chapter 12 : day 2

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Chapter 12 : Day 2. IDEAL GAS LAW. Using KMT to Understand Gas Laws. Recall that KMT assumptions are Gases consist of molecules in constant, random motion. P arises from collisions with container walls. No attractive or repulsive forces between molecules. Collisions elastic. - PowerPoint PPT Presentation

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11

Chapter 12 : Day 2

•IDEAL GAS LAWIDEAL GAS LAW

22

Using KMT to Understand Gas Using KMT to Understand Gas LawsLaws

Recall that KMT assumptions are

• Gases consist of molecules in constant, random motion.

• P arises from collisions with container walls.

• No attractive or repulsive forces between molecules. Collisions elastic.

• Volume of molecules is negligible.

33Properties of Properties of GasesGases

Gas properties can be modeled using math. Model depends on—

• V = volume of the gas (L)

• T = temperature (K)

• n = amount (moles)

• P = pressure (atmospheres)

44

IDEAL GAS LAWIDEAL GAS LAW

Brings together gas Brings together gas properties.properties.

Can be derived from Can be derived from experiment and theory.experiment and theory.

P V = n R TP V = n R T

55Using PV = nRTUsing PV = nRT

How much N2 is req’d to fill a small room with a volume of 27,000 L to P = 745 mm Hg at 25 oC?

R = 0.082057 L•atm/K•molSolution

1. Get all data into proper UNITS V = 27,000 L

T = 25 oC + 273 = 298 K

P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm

66

Using PV = nRTUsing PV = nRTR = 0.082057 L•atm/K•mol

Solution

2. Now calc. n = PV / RT

n = (0.98 atm)(2.7 x 10 4 L)

(0.0821 L • atm/K • mol)(298 K)n =

(0.98 atm)(2.7 x 10 4 L)

(0.0821 L • atm/K • mol)(298 K)

n = 1.1 x 10n = 1.1 x 1033 mol (or about 30 kg of gas) mol (or about 30 kg of gas)

77Deviations from Deviations from Ideal Gas LawIdeal Gas Law

• Real molecules

have volume.

• There are intermolecular forces.

–Otherwise a gas could not become a liquid.

Fig. 12.20Fig. 12.20

88Deviations from Ideal Gas Deviations from Ideal Gas LawLaw

Account for volume of molecules and Account for volume of molecules and intermolecular forces with intermolecular forces with VAN VAN DER WAAL’S EQUATIONDER WAAL’S EQUATION..

Measured V = V(ideal)Measured P

intermol. forcesvol. correction

J. van der Waals, J. van der Waals, 1837-1923, 1837-1923, Professor of Professor of Physics, Physics, Amsterdam. Amsterdam. Nobel Prize 1910.Nobel Prize 1910.

nRTV - nbV2

n2aP + ----- )(

99Deviations from Ideal Gas Deviations from Ideal Gas LawLaw

Deviations from Ideal Gas Deviations from Ideal Gas LawLaw

ClCl22 gas has gas has aa = 6.49, = 6.49, bb = 0.0562 = 0.0562

For 8.0 mol ClFor 8.0 mol Cl22 in a 4.0 L tank at 27 in a 4.0 L tank at 27 ooC.C.

P (ideal) = nRT/V = 49.3 atmP (ideal) = nRT/V = 49.3 atm

P (van der Waals) = 29.5 atmP (van der Waals) = 29.5 atm

Measured V = V(ideal)Measured P

intermol. forces

vol. correction

nRTV - nbV2

n2aP + -----

1010

Gases are most ideal when:

1. higher temperatures = more motion

T2. lower pressures = fewer hits

P3 larger volumes = more space between

V4. less gas in space = fewer number

n

1111

IDEAL GAS LAWIDEAL GAS LAW

Brings together gas Brings together gas properties.properties.

Can be derived from Can be derived from experiment and theory.experiment and theory.

P V = n R TP V = n R T

1212

1313

Chapter 12 : Day 3

GAS DENSITYGAS DENSITY

1414

GAS DENSITYGAS DENSITYGAS DENSITYGAS DENSITY

High High densitydensity

Low densitydensity = mass/ volume

d = g/V

1515

IDEAL GAS LAWIDEAL GAS LAW

Brings together gas properties.

Can be used to determine molar mass of a gas

or determine the density of a gas

P V = n R T

1616

Molar Conversions

moles = mass/ molar mass

n = g/

1717GAS DENSITYGAS DENSITYPV = nRT

or PV = nRT

or

d and d and proportional proportional

n = P_V RT

g = P where g is the mass

V RT where is the molar mass

d = g = P V RT

1818

USING GAS DENSITYUSING GAS DENSITY

The density of air at 15 oC and 1.00 atm is 1.23

g/L. What is the molar mass of air?

1. Calc. moles of air.1. Calc. moles of air.

V = 1.00 LV = 1.00 L P = 1.00 atmP = 1.00 atm T = 288 KT = 288 K

n = PV/RT = 0.0423 moln = PV/RT = 0.0423 mol

2. 2. Calc. molar mass

mass/mol = 1.23 g/0.0423 mol = 29.1 g/mol

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