chapter 5: gases - schempc demosdemo.schempc.com/ks/uploads/chem - 1303/print.pdf · chapter 5:...

Chapter 5: Gases

Laws for gases are simpler than for liquids and solids. The behavior of gases is predictable.

5.1 Substances that exist as gases

Air - 78 % N2; 21 % O2; 1 % other (including CO2 which is

associated with global warming)

A. Elements and Molecules

• O2, H2, N2, F2, O3, all Group VIII (Noble gases)

• Ionic compounds are never gases at room temperature (24 °C at

1 atmosphere); electrostatic forces are too strong to allow ions to

separate

• molecular compounds that are small can be gases or have low

boiling points: CO, CO2, HCl, CH4, NH3, SO2, etc.

B. Characteristics of Gases

• assume volume and shape of container

• very compressible

• mix evenly and completely

• lower density than liquids and solids

5.2 Pressure of a Gas

Molecules of a gas are in rapid and constant motion.

A. Units of Pressure

SI Units

force = mass x acceleration

Newton (N) = Kg x m/s2

Pressure = force/area = 1 N/ m2 = Pascal (Pa)

Chapter 5 Chemistry 1303

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B. Atmospheric Pressure

• pressure exerted by Earth's atmosphere

• gas molecules have mass, thus gravity affects them

• 50 % of the atmosphere's molecules are within 4 miles of surface

• measured by barometer (Figure 5.3)

• 760 mm Hg at sea level

• standard is 1 atm at 0 °C

• 1 atm = 1.01325 x 105 Pa

1 mm Hg = 1 Torr

1 atm = 760 Torr = 760 mm Hg


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5.3 Gas Laws

3 simple laws : Boyle's, Charles', Avogadro's

A. Boyle's Law (at constant T and n)

• volume decreases as pressure increases

• V ∝ 1/P or PV = constant

• P1V1 = P2V2

• See Figure 5.6


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B. Charles' Law (Gay-Lussac's) (at constant P and n)

• volume increases as temperature increases

• V ∝ T or V / T = constant

• V1/V2 = T1/T2

• always use the Kelvin Scale for T

• Figure 5.9

• note all lines converge at absolute 0 or -273.15 °C

• 0 K = -273.15 °C


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Gay-Lussac's Law

C. Combined Gas Law (for constant n)

• PV / T = constant

P1V1 P2V2

T1 T2=


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D. Avogadro's Law (constant temperature and pressure)

• equal volumes of gas have the same number of molecules (atoms)

• V ∝ n where n = moles

See practice problems in book.

Fig 5.10. Volume relationships of gases. Volume relationship is the same as the

mole relationship.


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5.4 Ideal Gas Equation

Ideal gas - volume of molecules is negligible; applies in most temperature

and pressure ranges.

• combination of gas laws and Avogadro's Principle

• Standard Molar Volume (STP)

• Standard Temperature (0°C or 273 K)

• Standard Pressure (1 atm)

• 1 mole of any gas occupies 22.4 L

• or 22.4 L = 1 mole but ONLY at STP

PV = nRT

where R = "universal gas constant"

= 0.0821 L⋅⋅⋅⋅atm / mole⋅⋅⋅⋅K memorize

Where does R come from?

R = PV/nT (Ideal gas law rearranged)

R = (1 atm)(22.414 L) / (1 mol)(273.15K) = 0.082057 L⋅⋅⋅⋅atm / mole⋅⋅⋅⋅K

{ Useful in many different kinds of calculations involving gases!!! }


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Example:

At STP, the density of a certain gas is 4.29 g/L. What is the molecular mass of the gas?

4.29 g

x 22.4 L

= 96.0 g / mole L mole

Example:

What is the density (g/L) of UF6 at 779 mm Hg and 62 °C?

PV = nRT and d = m/V

n/V = P/RT

i.e.,

n

V

P

RT=

and n = m / ` where ` = molar mass

m/ ` V = P/RT

i.e.,

M=

RT

P

V

m

m/V = d = P ` / RT

i.e.,

M=

RT

P

V

md =

779 mm Hg x x 352 g

mol

1 atm

760 mm Hg

335 Kx0.0821 L atm

mole K

=V

m

= 13.1 g / L

Note: alternate way will be done in class


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5.5 Gas Stoichiometry

Example:

Acetylene (welding gas), C2H2, is produced by hydrolysis of calcium carbide.

CaC2 (s) + 2 H2O → Ca(OH)2 (s) + C2H2 (g)

Starting with 50.0 g of CaC2, what is the theoretical yield of acetylene in liters, collected at 24°C and a pressure of 745 Torr?

CaC2 (s) + 2 H2O → Ca(OH)2 (s) + C2H2 (g)

1st find yield in moles:

50.0 g CaC2 x x = 0.780 mole C2H2

1 mole C2H2

1 mole CaC264.10 g CaC2

1 mole CaC22

now use ideal gas law to find volume of C2H2:

PV = nRTP

V = nRT

V =(0.780 mole) x (0.0821 L atm / mole K) x (297 K)

(745 torr) x (1 atm / 760 torr)= 19.4 L


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3 A + 2 B →→→→ 4 C + D


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5.6 Dalton's Law of Partial Pressures

A. For a mixture of gases: PT = Pa + Pb + Pc + . . . . .

PT = Pa + Pb = naRT / V + nbRT / V

PT = (na + nb) (RT / V)

Pa =

naRT / V =

na = Xa

Pt (na + nb)RT / V na + nb

Xa = mole fraction ( dimensionless quantity)

Xa = moles of one component / total moles present

Pi = Xi.PT


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Example:

A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of

ethane, 0.116 moles of C3H8. If the total pressure of the gas is 1.37

atm., what is the partial pressure of each gas?

PCH4 = XCH4

.PT

X CH4 = 8.24 moles / (8.24 + 0.421 + 0.116) moles = 0.940

X C2H6 = 0.421 moles / 8.777 moles = 0.048

X C3H8 = 0.116 moles / 8.777 moles = 0.0132

PCH4 = 0.940 x 1.37 atm = 1.29 atm

PC2H6 = 0.048 x 1.37 atm = 0.0658 atm

PC3H8 = 0.0132 x 1.37 atm = 0.0181 atm


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B. Gases are often prepared and collected over water:

Ptotal = Pgas + Pwater

where Pwater = vapor pressure of water (depends on temperature)

e.g., at 25°C, Pwater = 23.8 Torr

at 50°C, Pwater = 92.5 Torr


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Pressure of water vs temperature


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Example:

A sample of N2 gas was prepared and collected over water at 15°C. The total pressure of the gas was 745 Torr in a volume of 310 mL. Calculate the mass of N2 in grams.

Ptotal = Pgas + Pwater

745 Torr = Pgas + 12.8 Torr

Pgas = 732.2 Torr

PV = nRT

n =(0.0821 L atm / mole K) x (288 K)

(732.2 torr) x (1 atm / 760 torr) x (0.310 L)= 0.0126 mole N2

RTn =

PV

mass N2 = (0.0126 mole N2) x (28.0 g N2 / mole N2) = 0.354 g N2


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5.7 Kinetic Molecular Theory

Kinetic Theory of Gases --

Basic Postulates:

A gas consists of

• a very large number of very small particles

• separated by large distance

• in constant random motion

• undergo perfectly elastic collisions with each other and the

container walls

• with no attraction or repulsion between the particles

Temp ∝ average KE

KE = ½ m u2 = C T

where u = average velocity of all molecules, T is temperature in Kelvin

The kinetic theory "explains" the gas laws, pressure, etc. based on motion and kinetic

energy of gas molecules.

e.g., Boyle's Law (P ∝ 1 / V) -- at constant Temp (same average KE)

If volume of container is reduced, there are more gas particles per unit

volume, thus, more collisions with the container walls per unit area. ∴∴∴∴

higher pressure


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Root Mean Square

• average speed of molecule

• inversely proportional to square root of molar mass of molecule

• directly proportional to square roof of the temperature

urms = 3 R T

molar mass


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Graham's Law of Diffusion

• gradual mixing of molecules of one gas with molecules of another

• results from motion of molecule

• lighter gases move faster (urms)

• Graham's law states that under the same conditions of temperature and pressure, rates of diffusion for gases are inversely proportionhal to the sqaure roots fo their molar masses.

Gas Effusion

r1

r2

M2

M1 √ =


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5.8 Deviations from Ideal Gas Laws (Real Gases)

For real gases, small corrections can be made to account for:

• Actual volume of the gas particles themselves, and

• Intermolecular attractive forces Due to molecules getting closer together and slowing down: circumstances where intermolecular forces become important (e.g., at high pressure and low temperatures) One common approach is to use the Van der Waals' Equation:

P +

na2

V2(V - nb) = nRT

Where, a and b are empirical parameters that are dependent on the specific gas. These are experimentally determined. (see Table 5.4).

a ≈ intermolecular attractive forces b ≈ molecular size


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5.68 A sample of zinc reacts with HCl according to the equations

Zn + 2 HCl → ZnCl2 + H2

The H2 is collected over water and has a volume of 7.80 L and

pressure is 0.980 atm. Calculate the amount of Zn consumed.

Vapor pressure of H2O at 25 °C is 23.8 mm Hg.


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5.70 A sample of NH3 is completely decomposed to N2 and H2 over

heated iron wool. If the total pressure is 866 mm Hg, calculate

the partial pressure of N2 and H2.


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5.132 Air entering the lungs ends up in tiny sacs called alveoli. It is

from these alveoli that O2 diffuses into the blood. The average

radius of alveoli is about 0.0050 cm and the air inside contains

14 % O2. Assuming that the pressure in the alveoli is 1.0 atm

and the temperature is 37 °C, calculate the number of O2

molecules in one of the alveoli. (Hint: The volume of a sphere is

V = 4/3 πr3.)


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Periodic Table of the Elements

IA Periodic Table of the Elements VIIIA

(1) (18)

1 2

1 H IIA IIIA IVA VA VIA VIIA He1.0080 (2) (13) (14) (15) (16) (17) 4.0026

3 4 5 6 7 8 9 10

2 Li Be B C N O F Ne6.9410 9.0122 10.811 12.011 14.007 15.999 18.998 20.179

11 12 13 14 15 16 17 18

3 Na Mg IIIB IVB VB VIB VIIB. . . . . . . . . . VIIIB . . . . . . . . . . IB IIB Al Si P S Cl Ar22.990 24.305 (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) 26.982 28.086 30.974 32.066 35.453 39.948

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr39.098 40.078 44.956 47.880 50.942 51.996 54.938 55.847 58.933 58.690 63.546 65.380 69.723 72.610 74.922 78.960 79.904 83.800

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe85.468 87.620 88.906 91.224 92.906 95.940 98.907 101.07 102.91 106.42 107.87 112.41 114.82 118.71 121.75 127.60 126.90 131.29

55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

6 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn132.91 137.33 138.91 178.49 180.95 183.85 186.21 190.20 192.22 195.09 196.97 200.59 204.38 207.20 208.98 208.98 209.99 222.02

87 88 89 104 105 106 107 108 109 110 111 112 114

7 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Uuu Uub Uug223.02 226.03 227.03 (261) (262) (266) (264) (270) (268) (281)

58 59 60 61 62 63 64 65 66 67 68 69 70 71

Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu140.12 140.91 144.24 145.91 150.36 151.97 157.25 158.93 162.50 164.93 167.26 168.93 173.04 174.97

90 91 92 93 94 95 96 97 98 99 100 101 102 103

Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr232.04 231.04 238.03 237.05 244.06 243.06 247.07 247.07 242.06 252.08 257.10 258.10 259.10 260.11

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