal gas equation, is usually

expressed in the following form:

nRT PV P is pressure (in atm)V is volume (in liters)n is number of atoms (in moles)R is universal gas constant 0.0821 L.atm/K.molT is temperature (in Kelvin)

A Problem to Consider

An experiment calls for 3.50 moles of chlorine, Cl2. What volume would this be if the gas volume is measured at 34°C and 2.45 atm?

PnRT V since

atm 2.45K) )(307 1mol)(0.082 (3.50 Kmol

atmL

V then

L 36.0 V then

Molecular Weight DeterminationIn Chapter 3 we showed the relationship

between moles and mass.

mass molecular massmoles

or

mMmn

Molecular Weight DeterminationIf we substitute this in the ideal gas

equation, we obtain

RT)(PVmM

mIf we solve this equation for the molecular mass, we obtain

PVmRT Mm

A Problem to ConsiderA 15.5 gram sample of an unknown gas

occupied a volume of 5.75 L at 25°C and a pressure of 1.08 atm. Calculate its molecular mass.

PVmRT M Since m

L) atm)(5.75 (1.08

K) )(298g)(0.0821 (15.5 M then Kmol

atmL

m

g/mol 61.1 Mm

Density DeterminationIf we look again at our derivation of the

molecular mass equation,

RT)(PVmM

mwe can solve for m/V, which represents density.

RTPM

D Vm m

A Problem to ConsiderCalculate the density of ozone, O3 (Mm =

48.0g/mol), at 50°C and 1.75 atm of pressure.

RTPM

D Since m

K) )(323(0.0821g/mol) atm)(48.0 (1.75

D thenKmol

atmL

g/L 17.3 D

Stoichiometry Problems Involving Gas Volumes

Suppose you heat 0.0100 mol of potassium chlorate, KClO3, in a test tube. How many liters of oxygen can you produce at 298 K and 1.02 atm?

)g(O 3 KCl(s) 2 (s)KClO 2 23

• Consider the following reaction, which is often used to generate small quantities of oxygen.

Stoichiometry Problems Involving Gas Volumes

First we must determine the number of moles of oxygen produced by the reaction.

3

23 KClO mol 2

O mol 3 KClO mol 0100.0

2O mol 5001.0

Stoichiometry Problems Involving Gas Volumes

Now we can use the ideal gas equation to calculate the volume of oxygen under the conditions given.

PnRT V

atm 02.1K) )(298 0821.0)(O mol (0.0150 Kmol

atmL2V

L 0.360 V

Partial Pressures of Gas MixturesDalton’s Law of Partial Pressures: the

sum of all the pressures of all the different gases in a mixture equals the total pressure of the mixture.

....PPPP cbatot

Partial Pressures of Gas MixturesThe composition of a gas mixture is often

described in terms of its mole fraction.

tot

A

tot

AA P

Pnn

Aof fraction Mole

– The mole fraction, , of a component gas is the fraction of moles of that component in the total moles of gas mixture.

Partial Pressures of Gas MixturesThe partial pressure of a component gas,

“A”, is then defined as

totAA P P – Applying this concept to the ideal gas

equation, we find that each gas can be treated independently.RTn VP AA

A Problem to ConsiderGiven a mixture of gases in the atmosphere

at 760 torr, what is the partial pressure of N2 ( = 0 .7808) at 25°C?

torr) (760 (0.7808) P then2N

torr 593 P2N

totNN P P since22

Collecting Gases “Over Water”A useful application of partial pressures

arises when you collect gases over water.

– As gas bubbles through the water, the gas becomes saturated with water vapor.

– The partial pressure of the water in this “mixture” depends only on the temperature.

A Problem to ConsiderSuppose a 156 mL sample of H2 gas was

collected over water at 19oC and 769 mm Hg. What is the mass of H2 collected?

– First, we must find the partial pressure of the dry H2.

0HtotH 22P P P

A Problem to ConsiderSuppose a 156 mL sample of H2 gas was

collected over water at 19oC and 769 mm Hg. What is the mass of H2 collected?– The vapor pressure of water at 19oC as 16.5 mm Hg.

Hg mm 16.5 - Hg mm 697 P2H

Hg mm 527 P2H

A Problem to ConsiderNow we can use the ideal gas equation,

along with the partial pressure of the hydrogen, to determine its mass.

atm 989.0 Hg mm 527 P Hg mm 760atm 1

H2

L 0.156 mL 156 V K 292 273) (19 T

? n

A Problem to Consider

From the ideal gas law, PV = nRT, you have

)K 292)( (0.0821L) atm)(0.156 (0.989

RTPV

nKmol

atmL

mol 0.00644 n – Next,convert moles of H2 to grams of H2.

22

22 H g 0.0130

H mol 1H g 2.02

H mol 0.00644

Kinetic-Molecular Theory A simple model based on the actions of individual atoms

Volume of particles is negligibleParticles are in constant motionNo inherent attractive or repulsive forcesThe average kinetic energy of a collection

of particles is proportional to the temperature (K)

Molecular Speeds; Diffusion and EffusionThe root-mean-square (rms) molecular

speed, u, is a type of average molecular speed, equal to the speed of a molecule having the average molecular kinetic energy. It is given by the following formula:

mM3RT

u

Molecular Speeds; Diffusion and Effusion

Diffusion is the transfer of a gas through space or another gas over time.

Effusion is the transfer of a gas through a membrane or orifice. – The equation for the rms velocity of gases

shows the following relationship between rate of effusion and molecular mass.

mM1

effusion of Rate

Molecular Speeds; Diffusion and EffusionAccording to Graham’s law, the rate of

effusion or diffusion is inversely proportional to the square root of its molecular mass.

Agas of MB Gas of M

B"" gas of effusion of RateA"" gas of effusion of Rate

m

m

A Problem to ConsiderHow much faster would H2 gas effuse

through an opening than methane, CH4?

)(HM)(CHM

CH of RateH of Rate

2m

4m

4

2

8.2g/mol 2.0g/mol 16.0

CH of RateH of Rate

4

2

So hydrogen effuses 2.8 times faster than CH4

Real GasesReal gases do not follow PV = nRT

perfectly. The van der Waals equation corrects for the nonideal nature of real gases.

a corrects for interaction between atoms.

b corrects for volume occupied by atoms.

nRT nb)-V)( P( 2

2

Van

Real GasesIn the van der Waals equation,

where “nb” represents the volume occupied by “n” moles of molecules

nb)-V( becomesV

Real GasesAlso, in the van der Waals equation,

where “n2a/V2” represents the effect on pressure to intermolecular attractions or repulsions.

)P( becomes P 2

2

Van

Values of van der Waals constants for various gases can always be referred from.

A Problem to ConsiderIf sulfur dioxide were an “ideal” gas, the

pressure at 0°C exerted by 1.000 mol occupying 22.41 L would be 1.000 atm. Use the van der Waals equation to estimate the “real” pressure.The constants a and b for SO2

a = 6.865 L2.atm/mol2

b = 0.05679 L/mol

A Problem to ConsiderFirst, let’s rearrange the van der Waals

equation to solve for pressure.

2

2

V

an -

nb-VnRT

P

R= 0.0821 L.

atm/mol. KT = 273.2 KV = 22.41 L

a = 6.865 L2.atm/mol2

b = 0.05679 L/mol

A Problem to Consider

The “real” pressure exerted by 1.00 mol of SO2 at STP is slightly less than the “ideal” pressure.

2

2

V

an -

nb-VnRT

P

L/mol) 79mol)(0.056 (1.000 - L 22.41

)K2.273)( 06mol)(0.082 (1.000 P Kmol

atmL

2mol

atmL2

L) 41.22(

) (6.865mol) (1.000-

2

2

atm 0.989 P