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Chapter 10:

Gases

Overview

- Pressure
- Barometer & Atmospheric Pressure
- Standard Conditions
- Gas Laws
- Boyle’s Law
- Charles’ Law
- Avogadro’s Law
- Ideal Gas Law

Gas Laws under Two Conditions

- Gas Densities
- Darlton’s Law of Partial Pressure
- Kinetic Molecular Theory
- Molecular Effusion/Diffusion
- Graham’s Law
- Deviation from Ideality

Characteristics

- Solids
- have own shape and volume
- particles close together with strong interaction
- Liquids
- have own volume but assume shape of container
- particles farther apart but have moderate interaction
- Gases
- assume shape and volume of container
- particles far apart with little/no interaction
- highly compressible

Pressure

- P = F/A
- Force in Newtons
- Area in m2
- Barometer
- P in N/m2 = Pascal unit
- 1 x 105 N/m2 = 1 x 105 Pa or 100 kPa
- Standard Pressure
- 1 atm = 760 mm Hg = 1.01325 x 105 Pa = 101.325 kPa (or torr)

force of the atmosphere

force of the column

h

when atmospheric force equals the force of the column the atmospheric pressure is measured as “h”

Gas Laws

- Boyle’s Law
- P µ 1/V constant T, n
- volume increases as pressure decreases
- Charles’ Law
- V µ T constant P, n
- volume increases as temperature increases
- Avogadro’s Law
- V µ n constant P, T
- volume increases as moles of gas (n) increases

Ideal Gas Law

- combines all gas laws PV = nRT
- R = 0.0821 L-atm mol-K
- any volumes must be in liters
- any temperatures must be in kelvin
- any pressures must be in atmospheres
- STP or SC -- standard temperature/pressure
- P = 1 atm (same as 760 mm Hg)
- T = 273 K (same as 0° C)

Problem 10.3: A flashbulb contains 2.4 x 10 -4 mol of O2 gas at 1.9 atm and 19°C . What is the volulme?

- PV = nRT or V = nRT P
- V = 2.4x10 -4 mol x 0.0821 L-atm x 292 K mol-K 1.9 atm

V = 3.0 x 10 -3 L or 3.0 mL or 3.0 cm3

Gas Laws Under Two Conditions

- P1V1 = P2V2 T1 T2
- Problem 10.4: Pressure in a tank is kept at 2.20 atm. When the temp. is -15°C the volume is 28,500 ft3. What is the volume is the temp. is 31°C
- P1 = P2 = 2.20 atm T1 = 258 K T2 = 304 K V1 = 28,500 ft3
- V2 = P1 V1 T2 P2 T1
- V2 = 28,500 ft3 x 304 K = 258 K

33,600 ft3

Gas Densities

- n = P from PV = nRT V RT
- n = moles x g/mol = g = d = PMMV L L RT
- d = PMM RT

(atm)g mol L atm ( K)mol K

Dalton’s Law of Partial Pressures

- total pressure of a mixture = sum of each partial pressure
- PT = P1 + P2 + P3 . . . .
- each partial pressure = the pressure each gas would have if it were alone
- P1 = n1RT P2 = n2RT P3 = n3RT V1 V2 V3
- PT = n1RT+ n2RT + n3RT = (n1 + n2 + n3) RT V1 V2 V3 V

volumes are the same

P1 = n1 therefore P1 = n1 PTPT nT nTn1 = X1 mole fractionnTP1 = X1 PT

Kinetic Molecular Theory

- Gases consist of particles in constant, random motion
- Volume of gas particles is negligible
- Attractive and repulsive forces are negligible
- Average kinetic energy is proportional to temperature
- Collisions are elastic

molecular speed

- u = root mean square speed or speed of molecule with average kinetic energy
- R is the gas constant (8.314 J/mol-K), T is temp. in K & MM is molar mass
- What is the rms speed of an He atom at 25°C?
- u = (3 x 8.314 kg-m2/s2-mol-K x 298 K)1/2 ( 4.00 x 10 -3 kg/mol )
- u =

1.36 x 103 m/s

Effusion/Diffusion

- small molecules will effuse/diffuse faster than large molecules
- effusiondiffusion

Graham’s Law

- where r is rate of speed & MM is the molar mass
- Problem 10.14: Calculate the ratio of the effusion rates of N2 and O2.

rN2 = 1.07 rO2

Deviation from Ideality

- Occurs at very high pressure or very low temperature
- Correction due to volume
- ideal law assumes molecules have no volume
- for molecules which are far apart, this is a good assumption
- must correct for the volume of the molecules themselves

Correction due to attraction of molecules

- ideal law assumes the molecules have no attraction to each other
- for molecules which are far apart, this is a good assumption
- must correct for actual attraction of molecules

correction for molecular volume

correction for molecular attraction

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