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6 Gases

A gas expand to occupy the entire volume it is placed in. Molecules in a gas translate freely between collisions, and they all behave alike regardless of their type.

What are some of the properties of gases?

Pressure, temperature, heat capacity, volume, density, molar volume, color, average speed of molecules, solubility (in water or other liquid), absorption, compressibility, gas-liquid equilibrium, composition, identity (compound or element), chemical properties, combustibility, stability

Which of these properties are intensive properties, and which are extensive properties?

6 Gases

Announcement

Appointments for Winter '04 enrolment are now posted on your QUEST account – Each student has only a three day appointment time. If missed, you will have to wait until all students have completed their appointments. Open enrolment begins November 3rd, but courses could be full by then

CHECK YOUR QUEST ACCOUNT FOR YOUR APPOINTMENT as soon as possible.

6 Gases

Pressure

Pressure: force per area (1 Pascal = 1 N m–2)

Liquid pressure (explain these formulation in terms of physics)

F W g * m g * V * d g * h * A * d P = --- = ---- = ---------- = --------------- = ----------------- = g * h * dA A A A A

These equivalences are useful for unit conversions:

1 atm = 101.325 kPa = 76 cm Hg = 760 mm Hg (torr in honor of Torricelli) = 1.01325 b = 1013.25 mb (bar & m bar) = 14.6960 pounds / sq. inch

6 Gases

Torricelli’s Barometer

Barometric pressure

Explain Torricelli’s work(1608 - 1647)

P = g h d

1 atm

= 0.76 m Hg

13594 kg1 m3 Hg

9.80665 N1 kg

= 101317 N m–2 (Pascal)

= 101.32 k Pa

6 Gases

Torricelli Mercury Barometer

Evangelista Torricelli invented the Torricelli Mercury Barometer in 1644. He used a long glass tube, closed at the upper end, open at the lower and filled with mercury.

6 Gases

Pump Water from a Well

1

0.

3

3m

Water

The specific gravity of mercury is 13.5939. If water is used for a barometer, what is the height of water corresponding to 1.00 atm?

Solution:

76 cm Hg

13.5939 g1 cm3 Hg

1 cm3 H2O 1 g

= 1033.14 cm H2O

= 10.33 m H2O

Explain water pump and depth of well

Water water everywhere!What about a diver under water? Be sure to get that during lecture.

6 Gases

Pressure of Skater

A skater weighing 70 kg stands on one foot and the contact between the blade and ice is 1 cm2.What is the pressure in torr sustained by the ice?

Solution:

70 kg1 cm2

1 m3 Hg 13594 kg

1000 mm1 m

10000 cm21 m2

= 51493 mm Hg

= 51493 torr

= 67.8 atm

6 Gases

Avogadro’s Law

The ABCD of gas laws are Avogadro’s, Boyle’s, Charle’s & Dalton’s laws of gases.

Avogadro’s hypothesis: proposed in 1811at the same temperature T and pressure P, equal volumes contain equal amounts of gases in moles n.

at the same temperature T and pressure P, the volume V of a gas is proportional to the number of molecules or number of moles n.

VP,T = k n (k, a constant, 22.4 L at STP)

Explain Avogadro’s hypothesis and implications

Avogadro’s scientific contributions to science will be given.

6 Gases

Explain Avogadro’s law in your language

Boyle’s Law

Robert Boyle, (1621-91)

Mathematical aspects of PV product will be discussed

For a certain amount (constant n) of gas at constant temperature T, the volume V times the pressure P is a constant.

P V n, T = constant

= P1 V1 = P2 V2

How do you graph the Boyles law?

State the law in another way.

What curves are P-V plots?

P

T2n1

• P1 V1

T1n1

• P2 V2

V

6 Gases

Charle’s Law(law of Charles-Gay-Lussac)

For a certain amount of gas at constant pressure, its volume, V, is directly proportional to its temperature T in Kelvin.

Vn, P = b T (b is a constant)or Pn, V = b T (b is a constant)

State the Charle’s law in another wayV1V2 ----- = ----- = b, V1T2 = V2 T1T1T2

Vn, Por Pn, V

T

Charles, Jacques-Alexandre-César (1746-1823, top) first to ascend in a H2-balloon, developed this law in 1787. Later Joseph Louis Gay-Lussac (lower) published a paper citing Charles’s law.

6 Gases

General Gas Equation

Combining ABC gas laws, we have

P1 V1T1

P2 V2T2

=

= n R

Subscripts 1 and 2 refer to different conditions for the same quantities of gases (n).

Experiments show that one mole of gas at STP occupies 22.4 L.

6 Gases

The Ideal Gas Equation

The ABC laws of gases can be combined into one and the result is an ideal gas equation. A+B+C ideal

P V = n R T

1 atm * 22.4140 L R = ---------------------------- = 0.082057 L atm mol–1 K–1 1 mol 273.15 K

101.325 kPa * 22.414 L R = ----------------------------------- = 8.3145 L kPa mol–1 K–1 1 mol 273.15 K

Please confirm that 1 kPa L = 1 J (1 L = 1e-3 m3)

6 Gases

Dalton’s law

The Partial pressurePi is the pressure of a component in a mixture as if others don’t exist in that system – due to the fact all gases behave as if they are independent of each other.

Pi = ni

Dalton’s law:

The total pressure Ptotal, of a mixture of gases is the sum of the partial pressures of the components.

Ptotal = P1 + P2 + … + Pn

= (n1 + n2 + … + nn) (Vis common to all)

R TV

correction

R TV

6 Gases

Application of the Ideal gas Law

Parameters of the ideal gas law: P, V, T, n and a constant R

idealP V = n R T Ideal gas law A+B+C

At constant n and T, P1V1 = P2V2 Boyles law

At constant n and VP= (n R / V) TP1 / T1 = P2 / T2 Charles law

At constant n and PV = (n R / P) TV1 / T1 = V2 / T2ditto

At constant P and TV = (R T / P) nV = k n Avogardro’s law

6 Gases

Gas Densities

Evaluate the density of O2 (molar mass M = 32.0) at 300 K and 2.34 atm.

Hint: Density d of a gas with mass m (= n M) and volume V

m n M n dd = ----- = ------ ---- = ------V V V M

Thus n R T d R TP = ------------- = -----------V M

d = P M / R T

Find relationship between density d and M.

Manipulate symbols to get a useful formula before you calculate the quantities.

Include and work out the units plsed=2.34 *32.0 / (0.08205*300) = ___

6 Gases

Reactions Involving Gases

How much NaN3 is required to produce 12.0 L N2 gas at 302 K and 1.23 atm for the air bag in your designed Autotie?

Solution:Equations: 2 NaN3 = 2 Na + 3 N2; n = V (P / R T )

N2 1.23 atm mol K 0.08205 L atm * 302 K

(23+3*14) g NaN31 mol NaN3

2 mol NaN33 mol N2

12.0 L

= ___

Work out N2 volume for 51 g NaN3 used under the same condition.

6 Gases

- reaction involving gases

NO is made from oxidizing NH3 at 1123 K with platinum as a catalyst. How many liter of O2 at 300 K and 1 atm is required for each liter of NO measured also at 300 K and 1 atm?

Solution:

The reaction is: 4 NH3 + 5 O2 4 NO + 6 H2OSince n = ( P/RT ) V molar relationships are the same as volumetric relationship, providing T and P are the same.

1 L NO

5 mol O24 mol NO

Complicated problem may have a simple solution,

= 1.25 L O2

Further oxidation of NO leads to NO2, which is used to make HNO3, a valuable commodity.

6 Gases

How much H2O is produced?

A Mixture of Gases

What is the pressure exerted by the gas when 1.0 g of H2, 2.0 g of O2, and 0.1 g of CO2 are all confined in a 10.0 L cylinder at 321 K?

Solution:

Dalton’s law : ntotal = ini (count molecules non-discriminately)

P = n R T / V;

1 mol2 g H2

1 mol32 g O2

1 mol44 g CO2

n =

= 0.585 mol

1.0 g H2 + 2.0 g O2 + 0.1 g CO2

8.314J * 321 K 10 L mol K

P = 0.585 mol

= 126 kPa (note 1 J = 1 kPa L)

Calculate partial pressures of each gas plse

6 Gases

Vapor pressure

The (saturated) vapor pressure is the partial pressure that is at equilibrium with another phase.

Vapor pressure of ice

Vapor pressure of water

ExplainStructure of water moleculeHydrogen bondingStructure of waterStructure of iceVapor pressure of ice and waterRelative and absolute humidity

6 Gases

Collecting Gas Over Water

When 1.234 g of a sample containing Ag2O is heated, 40.6 mL of O2 is collected over water at 296 K, and the atmosphere is 751 mmHg. Vapor pressure of water at 296 K is 21.1 mmHg. What is the percentage of Ag2O in the sample?

Solution: Ag2O = 2 Ag + 0.5 O2n = P V / R T; R = 0.08205 L atm mol-1 K-1 P = (751 – 21.1) mmHg / 760 mmHg = 0.961 atm (PO2 = Ptotal – Pwater) V = 40.6 mL = 0.0406 L

Mass of Ag2O =

231.7 g Ag2O1 mol Ag2O

0.961 atm*0.0406 L O20.08205 L atm mol-1 K-1*296 K

1 mol Ag2O0.5 mol O2

= 0.744 g Ag2O

Percentage of Ag2O = 0.744 g Ag2O / 1.234 g = 0.603 Ag2O = 60.3 % Ag2O

6 Gases

Assumptions of Kinetic-molecular Theory

1. Gas is composed of tiny, discrete particles (molecules or atoms).

2. Particles are small and far apart in comparison to their own size.

3. Ideal gas particles are dimensionless points occupying zero volume.

4. Particles are in rapid, random, constant straight line motion.

5. There is no attractive force between gas molecules and between molecules and the sides of the container.

6. Molecules collide with one another and the sides of the container.

7. Energy is conserved but transferred in these collisions.

8. Energy is distributed among the molecules in a particular fashion known as the Maxwell-Boltzmann Distribution.

6 Gases

Kinetic-molecular Theory of Gases

For N gas molecules, molecule mass = m, molecular mass = Mspeed = u, average speed = u , Avogadro’s number Nvolume = V, temperature = T, Pressure = P,

Kinetic energy = ½ m u 2

Collision frequency u N / V

Pressure (m v) (u) (N / V) (N / V) m u2= 1/3 (N / V) m u2(1/3 due to 3-Dimensional space)

P V = 1/3N m u2 = n R T (Meaning of T)

Thus, 3 R T = NAm u2 correction

Furthermore, u2 = 3 R T / M (Temperature and speed)

Explain the significances of and apply these formulas for sciences

6 Gases

Molecular Speeds

- Distributions of speed of various gases will be demonstrated using a simulation program, and for each gas, three speeds are indicated.
- In the following: m = mass of a molecule, M = molar mass,R = gas constant, and k = R / Navogadro = Boltzmann constant.
- The most probable speed ump = (2 k T / m)1/2 = (2 R T / M)1/2

The root-mean-square speed urms = ( 3 k T / m)1/2 = (3 RT / M)1/2

The mean or average speeduave = ( 8 k T /π m)1/2 = (8 RT / π M)1/2

6 Gases

Diffusion of Gases

- All gases move together because they are subjected to the same pressure head.
- Different gases diffuse at different rates
- Diffusion contributes to net movement of O2 and CO2 across the alveolar-capillary membrane (breathe).
- Constant molecular motion.
- Diffusion from higher to lower concentration regions.
- Since uave = ( 8 k T /π m)1/2 = (8 RT / π M)1/2, (slide 24)
- 1Diffusion rate -------- Graham’s law

M

Discuss diffusion rates of H2, He, CH4, N2, O2, CO2, 235UF6, 238UF6at lecture

6 Gases

Diffusion problems

Problems are usually to compare diffusion or effusion rates in the following terms: urms = ( 3 k T / m)1/2 =

uave = ( 8 k T /π m)1/2 =

diffusion rateeffusion time for same amount

distance traveled by molecules in certain periodamount of gas effused

3RTM

8RT πM

1M

M

6 Gases

Comparing Effusion Rates

If 1e20 N2 molecules effuse from an orifice in 1.0 min, how many H2 molecules will effuse the same orifice at the same condition (T P)? How many minutes will be required for the same number of H2 molecules to effuse?

H2 effusion rate = MN2 / MH2 (N2effusion rate)

= (28 / 2 )1e20 molecules/min)

= __________ figure out value and units

Time for 1e20 H2 molecules to effuse = (2 / 28 ) 1.0 min

= __________

What is the effusion rate ratio of N2 and H2 or any two gases?

6 Gases

P2

CO2

A

P1

T

Effusion and LifeBreathing chemistry is complicated, and we can only scratches the surface!

6 Gases

The van der Waals equation for real gases

Gases tend to behave ideally at high T and low P. Required T and P for ideality depends on gas properties and molar mass, and van der Waals proposed correction terms for the ideal gas equation for real gases.

(P + ) (V – n b) = nR T

Correction for intermolecular forces

Correction for volume of molecules

n 2aV 2

where a and b are gas-dependant constants.

Gas aL2 atm mol-2b L mol-1 He 0.0341 0.0237 Ne 0.211 0.0171 N2 1.39 0.0391 O2 1.36 0.0318 CO2 3.59 0.0427 Cl2 6.49 0.0562

Explain the meaning of vdW eqn

Note units for a and b

6 Gases

Application of van der Waal’s Equation

What is the pressure of Cl2 at 300 K occupying 20.0 L according to vdW and ideal gas laws?

Solution: Look up data for Cl2, a = 6.49 L2 atm mol-2, b = 0.0562 L mol-1,n = 1 mol, R = 0.08205 L atm K-1 mol-1

n R TV – n b

n 2a V2

P = -

12 mol2 * 6.49 L2 atm mol-220.02 L2

1 mol * 0.08205 L atm mol-1 K-1*300 K20.0 L – 1 mol 0.0562 mol-1 L

= -

= _____________________

Please calculate the results, and P from the ideal gas law.

6 Gases

Problems related to van der Waal’s Equation

What is the molar volume of Cl2 at 300 K and 1 atm according to vdW?

Solution: Look up data for Cl2, a = 6.49 L2 atm mol-2, b = 0.0562 L mol-1,n = 1 mol, R = 0.08205 L atm K-1 mol-1

n R TP + n 2a / V2

V = + n b = 24.615 / (1+0.013) = 24.299 L

try V = 22 L= 24 Ltry V = 24.434 Land calculate V

= 24.625 / (1+ 0.011) = 24.434 L

= ?

Calculate P for a definite volume is easier, and using the successive method for V is interesting, but it’s a challenge.

6 Gases

Find out how engineers deal with real gases.

Volume of vdW eqn

What is the volume of occupied by 132 g CO2 gas at 12.5 atm and 300 K?

Solution:Solve volume of van der Waals equation for V

R T + b PP

n2 aP

n2 a bP

V 3 – n ( ) V 2 + ( ) V – ( ) = 0 Derive please

a = 3.59 L2 atm mol-2, b = 0.0427 L mol-1,n = 1 mol, R = 0.08205 L atm K-1 mol-1

This is similar to problem 6 –106, a question for practicing the successive approximation method.

6 Gases

Successive Method again

What is the volume of occupied by 132 g CO2 gas at 12.5 atm and 300 K?

Solution:a = 3.59 L2 atm mol-2, b = 0.0427 L mol-1, T = 300 Kn = 132/44 = 3 mol, R = 0.08205 L atm K-1 mol-1

n R TP + n 2a / V2

V = + n b

= 3*0.08205*300 / (12.5 + 32* 3.59 / 12) + 3*.0427 = 1.78 L= 73.845 / (12.5 + 32*3.59 / 32) + 0.128 = 4.7 L= 73.845 / (12.5 + 32*3.59 / 42) + 0.128 = 5.2 L= 73.845 / (12.5 + 32*3.59 / 52) + 0.128 = 5.5 L= 73.845 / (12.5 + 32*3.59 / 5.62) + 0.128 = 5.58 L

6 Gases

Molecular Formula of Gas

Combustion of 1.110 g hydrocarbon produces 3.613 CO2 and 1.109 g H2O. A 0.288 g sample of the same has a volume of 131 mL at 298 K and 753 mmHg. Find the molecular formula.

Solution:

C : H = : = 0.0821 : 0.123 = 1 : 1.5 = 2 : 3

3.61344

1.109*218

Empirical formula is C2H3

d R TP

(0.288 g / 0.131 L–1) * 0.08205 L atm mol–1 K –1 * 298 K(753 / 760) atm

M = =

= 54.3_______ work out units

C4H6 has a molar mass of 54.3. Confirm and conclusion please!

6 Gases

A 2-step problem, similar to one in Advanced Exercises

ROOMS FOR TEST #1 (CHEM 120) – Wed., Oct. 8th

Write the testduring your regular lecture time (10:30 am) on Wed., Oct. 8th. Go to DC 1350 and ESC 146/149 according to your Surnames

Surnames Room(s)

For 8:30 class A – L DC 1350 M – Z ESC 146 & 149For 9:30 class A – Mo DC 1350 Mu – Z ESC 146 & 149

10:30 am A – Ma DC 1350

(For you) Mc – Z ESC 146 & 149 (First Year Chem Lab)

For 11:30 class A – Ma DC 1350 Mc – Z ESC 146 & 149

6 Gases

Regarding Test 1

·Do not enter the room until directed to do so by the proctors. We need space and time to set out the test booklets and computer answer cards.

·Bring a calculator and a pencil for filling out the computer answer card.

·Do NOT bring your own scrap paper or periodic table. All work must be done on the test booklet. A periodic table will be supplied.

6 Gases

Some concepts to review

Convert between mass and mole and vice versa

Find empirical and molecular formulas

Figure out limiting and excess reagent, calculate theoretical and percent yields

Calculate concentrations in molarity, mass percentage, etc even when solutions are combined (dilution)

Analyze binary mixture: extra problems B2 and B3 (handout page 8)

Figure out the net ionic reaction equations

Balance redox reaction equations (figure out oxidation states, balance half-reaction equations and balance equations)

6 Gases

Concepts to review – cont.

Apply ideal gas low to various problems

Calculate stoichiometric quantities using on gas law and reaction equations.

Apply Dalton’s partial pressure equation

Compare effusion or diffusion rates of gases

6 Gases

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