Properties of Gases

Chapter 6 – Properties of gases lecture notes

1. Gas phase Gases have neither definite shape or volume 1) volume changes with pressure 2) volume changes with temperature 3) gases are miscible 4) gases are generally MUCH less dense than liquids

2. Atmospheric pressure 1 atm = 760 torr (mm of Hg) F= ma F = force m = mass a = acceleration P = F/A - ma/A P = pressure A = area

1 Pa (Pascal) = 1kg/m-s2 : 1 kg per meter-second squared

1 atm = 101,325 Pa : 1 atm = 101.325 kPa 1 atm = 1013.25 mbar 1 kPa = 10 mbar

Sample exercise:
The mass of the earth’s atmosphere is estimated to be 5.3 x 1018 kg
Surface are is 5.1 x 1014 meters2
Acceration due to gravity is 9.8 m/s2 .
Calculate the average pressure in kPa ( supplemental - convert to atm) F = ma P = F/A P = ma/A = (5.3 x 1018 kg ) (9.8 m/s2)/ (5.1 x 1014 meters2) P = 1.0 x 105 Pa = 1.0 x 102 kPa = 100 kPa 100kPa /101.325 kPa/atm = 0.99 atm

1 atm = 101.325 kPa : 1 atm = 14.7 lbs/in2 : 1 atm = 29.92 in of Hg

Practice exercise (p242) Calculate the pressure exerted on a tabletop by a cube of iron that is 1.00 cm on a side and weighs 7.87 gm

P = F/A = (m)(a)/(A) = (7.87 x 10-3 kg) (9.8 m/s2)/ (1 x 10-2 m)2 P = (77 x 10-3)/ (1 x 10-4 m2) = (770 x 101 kg/ m-s2) = 0.00076 atm

Practice exercise (p 244) Atmospheric pressure in the eye of Katina was 90.2 kPa calculate in atm, mmHg and torr

Manometer – “U” shaped tube --- use Power Point diagram

Sample exercise (p245) With a closed manometer, the difference in the heights is a direct measure of the gas pressure.

Practice exercise (p245)
References sample exercise calcium carbonate decomposed to carbon dioxide
(h = 143.7 mm into vacuum…..exterior pressure = 760 mm Hg
Where would the mecury column be at the end of this experiment and what is (h
In an open manometer the difference in height represents the difference between the pressure in the flask and the atmospheric pressure 760-144 = 616

3. The Gas Laws

Boyle’s law P ( 1/V PV = constant where T and n are fixed)

P1V1 = P2V2

Practice exercise (p. 249) A diver exhales 3.5 L of air 20M below the surface of the water where the sum of water and air pressure is 3 atm. What is the size of the air bubble at the surface.

(3 atm)(3.5 L) = (1 atm)( V2 ) 10.5 L = V2

Charle’s Law

V ( T where P and n are fixed V = constant x T (T MUST be in Kelvin) V/T = constant

V1 / T1 = V2 / T2

Practice exercise (p 251) If the temp of a gas in a piston increases from 245 to 560 C, what is the ratio of the initial volume to the final volume

V1 / (245+ 273) = V2 / (560+273)

V1 / V2 = (245 + 273)/ (560 + 273) = 518/ 833 = 0.62

Avogadro’s Law

V ( n where P and T are fixed

Amonton’s Law

P ( T or P/T = constant when n and V are fixed

Practice exercise (p 254)

Tire pressures are at 28 psi (T = 68 F) T ( 140 F what is resulting pressure

4. Ideal Gas Law

PV = nRT R is universal gas constant R = PV/nT R = atm –Liters /(moles – oK)

R = 0.0826 (L-atm)/(mole – oK) R = 8.314 (kg-m2)/ (s2 –mole – oK) R = 8.314 (J )/ (mole – oK) R = 8.314 (m3 – Pa)/ (mole - oK) R = 62.37 (L – torr) / (mole - oK)

Practice exercise (p257)

A balloon with 100.0 L of He T = 20 oC P = 755 torr Calculate the volume at 28.0 torr and –45 oC

P1V1 / T1 = P2V2 / T2

(755) (100.0)/ (20 + 273) = (28.0) (V2 )/( -45 + 273)

Practice exercise (p268)

Using information from sample exercise calculate the volume of oxygen delivered at – 38 oC and 0.35 atm. P1V1 / T1 = P2V2 / T2 (2025 psi) ( I atm/14.7 psi)(5.90L) / (25 + 273) = (0.35 atm) (V2) / ( -38 + 273)

sample exercise – combining stoiciometry with ideal gas law NaClO3 (s) + Fe(s) ( O2 (g) + NaCl(s) + FeO(s)

How many grams of are needed to procude 125 liter of oxygen at 1.00 atm and 20.0

STP 0 oC (273 oK ) -- 1 atm pressure -- 1 mole of gas 22.4 L

5. Gas Density

At STP, gas density = molar mass( g / L)/ 22.4 L) = g/L

concept test (p262) which of the following gases has the highest density at STP CH4, Cl 2 Kr, C3H8

PV = n RT n/V = P/RT n = g / M.W.

g/V = density (d) = (M.W.)(P) / (RT)

Sample exercise (p264)

Density = 0.65 g/L at 25 C and 757 mm Hg What is the molar mass?

Practice exercise (p265)

HCl + NaHCO3 ( gas??? Density of gas = 1.81 g/L at 1.00 atm and 23 C What is the molar mass? What is the gas??

1.81 g/l = (M.W.) (1.00 atm) / (0.08207 L-atm/Mole – K)( 296K) 1.81 (0.08207)(296) = mw = 44 gm /mole CO2

6. Dalton’s Law and Mixtures of Gases

Ptotal = P1 + P2 + P3 + P4 + …..

Mole fraction = Xx = nx / n total

1) unlike molarity, mole fractions have no units 2) ulike molarity, mole fractions are MASS based 3) mole fractions of all components must add = 1.000

Partial pressure is directly related to mole fraction

Sample exercise (p266)

Scuba divers trimix is 11.7 % He, 65.2 % N2 and 32.1 % O2 by weight Calculate mole fractions for each 11.7/ 4 = 2.92 65.2/28 = 2.01 32.1/32 = 1.00

2.92 / 5.93 = .492 2.01/ 5.93 = .339 1.00/5.93 = .169

laboratory experiment using partial pressures to determine the quantity of oxygen generated by reactions,

Sample exercise (p269) KClO3 (s) ( O2 (g) + ??????
92 ml of gas collected T = 25 C P = 756 torr
What is mass of oxygen collected

PO2 = Ptotal – P H2O = 756 – 24 = 732 torr

(732/760) ( 92ml/ 1000 ml/L) = (x g / 32 g/mole) (0.08206 L-atm/mole K) (25 + 273)

solving for x = 0.116 g oxygen
Practice exercise (p270)
27 mol of h2 are collected over water at 25C P = 761 torr How many g of H2? P H2O = 24

7. Kinetic Molecular Theory of Gases and Graham’s Law Basic assumptions 1) gas molecules have tiny volumes compared to overall volume – large intermolecular distances 2) gas molecules are in constant, random motion 3) the motion is related to the average kinetic energy which is proportional to the absolute temperature of the gas. All populations of gas molecules at the same temp have the same kinetic energy 4) gas molecules continuously collide with each other and container walls. Collisions are elastic – there is no transfer of kinetic energy to the walls, therefore the average kinetic energy does not change as long as there is no change in temperature 5) each molecule acts independently of all others in the sample. No attraction repulsion except at very short distances.

Molecular speeds and Kinetic Energy

K.E. = 1/ 2 m v2

and

vRMS = (3RT/M)1/2

Sample exercise (p274)
Calculate the RMS speed of nitrogen at 300.0K Express the anser in (a) meters per second (b) MPH

V = [ (3)( 8.314 kg-m2/ s2 –mole – oK)(300.0 K))/ (0.02801 kg/mole)] 1/2

V = 516.9 m/s = 1156 MPH

Practice exercise (p275) Calculate the RMS speed of He at 300.0K Compare to nitrogen

Concept test (P275) Compare RMS speed for H2, CO2, Ar, SF6, UF6 and Kr.

Review P275 comparison of kinetic energy and atomic mass - use example of bullets

9mm weight is 0.3 oz (approx 8 gm) speed approx 300 m/s
K.E. = ½ (.008 kg)( 300 m/s) 2
K.E. = 0.004 x 90000 = 360 kg-m2/sec2

Figure weight of object at 100 MPH ( roughly 30 m/sec) 360 = (0.5) (m) (30 m/sec)(30 m/sec) = 720 = m(900 m2/sec2) m = 7.2 kg (approx 16 lbs ) at 100 MPH baseball weighs 5.25 oz = 142 g

Graham’s Law of Effusion and Diffusion

Effusion – leaking out through small orfice Related to molar mass of gas

Rate of effusion is inversely proportional to molar mass

rx / ry = (My / Mx) ½

diffusion – spread of one substance into another

Sample exercise 6.16 (p277) A gas diffuse 2.92 times faster than He What is the molar mass of the gas

RHe / ry = (My / MHe) ½ 2.92 = (My / 4.003 g/mole) ½ 8.52 = My / 4.003 g/mole My = 34.4 g/mole

Practice exercise (p278) Helium diffuses 3.16 times faster than what noble gas RHe / ry = (My / MHe) ½ 3.16 = (My / 4.003 g/mole) ½ 9.99 = My / 4.003 g/mole My = 40.0 gm/mole Ar

8. Real Gases

Deviations from ideality

What are ideal assumptions?

Basic assumptions 1) gas molecules have tiny volumes compared to overall volume – large intermolecular distances 2) gas molecules are in constant, random motion 3) the motion is related to the average kinetic energy which is proportional to the absolute temperature of the gas. All populations of gas molecules at the same temp have the same kinetic energy 4) gas molecules continuously collide with each other and container walls. Collisions are elastic – there is no transfer of kinetic energy to the walls, therefore the average kinetic energy does not change as long as there is no change in temperature 5) each molecule acts independently of all others in the sample. No attraction repulsion except at very short distances.

Where do these fall apart a) high pressures putting molecules close together and volume of molecules approaches volume of container and attractive forces manifest themselves b) low temps when speeds are low and interactions are not elastic attraction/repulsion effects appear

van der Waals equation ( P + n2a/V2) (V- nb) = nRT

practice exercise (p281)

1. mole of He in 1.00 L container at 300.0K calculate P using van der Waal’s equation and compare to results for nitrogen which yielded a P of 24.2 atm rather than 24.6

P = nRT /( V – nb) – (n2a/V2) for He a = 0.0341 L2-Atm/mole2 b = 0.02370 L/mole

P = 1.0 mole(0.08206 L-atm/mole K)(300.0K) /( 1.00 L – 1.0 mole x 0.02370 L/mole) - (1.0 mole)2 (0.0341 L2-Atm/mole2 ) / (1.00 L)2

P = (24.618)/( 0.9763) – 0.0341 P = (25.286) – 0.0341 = 25.3 atm ideal = 24.6

References: P1V1 / T1 = P2V2 / T2 (2025 psi) ( I atm/14.7 psi)(5.90L) / (25 + 273) = (0.35 atm) (V2) / ( -38 + 273)