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Dalton’s LawIdeal Gas Law(p. 322-325, 340-

346)

Ch. 10 & 11 - Ch. 10 & 11 - GasesGases

A.A. Dalton’s Law of Partial Dalton’s Law of Partial PressuresPressures

A.A. Dalton’s Law of Partial Dalton’s Law of Partial PressuresPressures

Total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.• In the absence of a

chemical reaction

PT = P1+ P2+ P3+…

Practice ProblemPractice ProblemPractice ProblemPractice Problem

A mixture of four gases exerts a total pressure of 1200 mm Hg. Gases A and B each exert 420 mm Hg. Gas C exerts 280 mm Hg. What pressure is exerted by gas D?

PT = P1 + P2 + P3 + P4

1200 = 420 mm Hg + 420 mm Hg + 280 mm Hg + P4

1200 = 1120 mm Hg + P4

P4 = 80 mm Hg

B. Vapor pressure of B. Vapor pressure of waterwaterB. Vapor pressure of B. Vapor pressure of waterwater

Gases are often collected in lab by water displacement and are mixed with water vapor

Patm = Pgas + PH2O

To determine the pressure of the gas collected – subtract the vapor pressure of the water at that temperature from the current atmospheric pressure

C. Ideal Gas LawC. Ideal Gas LawC. Ideal Gas LawC. Ideal Gas Law

The mathematical relationship among pressure, volume, temperature and the number of moles of a gas.

Derived by combining the gas laws.

PV=nRT

PV

TVn

PVnT

D. Ideal Gas ConstantD. Ideal Gas ConstantD. Ideal Gas ConstantD. Ideal Gas Constant

= k

IDEAL GAS CONSTANTR=0.0821 Latm/molKR=8.315 dm3kPa/molK

= R

Merge the Combined Gas Law with Avogadro’s Principle:

GIVEN:

P = ? atm

n = 0.412 mol

T = 16°C = 289 K

V = 3.25 LR = 0.0821Latm/molK

WORK:

PV = nRT

P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K

P = 3.01 atm

E. Ideal Gas Law ProblemsE. Ideal Gas Law ProblemsE. Ideal Gas Law ProblemsE. Ideal Gas Law Problems Calculate the pressure in atmospheres of

0.412 mol of He at 16°C & occupying 3.25 L.

GIVEN:

V = ?

n = 85 g

T = 25°C = 298 K

P = 104.5 kPaR = 8.315 dm3kPa/molK

E. Ideal Gas Law ProblemsE. Ideal Gas Law ProblemsE. Ideal Gas Law ProblemsE. Ideal Gas Law Problems

Find the volume of 85 g of O2 at 25°C and 104.5 kPa.

= 2.7 mol

WORK:

85 g 1 mol = 2.7 mol

32.00 g

PV = nRT(104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K

V = 64 dm3

F.F. Finding Molar Mass Finding Molar Mass from the Ideal Gas Lawfrom the Ideal Gas Law

F.F. Finding Molar Mass Finding Molar Mass from the Ideal Gas Lawfrom the Ideal Gas Law

n = mass n = m Molar mass M

PV = mRT OR M

PV=nRT

M = mRT PV

G.G. Finding Density Finding Density from the Ideal Gas Lawfrom the Ideal Gas Law

G.G. Finding Density Finding Density from the Ideal Gas Lawfrom the Ideal Gas Law

D = mass or D = m volume V

M = DRT P

M = mRT PV

D = MP RT

PRACTICE PROBLEMSPRACTICE PROBLEMSPRACTICE PROBLEMSPRACTICE PROBLEMS At 28°C and 0.974 at, 1.00 L of gas has

a mass of 5.16 g. What is the molar mass of this gas?

P = 0.974 atm V = 1.00 L T = 28°C = 273 = 301 K m = 5.16 g

M = mRT PV= (5.16 g) (0.0821 L∙atm/mol∙K) (301K) (0.974 atm)(1.00 L) = 131g/mol

PRACTICE PROBLEMSPRACTICE PROBLEMSPRACTICE PROBLEMSPRACTICE PROBLEMS What is the density of a sample of

ammonia gas, NH3, if the pressure is 0.928 atm and the temperature is 63.0°C?

P = 0.928 atm T = 63.0°C + 273 = 336 K

M = 17.034 g/mol R = 0.0821 L∙atm/mol∙K

D = MP = (17.034 g/mol)(0.928 atm) RT (0.0821L∙atm/mol∙K)((336 K)

= 0.572 g/L NH3

Homework AssignmentHomework AssignmentHomework AssignmentHomework Assignment

Workbook.Complete problems #1 and 2

on pp. 173-174, 1-2 p. 178, 1-2 p. 180

Practice Test Part 2Practice Test Part 2Practice Test Part 2Practice Test Part 2

P. 181 – 182 #1-7, 12, 18 - 21

Practice test Part 2Practice test Part 2Practice test Part 2Practice test Part 2

Workbook p. 182 #1-7