chemistry 161 chapter 5 chem.hawaii/bil301/welcome.html

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CHEMISTRY 161 Chapter 5 www.chem.hawaii.edu/Bil301/welcome.html

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CHEMISTRY 161 Chapter 5 www.chem.hawaii.edu/Bil301/welcome.html. REVISION. Boyle’s Law Gay-Lussac’s Law Avogadro’s Law. p ∞ 1/V. V ∞ T. n ∞ V. 1. IDEAL GAS EQUATION. (1) p ∞ 1/V. V ∞ 1/p. (2) V ∞ T. V ∞ T. (3) n ∞ V. V ∞ n. V ∞ T × n / p. p × V = const × n × T. - PowerPoint PPT Presentation

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Page 1: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

CHEMISTRY 161

Chapter 5

www.chem.hawaii.edu/Bil301/welcome.html

Page 2: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

REVISIONBoyle’s Law

Gay-Lussac’s Law

Avogadro’s Law

p ∞ 1/V

n ∞ V

V ∞ T

Page 3: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

(1) p ∞ 1/V

p × V = const × n × T

(2) V ∞ T

1. IDEAL GAS EQUATION

(3) n ∞ V

V ∞ 1/p

V ∞ T

V ∞ n

V ∞ T × n / p

Page 4: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

p × V = const × n × T

p × V = R × n × T

p × V = n × R × T

ieal gas equation

Page 5: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

p × V = n × R × T

[R] = [p] × [V] / [n] / [T]

Pa = N/m2 m3 mol K

[R] = N × m / mol / K

[R] = J / mol / K

Page 6: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

R = 8.314 J / mol / K

[R] = J / mol / K

ideal gas constant

Page 7: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

p × V = n × R × T

2. MOLAR VOLUME

What is the volume of 1 mol of a gas at

273.15 K (0oC) and 1 atm (101,325 Pa)?

standard temperature and pressure

(STP)

V = 22.4 l

Page 8: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

p × V = n × R × T

the molar volume at standard pressure and temperature is independent on the gas type

V = 22.4 l

Vm = 22.4 l

Page 9: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

3. STOICHIOMETRY

NaN3(s) → Na(s) + N2(g)

How many liters of nitrogen gas are produced in the decomposition of 60.0 g sodium azide at 80oC and 823 torr?

1. Balancing

2. Mole ratios

3. Convert grams into moles

4. Convert moles into liters

Page 10: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html
Page 11: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

4. DENSITY CALCULATION

p × V = n × R × T

ς = m / V

relate the moles (n) to the mass (m) via the molecular weight (M)

n = m / M m = n × M

V = n × R × T / p

ς = p × M / (R × T)

Page 12: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

5. DALTON’S LAW

Dalton

(1801)

pure gases

gas mixtures

(atmospheres)

Page 13: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

DALTON’S LAW

the total pressure of a gas mixture, p, is the sum of the

pressures of the individual gases (partial pressures) at a

constant temperature and volume

p = pA + pB + pC + ….

Page 14: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

pA × V = nA × R × T pA = nA × R × T / V

pB × V = nB × R × T

p × V = n × R × T

pB = nB × R × T / V

p = pA + pB

p = (nA + nB) × R × T / V

p × V = n × R × T

Page 15: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

pA = nA × R × T / V

p × V = (nA + nB) × R × T

pA / p = nA /(nA + nB) = xA

mole fraction

x < 1

pA = xA × p

Page 16: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

A gas mixture contains 4.46 moles of neon, 0.74 moles of Ar, and 2.15 moles of xenon. Calculate

the partial pressure of each gas if the total pressure is 2.0 atm.

Page 17: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

2 KClO3 → 2 KCl + 3 O2

Page 18: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

SUMMARY

p × V = n × R × T

1. ideal gas equation

R = 8.314 J / mol / K

Vm = 22.4 l

2. molar volume

Page 19: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

ς = p × M / (R × T)

3. Density of gases

4. Dalton’s Law

p = Σ pii=1

n

Page 20: CHEMISTRY 161 Chapter 5 chem.hawaii/Bil301/welcome.html

Homework

Chapter 5, p. 166-179problems