chapter 3 - the gaseous state

CHM 160CHAPTER 3

THE GASEOUS STATE

2

The Nature of Gases

• Gases are compressible.• Gases have low density.• Gases mix thoroughly.• A gas fills a container uniformly and

completely.• A gas exerts pressure uniformly on all

sides of a container.

Force exerted per unit Force exerted per unit area of surface by area of surface by molecules in motion.molecules in motion.

P = Force/unit area– 1 atmosphere = 14.7 psi– 1 atmosphere = 760 mm Hg– 1 atmosphere = 29.92 in Hg– 1 atmosphere = 101,325 Pascals– 1 Pascal = 1 kg/m.s

PressurePressure

4

The Pressure of a Gas

• Pressure is the force per unit area.

• Originally measured with a barometer.

P(Pressure)F(Force)

A(Area)=

SI unit of pressure: kg/m/s2

= Pascal

5

Units for Pressure

• Pressure is measured in a variety of units, since the SI unit is cumbersome.

• One of the most common units is fractions of atmospheric pressure at sea level – the atmosphere (atm).

• One standard atm is defined as the pressure that will support 760 mm of Hg (mmHg or torr).

iiff VPVP Boyle’s LawBoyle’s Law:: The The

volume of a sample volume of a sample of gas at a given of gas at a given temperature varies temperature varies inversely with the inversely with the applied pressure. applied pressure.

V 1/P

(constant moles and T)

7

Boyle’s Law of Pressure and Volume

• There is an inverse relationship between pressure and volume at constant temperature.

V =constant

Por PV = constant (P1V1 = P2V2)

8

Boyle’s Law

• At constant temperature and number of moles of gas, we can calculate the effect of changes in pressure or volume on a gas.

Inside an car engine, the volume of a cylinder is 475 mL, whenthe pressure is 1.05 atm. When the gas is compressed, the pressureincreases to 5.65 atm (at the same temperarure). What is the volumeof the compressed gas?

SolutionInitial conditions Final conditionsV1 = 475 mLP1 = 1.05 atm

V2 = ?P2 = 5.65 atm

P1V1 = P2V2

V2 = 475 mL x1.05 atm

5.65 atm= 88.3 mL

9

Charles’, Gay-Lussac’s and Avogadro’s Law

• Charles found that the volume of a gas at constant pressure was directly proportional to temperature.

• A plot of V vs. T is linear.

10

Absolute Temperature

• If Celsius degrees are used for temperature, extrapolating to zero volume yields a temperature of -273.15 C.

11

Absolute Temperature

• A gas cannot be cooled infinitely (it condenses to a liquid or a solid).

• The zero volume temperature does have significance, since it is theoretically the lowest possible temperature.

• At this temperature, all translational motion (point to point motion) ceases.

12

The Kelvin Scale

• Begins at 0 K (-273.15 C), known as absolute zero.

• Kelvin scale has no negative values.• Magnitude of the Kelvin and Celsius

degrees are the same, hence the conversion between the two is simple.

(K = C + 273.15)

13

Charles’ Law for V and T

The volume of a gas is directlyproportional to the temperature(in K) at constant prssure.

V = constant T or VT

= constant

V1

T1

V2

T2=

example

• A 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.

i

i

f

f

TV

TV

using

)K278()K359)(L8.3(

TTV

fi

fiV

L 9.4Vf

15

Gay-Lussac’s Law

• Pressure is directly proportional to the Kelvin temperature at constant volume.

P = constant T orP1

T1

P2

T2=

example

• 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

T

P using

)K298()K1473)(atm4.1(

TTP

fi

fiP

atm9.6Pf

17

Charles’ and Gay-Lussac’s Laws

• Both laws follow from the kinetic molecular theory of gases.

• If T is increased, the speed of the gas molecule increases.

• The gas molecule travels farther in a given time and strikes other objects with more force.

• Hence either V or P must increase.

18

Combined Gas Law

• Taking Boyle’s, Charles’ and Gay-Lussac’s Laws and combining them yields the combined gas law.

• Note that T must be in K, and P1, P2 and V1, V2 must be in the same units.

PV

T= constant P1V1

T1

P2V2

T2=or

example

• A 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?

f

ff

i

iiTVP

TVP

using

)K 303)(Hg mm 800()K 473)(L 5.4)(Hg mm 650(

TPTVP

Vif

fiif

L7.5Vf

20

Standard Temperature and Pressure

• It is convenient for comparing gas properties to have a set of standard conditions.

• Standard conditions show up in many areas of chemistry.

• For gases– Standard temperature is 273 K or 0 C– Standard pressure is 760 torr or 1 atm

21

Avogadro’s Law

• Equal volumes of gases at the same pressure and temperature contain equal numbers of molecules.or

• The volume of a gas is proportional to the number of molecules (moles) of gas present at constant pressure and temperature.

• V n or V = k × n

example

• A sample of fluorine gas has a volume of 5.80 L at 150.0oC and 10.5 atm of pressure. How many moles of fluorine gas are present?

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

TPTVP

Vistd

stdiiSTP

LSTP 3.39V

First, use the combined empirical gas law to determine the volume at STP.

• Since 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

Malone and Dolter - Basic Concepts of Chemistry 8e 24

Summary Chart

Gas Laws

Moles

Volume

Temperature Pressure

V = constant x n

V =constant

P

PT = constant

V = constant x T

25

Pioneers of the Gas Laws

Lavoisier Priestley

BoyleGay-Lussac Avogadro

The Ideal Gas Law

• From the empirical gas laws, we see that volume varies in proportion to pressure, absolute temperature, and moles.

Law sBoyle' 1/PV

Law Charles' TV abs

Law sAvogadro' n V

27

The Ideal Gas Law

• We can combine the three lawsV 1/P; V T; V n to yieldV nT/P

• If we introduce a constant R and rearrange, we get the ideal gas law (ideal in that it assumes the kinetic molecular theory)

PV = nRT

• where R = 0.082057 L atm/K mol

• Thus, 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)

29

Calculation of Volume Using the Ideal Gas Law

• What is the volume of 1.00 mol of gas at STP?

Solution

PV = nRT or V =nRTP

V = 1.00 mol x 0.0821 L atm/K mol x 273 K

1.00 atm= 22.4 L

30

Dalton’s Law of Partial Pressures

• The total pressure of a gas in a system is the sum of the partial pressure of each component gasPtotal = P1 + P2 + P3 + …where P1 is the pressure due to gas 1, etc.

• For example, 21 % of the molecules in the atmosphere, therefore 21 % of the volume and the pressure of the atmosphere, is due to oxygene. g.

P(O2) = 0.21 × 760 torr = 160 torr

Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures:: the the sum of all the pressures of all the sum of all the pressures of all the different gases in a mixture equals different gases in a mixture equals the total pressure of the mixture. the total pressure of the mixture.

32

Calculation of Total Pressure of a Mixture of Gases

• What is the pressure (in atm) exerted by 12.0 g of N2, and 12.0 g of O2 in a 2.50 L flask at 25 oC?

Solution

1.Find the number moles of N2 and O2.2. Use the total moles and calculate P from the ideal gas law.

n(O2) = 12.0 g x1 mol O2

32.00 g O2

= 0.375 mol O2.

Likewise, n(N2) =0.428 mol N2

P = nRT

V= 0.803 mol x 0.0821 L atm

K molx 298 K

2.50 L

= 7.86 atm

example

• Suppose a 156 mL sample of H2 gas was collected over water at 19oC and 769 mm Hg. What is the mass of H2 collected?

Hg mm 16.5 - Hg mm 697 P2H

Hg mm 527 P2H

– Table lists the vapor pressure of water at 19oC as 16.5 mm Hg.

• Now 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

• 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

36

The Molar Volume and Density of a Gas

• Molar volume of a gas (properties that are independent of the identity of the gas)– 22.4 L at STP (ca 6 gallons)– Volume of one mole of gas– Avogadro’s number (6.022 × 1023 molecules)

• One molar mass (depends on the gas)• Gas densities are on the order of g/L.

Molecular Weight Determination

• the relationship between moles and mass.

mass molecular massmoles

or

mMmn

• If 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

example

• A 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 determination

• If we look again at our derivation of the molecular mass equation,

RT)(PVmM

m

RTPM

D Vm m

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

example

• Calculate 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

42

Calculating Mass from Volume at STP

• What is the mass of 4.55 L of O2 measured at STP?

Solution

Given: L of O2 Required: g of O2

Volume (STP) Moles Mass

4.55 L x1 mol

22.4 Lx

32.00 g

mol= 6.50 g

43

Stoichiometry Involving Gases

• The ideal gas law allows us to convert a given volume of gas at a specified temperature and pressure into moles of gas.

• For example, what mass of NaBH4 is needed to produce 50.0 L of H2 at STP?

NaBH4(s) + 2H2O(l) NaBO2(aq) + 4H2(g)

Stoichiometry 1 NaBH4: 4 H2

Given: L of H2 Requested: g of NaBH4

Solution

44

Stoichiometry Involving Gases…Cont’d

Volume H2 Mol H2 Mol NaBH4 Mass NaBH4

50.0 L (STP) x1 mol H2

22.4 L (STP)x

1 mol NaBH4

4 mol H2

37.83 g NaBH4

mol NaBH4

x

= 21.1 g NaBH4

45

Stoichiometry Involving Gases

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.

mM1

effusion of Rate

–The equation for the velocity of gases shows the following relationship between rate of effusion and molecular mass.

• According 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

example

• How 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 Gases

• Real gases do not follow PV = nRT perfectly. The van der Waals equation corrects for the nonideal nature of real gases.

nRT nb)-V)( P( 2

2

Van

a corrects for interaction between atoms.

b corrects for volume occupied by atoms.

In the van der Waals equation,

nb)-V( becomesV

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

• Also, in the van der Waals equation,

)P( becomes P 2

2

Van

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

example

• If 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.

Table lists the following values for SO2a = 6.865 L2.atm/mol2b = 0.05679 L/mol

• First, 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. K

T = 273.2 K

V = 22.41 L

a = 6.865 L2.atm/mol2

b = 0.05679 L/mol

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 The “real” pressure exerted by 1.00 mol of SOThe “real” pressure exerted by 1.00 mol of SO22

at STP is slightly less than the “ideal” at STP is slightly less than the “ideal” pressure.pressure.