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CHEMISTRY 161 Chapter 11 Gases. Classification of Matter solid liquid gas.

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Presentation on theme: "CHEMISTRY 161 Chapter 11 Gases. Classification of Matter solid liquid gas."— Presentation transcript:

1 CHEMISTRY 161 Chapter 11 Gases

2 Classification of Matter solid liquid gas

3 1. Gases substances that exist in the gaseous phase under normal atmospheric conditions T = 25 o C p = 1 atm

4 HF, HCl, HBr, HI CO, CO 2 CH 4, NH 3, H 2 S, PH 3 NO, NO 2, N 2 O SO 2

5 Jupiter (H 2, He) Io (SO 2 )

6 Helix Nebula Orion Nebula

7 2. Pressure molecules/atoms of gas are constantly in motion Ar

8 Atmospheric Pressure

9 Standard Atmospheric Pressure pressure of the atmosphere is balanced by pressure exerted by mercury 760 mm at 273 K at sea level 1 atm = 760 mm Hg = 760 torr barometer Torricelli

10 pressure = force area p = F / A [p] = Nm -2 = kg m -1 s -2 = Pa SI units

11 manometer pressure measurement

12 3. Gas Laws 3.1. pressure p versus volume V 3.2. temperature T versus volume V 3.3. volume V versus amount n p, V, T, n

13 3.1. Boyle’s Law pressure – volume relationship (temperature is constant) Boyle (1627-1691)

14 p ∞ 1/V

15 p = const/V p × V = const p 2 × V 2 = constp 1 × V 1 = const p 1 × V 1 = p 2 × V 2

16 3.2. Gay-Lussac’s Law temperature – volume relationship (pressure is constant) Gay-Lussac (1778-1850)

17 V ∞ T

18 V = const’ ×T V/T = const’ V 2 / T 2 = constV 1 / T 1 = const’ V 1 / T 1 = V 2 / T 2

19 3.3. Avogadro’s Law amount – volume relationship (pressure and temperature are constant) Avogadro (1776-1856)

20 2 H 2 (g) + O 2 (g) → 2 H 2 O(l) n ∞ V

21 n = const’’ × V n/V = const’’ n 2 / V 2 = const’’n 1 / V 1 = const’’ n 1 / V 1 = n 2 / V 2

22 SUMMARY 3.1. Boyle’s Law 3.2. Gay-Lussac’s Law 3.3. Avogadro’s Law p ∞ 1/V n ∞ V V ∞ T

23 (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

24 p × V = const × n × T p × V = R × n × T p × V = n × R × T ideal gas equation

25 p × V = n × R × T [R] = [p] × [V] / [n] / [T] Pa = N/m 2 m3m3 molK [R] = N × m / mol / K [R] = J / mol / K

26 R = 8.314 J / mol / K [R] = J / mol / K ideal gas constant

27 p × V = n × R × T 2. MOLAR VOLUME What is the volume of 1 mol of a gas at 273.15 K (0 o C) and 1 atm (101,325 Pa)? standard temperature and pressure (STP) V = 22.4 l

28 p × V = n × R × T the molar volume at standard pressure and temperature is independent on the gas type V = 22.4 l V m = 22.4 l

29 3. STOICHIOMETRY NaN 3 (s) → Na(s) + N 2 (g) How many liters of nitrogen gas are produced in the decomposition of 60.0 g sodium azide at 80 o C and 823 torr? 1.Balancing 2.Mole ratios 3.Convert grams into moles 4.Convert moles into liters

30

31 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)

32 5. DALTON’S LAW Dalton (1801) pure gases gas mixtures (atmospheres)

33 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 = p A + p B + p C + ….

34 p A × V = n A × R × Tp A = n A × R × T / V p B × V = n B × R × T p × V = n × R × T p B = n B × R × T / V p = p A + p B p = (n A + n B ) × R × T / V p × V = n × R × T

35 p A = n A × R × T / V p × V = (n A + n B ) × R × T p A / p = n A /(n A + n B ) = x A mole fraction x < 1 p A = x A × p

36 2 KClO 3 → 2 KCl + 3 O 2

37 SUMMARY p × V = n × R × T 1. ideal gas equation R = 8.314 J / mol / K V m = 22.4 l 2. molar volume

38 ς = p × M / (R × T) 3. Density of gases 4. Dalton’s Law p = Σ p i i=1 n

39 1. Kinetic Molecular Theory of Gases Maxwell (1831-1879) Boltzmann (1844-1906) macroscopic (gas cylinder) microscopic (atoms/molecules)

40 Kinetic Energy of Gases physical properties of gases can be described by motion of individual gas atoms/molecules each macroscopic and microscopic particle in motion holds an energy (kinetic energy)

41 Assumptions of the Kinetic Theory of Gases 1.gases are composed of atoms/molecules which are separated from each other by a distance l much more than their own diameter d l d = 10 -10 m l = 10 -3 m….. few m molecules are mass points with negligible volume

42 2. gases are constantly in motion in random reactions and hold a kinetic energy gases collide and transfer energy (billiard ball model)

43 3. gases atoms/molecules do not exert forces on each other (absence of intermolecular interactions) F (inter) = 0 p (inter) = 0

44 Gas Diffusion

45 2. Distribution of Molecular Speeds Maxwell-Boltzmann distribution

46 3. Real Gases p × V = n × R × T (n = 1) deviation of ideal gas law at high pressures p ≈ 90 atm

47 p << 10 -10 atm North America Nebula

48 ideal gas law p V = n R T real gas law (van der Waals equation) (p + (a n 2 / V 2 ) ) (V – n b) = n R T corrected volume (volume occupied by molecules) corrected pressure (additional pressure/force from attraction)


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