PV = Nrt 13 © Prof

נושא 6 גזים

1 © Prof. Zvi C. Koren 19.07.10 Charles Avogadro Graham Torricelli Dalton

Boyle

Gay-Lussac Kelvin Maxwell Boltzmann 2 © Prof. Zvi C. Koren 19.07.10 Gas Laws: A Practical Application - Air Bags

Example: An automobile air bag is rapidly inflated by the explosion of

sodium azide (NaN3), which releases nitrogen gas when detonated:

2NaN3(s)  2Na(s) + 3N2(g) • How much sodium azide is required to inflate an air bag with a volume of 50.0 L? • Does the amount of sodium azide required depend on the temperature of the gas produced? • Does altitude affect an air bag's performance? You will be able to answer these questions at the end of this chapter.

3 © Prof. Zvi C. Koren 19.07.10 Air-Bag Chemistry The signal from the deceleration sensor ignites the gas-generator mixture by an electrical

impulse, creating the high-temperature condition necessary for NaN3 to decompose. Air-Bag Chemistry Reactants (Substances in red Gas-Generator Reaction Products are initially in the air-bag) Initial Reaction Triggered by Na(s) extremely reactive NaN3(s) Sensor. N2(g) inert K O(s) reactive Na 2 Second Reaction. Na2O(s) reactive KNO3(s) N2 inert

K2O Alkaline Final Reaction. Na2O silicate inert SiO2(s) (glass)

Write balanced chemical reactions for each of these steps.

4 © Prof. Zvi C. Koren 19.07.10 Properties of Gases: Gas Pressures, P The Mercury Barometer: Measuring Atmospheric Pressure Evangelista Torricelli (1608 – 1647, Italy): “We live submerged at the bottom of an ocean of air.”

Aristotle (4th century BCE): “Nature abhors a vacuum.”

5 © Prof. Zvi C. Koren 19.07.10 The Mercury Barometer: Measuring Atmospheric Pressure Evangelista Torricelli

vacuum The atmospheric pressure meniscus (or barometric) pressure can be described in terms of “mm Hg”

1 standard atmosphere is supported by a mercury column that is 760 mm Hg in height:

Patm PHg 1 atm  760 mm Hg

1 torr  1 mm Hg 760 mm Hg 600 mm Hg

6 © Prof. Zvi C. Koren 19.07.10 Pressure Units P = pressure SI (and other) Units f = force Energy: 1 J = 1 kg·m2/s2 A = area Force: 1 N = 1 kg·m/s2 (“kms”) w = weight Pressure: 1 Pa = 1 N/m2 = 1 kg/m·s2 m = mass 1 bar = 105 Pa g = acceleration due to gravity = 9.80665 m/s2 1 torr = 1 mm Hg h = height of column d = density = m/V J = Joule, N = Newton, Pa = Pascal f w mg h mgh P       dgh P = d·g·h A A A h V For the mercury barometer (at 0 oC): 3 3 2 3 4 3 dHg = 13.5951 g/cm = 0.0135951 kg/cm x (10 cm / 1 m) = 1.35951 x 10 kg/m h = 760 mm Hg = 76 cm Hg = 0.76 m Hg P = d · g · h 1 atm = (1.35951x104 kg/m3)·(9.80665 m/s2)·(0.76 m) = 1.01325x105 kg/m·s2 = 1.01325 x 105 Pa = 101.325 kPa = 1.01325 bar = 14.6960 lb/in2 (“PSI”, Pounds per Square Inch)

7  760 (exactly) mm Hg © Prof. Zvi C. Koren 19.07.10 The Mercury Manometer: Measuring Gas Pressure CLOSED OPEN Manometer Manometer

vacuum Patm

P P H Pgas PH gas g g

gas gas Pgas = PHg Pgas = Patm + PHg

8 © Prof. Zvi C. Koren 19.07.10 Gas Laws

• Changes in gas properties have been mathematically modeled for many years. • These mathematical models are collectively called gas laws. • They define the relationships between pairs of gas properties and experiments confirm these gas laws.

Chemists use four basic measurements when working with gases:

1. The quantity of the gas, n (in moles) 2. The temperature of the gas, T (in kelvins) 3. The volume of gas, V (in liters) 4. The pressure of the gas, P (in atmospheres)

9 © Prof. Zvi C. Koren 19.07.10 Boyle’s Law

Robert Boyle (1627 – 1691) England

P  V-1, [n,T]

P • V = k or P = k •V–1 , [n,T]

20 hyperbola P1V1 = P2V2 , [n,T] 18 16

piston 12 P 10

6 gas 4 2

0 0 5 10 15 20 25 30 35

10 V © Prof. Zvi C. Koren 19.07.10 Charles’s and Gay-Lussac’s Law Jacques Alexandre César Joseph Louis Charles Gay-Lussac (1746-1823) (1778 – 1850)

V  T, [n,P]

V V = k T, [n,P]  k , [n,P] T V V 1  2 , [n,P] V T1 T2 extrapolation “Absolute Zero” William Thomson T “Lord Kelvin” (1824 -1907), Irish -273 oC The title “Baron Kelvin” was given in honor of his achievements, and named after the River Kelvin, which flowed past his university in Glasgow, Scotland. 0 11 Absolute temperature scale is in kelvins: T(K) = t( C) + 273.15 © Prof. Zvi C. Koren 19.07.10 Avogadro’s Law Lorenzo Romano Amedeo Carlo Avogadro di Quareqa e di Carreto (1776 – 1854)

V  n, [P,T]

V V = k n, [P,T]  k , [P,T] n

V V 1  2 , [P,T] n1 n 2

n 12 © Prof. Zvi C. Koren 19.07.10 Ideal Gas Law Combining all three gas laws: Boyle: P V = k, [n,T]

+ Charles: P V T = k, [n]

+ Avogadro: P V  “Gas Constant” n T = k  R At STP (Standard Temperature and Pressure), 0oC (= 273.15 K) & 1 atm: 1 mole of an ideal gas occupies 22.414 L of volume.

PV (1 atm)(22.414 L) R    0.082057 Latm/ mol  K n T (1 mol)(273 .15 K) = 8.31 J/mol·K = 1.99 cal/mol·K = [energy/mol·K] PV = nRT 13 © Prof. Zvi C. Koren 19.07.10 mole

Mole Day is celebrated on Oct. 23rd from 6:02 am to 6:02 pm

14 Happy Mole Day to You!!! © Prof. Zvi C. Koren 19.07.10 Summary: Moles in Chemistry

# of items: Avogadro’s #

Mass: m/MW mole Gas: PV = nRT

Solution: MV

15 © Prof. Zvi C. Koren 19.07.10 Gas Laws and Chemical Reactions: Stoichiometry

Problem: Potassium superoxide is used to purify air in a spacecraft. o What mass of KO2 is required to recycle 500.0 mL of CO2 at 25.00 C and 750. mmHg?

4KO2(s) + 2CO2(g)  2K2CO3(s) + 3O2(g)

Solution: Stoichiometric Flow-Chart: mass KO2  moles KO2  moles CO2  (V, T, P) of CO2 ? MW rxn PV=nRT R = 0.082057 L·atm/mol·K  1 atm  P = 750. mmHg   = 0.98684 atm PV  760 mm Hg  n  n CO = 0.020168 CO 2  1 L  2 RT V = 500.0 mL   = 0.5000 L 1000 mL  T = 25.00 + 273.15 = 298.15 K  4 mol KO   71.0971 g KO   2   2  0.020168 mol CO2     = 2.87 g KO2  2 mol CO2   1 mol KO 2  16 © Prof. Zvi C. Koren 19.07.10 Gas Mixtures and Dalton’s Law of Partial Pressures John Dalton (1766-1844), England

1 3 2 1 2 1 2 2 2 3 2 1 2 1 3

Pi = partial pressure of gas i = niRT/V

Ptotal = ΣPi = P1 + P2 + P3 + … = n1RT/V + n2RT/V + n3RT/V + … = (n + n + n + …) RT/V 1 2 3 n RT = n RT/V P  total total total V n Pi niRT/V i    Xi  mole fraction of gas i  ΣX = 1 n i Ptotal n totalRT/V total

Pi = Xi•Ptotal

17 © Prof. Zvi C. Koren 19.07.10 Gas Density

From: PV = nRT m PV  n  MW RT

m P   MW V RT

P d   MW d  MW, [P,T] RT

For example: 1 atm d @ STP  (32.00 g / mol) O2 (0.0821 Latm/mol  K)(273 K)

Dumas Method

vent Measurements: Mass of vapor = 1.2189 g (measurements of empty and filled flask) Volume of flask = 510.0 mL (measured with water) Temperature of vapor = 100.0 oC Atmospheric pressure = 745 torr

MW = ?

19 © Prof. Zvi C. Koren 19.07.10 20 © Prof. Zvi C. Koren 19.07.10 Kinetic Molecular Theory (KMT) of Gases Tenets of KMT: 1. Gases consist of molecules whose separation is much larger than the volume of the molecules themselves. 2. The gas molecules are in continuous random and rapid motion. 3. The average kinetic energy of the gas molecules is determined by the absolute gas temperature. Notes: All gas molecules at the same T, regardless of mass, have the same average kinetic energy. Temperature is a bulk (macroscopic) property, and not one based on an individual (microscopic) molecule) 4. Gas molecules collide with each other and with the walls of the container, but they do so without loss of energy. These collisions are perfectly elastic.

Ideal Gas Properties: • No intermolecular forces • Molecular volume is negligible

2 Kinetic Energy for one molecule  KEmolecule = ½mv , m  mass of one molecule 2 Kinetic Energy for one mole  KEmole = ½Mv , M  MW, Molecular weight

2 Average Kinetic Energy for one mole  KEmole = ½Mv ,

But also: Combine these two

Average Kinetic Energy for one mole  KEmole  T = 3(½ RT) proof is not for this course

2 3(½ RT) = KEmole = ½Mv

____ 1/2 2 1/2  3RT  R = (J/mol·K) Graham’s Law ( v )    M MW = (kg/mol) of Diffusion   for 2 Gases vrms = root-mean-square velocity (m/s) at same T: ½ v1/v2 = (M2/M1) Thomas Graham (1805 – 1869), Scotland

1. Diffusion of NH3(g) vs. HCl(g)

James Clerk Maxwell Ludwig Eduard Boltzmann (1831 – 1879), Scotland (1844 – 1906), Austria

Why do molecules, all at the same T, have such a wide span of velocities?

MW Effects:

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