Ch5 z5e gases

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Chapter 5Chapter 5

The Gas LawsThe Gas Laws pp pp

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5.1 Pressure5.1 Pressure Force per unit area.Force per unit area. Gas molecules fill container.Gas molecules fill container. Molecules move around and hit Molecules move around and hit

sides.sides. Collisions are the force.Collisions are the force. Container has the area.Container has the area. Measured with a barometer.Measured with a barometer.

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BarometerBarometer The pressure of the The pressure of the

atmosphere at sea atmosphere at sea level will hold a level will hold a column of mercury column of mercury 760 mm Hg.760 mm Hg.

1 atm = 760 mm Hg1 atm = 760 mm Hg

1 atm Pressure

760 mm Hg

Vacuum

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ManometerManometer

Gas

h

Column of Column of mercury to mercury to measure measure pressure.pressure.

h is how much h is how much lower the lower the pressure is pressure is than outside. than outside.

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ManometerManometer h is how much h is how much

higher the gas higher the gas pressure is than pressure is than the atmosphere.the atmosphere.

h

Gas

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Units of pressureUnits of pressure 1 atmosphere = 760 mm Hg1 atmosphere = 760 mm Hg 1 mm Hg = 1 torr1 mm Hg = 1 torr 1 atm = 101,325 Pascals = 101.325 kPa1 atm = 101,325 Pascals = 101.325 kPa Can make conversion factors from these.Can make conversion factors from these. What is 724 mm Hg in kPa? . . .What is 724 mm Hg in kPa? . . . 96.5 kPa96.5 kPa In torr?In torr? 724 torr724 torr In atm?In atm? 0.952 atm.0.952 atm.

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5.2 The Gas Laws of Boyle, Charles, and Avogadro5.2 The Gas Laws of Boyle, Charles, and Avogadro

Boyle’s LawBoyle’s Law Pressure and volume are inversely Pressure and volume are inversely

related at constant temperature.related at constant temperature. PV= kPV= k As one goes up, the other goes As one goes up, the other goes

down.down. PP11VV11 = P = P22 V V22

GraphicallyGraphically

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1 atm

4 Liters

As the pressure on a gas increases

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2 atm

2 Liters

As the pressure on a gas increases the volume decreases

Pressure and volume are inversely related

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A Boyle’s Law Relationship

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V

P (at constant T)

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V

1/P (at constant T)

Slope = k

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PV

P (at constant T)

CO2

O2

22.4

1 L

atm Ne

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ExampleExample pp pp

20.5 L20.5 L of nitrogen at 25.0ºC and of nitrogen at 25.0ºC and 742 torr742 torr are are compressed to compressed to 9.80 atm9.80 atm at constant T. What at constant T. What is the is the new volumenew volume? Steps . . .? Steps . . .

PP11VV11 = = PP22 V V22

(20.5 L)(742 torr)(20.5 L)(742 torr) = = (9.80 atm x 760 torr/atm)(x)(9.80 atm x 760 torr/atm)(x) x = 2.04 Lx = 2.04 L

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You try it!You try it!

30.6 mL of CO2 at 740 torr is expanded at constant temperature to 750 mL. What is the final pressure in kPa? Ans. . .

4.02 kPa. Steps . . .

PP11VV11 = = PP22 V V22

((30.6 mL)(740 torr)30.6 mL)(740 torr) = =(750 mL) (x torr)(750 mL) (x torr) x = 30.2 torrx = 30.2 torr (30 torr)(101.325 kPa/ 760 torr) = (30 torr)(101.325 kPa/ 760 torr) = 4.02 kPa4.02 kPa

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Charles’ LawCharles’ Law Volume of a gas varies directly with Volume of a gas varies directly with

the absolute temperature at constant the absolute temperature at constant pressure.pressure.

V = kT (if T is in Kelvin)V = kT (if T is in Kelvin)

VV1 1 = V = V22

T T11 = T = T22

GraphicallyGraphically

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Tr 51A Charles’ Law Relationship

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V (

L)

T (ºC)

He

H2O

CH4

H2

-273.15ºC

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What would the final What would the final volumevolume be if be if 247 mL247 mL of of gas at gas at 22ºC22ºC is heated to is heated to 98ºC98ºC , if the , if the pressure is held constantpressure is held constant? Steps.? Steps.

VV11/T/T11 = = VV22/T/T22

(247 mL)/(22 + 273) K(247 mL)/(22 + 273) K = = (x mL)/(273 + 98) K(x mL)/(273 + 98) K x = 311 mlx = 311 ml

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Examples (do this)Examples (do this) At what temperature would 40.5 L of gas at At what temperature would 40.5 L of gas at

23.4ºC have a volume of 81.0 L at constant 23.4ºC have a volume of 81.0 L at constant pressure? pressure? a) 173 ºC b) 280 ºC c) 320 ºC d) 593 ºCa) 173 ºC b) 280 ºC c) 320 ºC d) 593 ºC

320. ºC 320. ºC (change K to ºC!)(change K to ºC!)

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Avogadro's LawAvogadro's Law At constant temperature and At constant temperature and

pressure, the volume of gas is pressure, the volume of gas is directly related to the number of directly related to the number of moles.moles.

V = k n (n is the number of moles)V = k n (n is the number of moles)

VV1 1 = V = V22

n n11 = n = n22

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Figure 5.10 p. 195Figure 5.10 p. 195Balloons Holding Balloons Holding 1.0 L 1.0 L of Gas at 25º C and 1 of Gas at 25º C and 1 atmatm. Each contains . Each contains 0.041 mol of gas, or 2.5 0.041 mol of gas, or 2.5 x 10x 102222 molecules, molecules, even even thoughthough their their masses masses are are different (different (equal equal numbers, different numbers, different massesmasses).).

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Gay- Lussac LawGay- Lussac Law At constant volume, pressure and At constant volume, pressure and

absolute temperature are directly absolute temperature are directly related.related.

P = k TP = k T

PP1 1 = P = P22

T T11 = T = T22

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Tr 52A Gay-Lussac’s Law Relationship

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Combined Gas LawCombined Gas Law pp pp

If the moles of gas If the moles of gas remain constantremain constant, , use this formula and cancel out the use this formula and cancel out the other things that don’t change.other things that don’t change.

PP1 1 VV11 = P = P22 V V22

.. T T11 T T22

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ExamplesExamples pp pp

A deodorant can has volume of 175 mL & A deodorant can has volume of 175 mL & pressure of 3.8 atm at 22ºC. What pressure pressure of 3.8 atm at 22ºC. What pressure if heated to 100.ºC?if heated to 100.ºC?

V is constant, so (PV is constant, so (P11VV11)/T)/T11 = (P = (P22VV22/T/T22)) (3.8)/(295) = (x)/373(3.8)/(295) = (x)/373 x = 4.8 atmx = 4.8 atm

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ExamplesExamples A deodorant can has volume of A deodorant can has volume of 175 mL175 mL & &

pressure of pressure of 3.8 atm3.8 atm at 22ºC. at 22ºC. What volume What volume of gas could the can release at of gas could the can release at 22ºC22ºC and and 743743 torr? torr? Answer . . . Answer . . .

680 mL680 mL Steps. . .Steps. . . ((3.83.8)()(175175)/(273 + 22) = ()/(273 + 22) = (743/760743/760)(x)/()(x)/(273 + 22273 + 22))

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5.3 Ideal Gas Law5.3 Ideal Gas Law pp pp

PV = nRTPV = nRT V = 22.41 L at 1 atm, 0ºC, n = 1 mole, V = 22.41 L at 1 atm, 0ºC, n = 1 mole,

what is R?what is R? R is the ideal gas constant.R is the ideal gas constant. R = 0.08206 L atm/ mol KR = 0.08206 L atm/ mol K R also = 62.396 L torr/k mol orR also = 62.396 L torr/k mol or

8.3145 J/k mol8.3145 J/k mol Tells you about a gas is Tells you about a gas is NOWNOW.. The The otherother laws tell you about a gas laws tell you about a gas

when it when it changeschanges ( (exam hint!exam hint!)). .

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Ideal Gas LawIdeal Gas Law An An equation of stateequation of state.. Independent of how you end up Independent of how you end up

where you are at. Does not depend where you are at. Does not depend on the path.on the path.

Given 3 measurements you can Given 3 measurements you can determine the fourth.determine the fourth.

An Empirical Equation - based on An Empirical Equation - based on experimentalexperimental evidence. evidence.

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Ideal Gas LawIdeal Gas Law A hypothetical substance - the ideal A hypothetical substance - the ideal

gasgas Think of it as a limit.Think of it as a limit. Gases only approach ideal behavior Gases only approach ideal behavior

at low pressure (< 1 atm) and high at low pressure (< 1 atm) and high temperature.temperature.

Use the laws anyway, unless told to Use the laws anyway, unless told to do otherwise.do otherwise.

They give good estimates.They give good estimates.

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A A 47.3 L47.3 L container with container with 1.62 mol1.62 mol of He of He heated until pressure reaches heated until pressure reaches 1.85 atm1.85 atm. . What is the temperature? Steps . . .What is the temperature? Steps . . .

PVPV = = nRnRTT TT = ( = (1.851.85)()(47.347.3)/(.)/(.08210821)()(1.621.62)) ==

658 K or 658 K or 385 ºC385 ºC

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Kr gas in a 18.5 L cylinder exerts a Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8ºC. What is pressure of 8.61 atm at 24.8ºC. What is the mass of Kr?the mass of Kr? Ans. . . Ans. . .

546 g546 g Steps . . . Steps . . . n = PV/RT = 6.51 moln = PV/RT = 6.51 mol Since MSince M = m/mol = m/n, m = M • n = m/mol = m/n, m = M • n Get M from periodic tableGet M from periodic table m = 83.8 • 6.51 = 546 gm = 83.8 • 6.51 = 546 g

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ExamplesExamples

A gas is 4.18 L at 29 ºC & 732 torr. What A gas is 4.18 L at 29 ºC & 732 torr. What volume at 24.8ºC & 756 torr? Ans. . .volume at 24.8ºC & 756 torr? Ans. . .

Use combined law. Why? . . .Use combined law. Why? . . . The conditions are The conditions are changingchanging x = 3.99 L (can convert torr to atm when x = 3.99 L (can convert torr to atm when

calculating, but pressure units cancel out so calculating, but pressure units cancel out so don’t need to for this problem).don’t need to for this problem).

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5.4 Gases and Stoichiometry5.4 Gases and Stoichiometry Reactions happen in molesReactions happen in moles At Standard Temperature and At Standard Temperature and

Pressure (STP, 0ºC and 1 atm) 1 Pressure (STP, 0ºC and 1 atm) 1 mole of gas occupies 22.41 L.mole of gas occupies 22.41 L.

If not at STP, use the ideal gas law to If not at STP, use the ideal gas law to calculate moles of reactant or calculate moles of reactant or volume of product.volume of product.

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Figure 5.11Figure 5.11A Mole of Any Gas Occupies a Volume of A Mole of Any Gas Occupies a Volume of

Approximately 22.4 L at STPApproximately 22.4 L at STP

How many particles are in each box?

6.022 x 1023 particles

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Can get mercury by the following:Can get mercury by the following:

What volume oxygen gas produced from What volume oxygen gas produced from 4.10 g of mercury (II) oxide (M = 217 4.10 g of mercury (II) oxide (M = 217 g/mol) at STP? Steps.g/mol) at STP? Steps.

Balance first! 2HgO Balance first! 2HgO 2Hg + O 2Hg + O22

Since at STP can then use stoich (spider)Since at STP can then use stoich (spider) 4.10 g HgO • 1 mol HgO/217g HgO • 1 mol O4.10 g HgO • 1 mol HgO/217g HgO • 1 mol O22/2 mol HgO • 22.4 L O/2 mol HgO • 22.4 L O22/1 mol O/1 mol O22

= 0.212 L or 212 mL= 0.212 L or 212 mL

(g)O + (l) Hg HgO 2heat

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Can get mercury by the following:Can get mercury by the following:

What volume oxygen gas can be produced What volume oxygen gas can be produced from 4.10 g of mercury (II) oxide at 400.ºC from 4.10 g of mercury (II) oxide at 400.ºC & 740. torr? (Hint: conditions are changed, & 740. torr? (Hint: conditions are changed, so whatso what equation do you then use?) . . . equation do you then use?) . . .

1st, find molar mass of mercury(II) oxide.1st, find molar mass of mercury(II) oxide. 2nd, balance reactions & spider under STP2nd, balance reactions & spider under STP 3rd, then use the combined law to convert.3rd, then use the combined law to convert. (1)(212)/273 = (740/760)(x)/(273 + 400)(1)(212)/273 = (740/760)(x)/(273 + 400) 0.537 L0.537 L

HgO heat Hg(l) + O2(g)

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ExamplesExamples Using the following reactionUsing the following reaction

calculate the mass of sodium hydrogen calculate the mass of sodium hydrogen carbonate (M = 84.01) that produces 2.87 carbonate (M = 84.01) that produces 2.87 L of carbon dioxide at 25ºC & 2.00 atm.L of carbon dioxide at 25ºC & 2.00 atm.

a.a. Not at STP so use PV=nRT; ans. . .Not at STP so use PV=nRT; ans. . .

b.b. 19.7 g19.7 g

NaCl(aq) + CO (g) +H O(l)2 2

NaHCO (s) + HCl 3

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ExamplesExamples Using the following reactionUsing the following reaction

If 27 L of gas are produced at 26ºC and 745 torr If 27 L of gas are produced at 26ºC and 745 torr

when 2.6 L of HCl are added what is the when 2.6 L of HCl are added what is the

concentration of HCl?concentration of HCl?

a.a. Use stoich to get moles HCl, then mole ratio Use stoich to get moles HCl, then mole ratio moles moles COCO22, then moles/L = concentration. Or, use Chas law , then moles/L = concentration. Or, use Chas law to correct to 24.2 L, then stoich to get 1.08 mol HCl = to correct to 24.2 L, then stoich to get 1.08 mol HCl =

1.08 mole CO1.08 mole CO22; so concentration is 0.415 ; so concentration is 0.415 MM

NaCl(aq) + CO (g) +H O(l)2 2

NaHCO (s) + HCl 3

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ExamplesExamples

Consider the following reactionConsider the following reaction

What volume NO at 1.0 atm & 1000ºC What volume NO at 1.0 atm & 1000ºC produced from 10.0 L of NHproduced from 10.0 L of NH33 & xs O & xs O22 at at

samesame temperature & pressure? temperature & pressure? T & P constant & cancel out so volume is T & P constant & cancel out so volume is

proportional to moles = 10 Lproportional to moles = 10 L

4NH3(g) + 5 O2(g) 4 NO(g) + 6H2O(g)

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You try it!You try it!

Consider the following reactionConsider the following reaction

What volume of OWhat volume of O22 at STP will be at STP will be

consumed when 10.0 kg NHconsumed when 10.0 kg NH33 is reacted? is reacted?

(Use “rounded” molar masses).(Use “rounded” molar masses). Ans. . . Ans. . . watch Kg! Ans. Rounded to watch Kg! Ans. Rounded to 16 500 L16 500 L (from (from

calc ans of 16 470)calc ans of 16 470)

O(g)6H +NO(g) 4 )(O 5 + (g)4NH 223 g

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The Same reaction with some twistsThe Same reaction with some twists pp pp

What mass of HWhat mass of H22O produced from 65.0 L OO produced from 65.0 L O22

& 75.0 L of NH& 75.0 L of NH33 measured at STP? . . . Ans. measured at STP? . . . Ans.

As before, but must find As before, but must find LR = O2 ans. 313 g313 g What volume of NO would be produced?What volume of NO would be produced? Above + mole ratio. Ans. = Above + mole ratio. Ans. = 52 L52 L What mass of NO is produced from 500. L of What mass of NO is produced from 500. L of

NHNH33 at 250.0ºC and 3.00 atm? Ans. at 250.0ºC and 3.00 atm? Ans. non-STP so use combined gas law to get to non-STP so use combined gas law to get to

STP (783 L), then “spider” Ans. = STP (783 L), then “spider” Ans. = 895 g895 g

4NH3(g) + 5 O2(g) 4 NO(g) + 6H2O(g)

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Gas Density and Molar MassGas Density and Molar Mass D = m/VD = m/V Let Let MM stand for molar mass stand for molar mass MM = m/n = m/n n= PV/RTn= PV/RT MM = = m m

PV/RT PV/RT MM = mRT = m RT = DRT = mRT = m RT = DRT

PV PV V P V P P P

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Finding M or D from ideal gas law M = m

(PV/RT) M = m RT

V P Since m/V = density (D) M = DRT = molar mass

P Density varies directly with M and P Density varies indirectly with T

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Quick Example pp

What mass of ethene gas, C2H4, is in a 15.0 L tank at 4.40 atm & 305 K? (steps)

PV = mRT/M and M = 28 g/mol m = PVM/RT m = (4.40 atm)(15.0 L)(28 g/mol)

(0.0821 L atm/mol K)(305 K) m = 73.8 g C2H4

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Another way to solve it (quick) pp

What mass of ethene gas, C2H4, is in a 15.0 L tank at 4.40 atm & 305 K? (steps)

PV = nRT, solve for n n = PV/RT n = (4.40 atm)(15.0 L)

(0.0821 L atm/mol K)(305 K) n = 2.64 mol C2H4

Use spider to get g C2H4 (M = 28 g/mol)

Answer is 73.8 g C2H4

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Quick Example A gas has a density of 3.36 g/L at 14 ºC

and 1.09 atm. What is its molar mass? Use M = DRT/P to get an answer of . . . 72.6 g/mol. Calculation . . . (3.36)(0.0821)(14 + 273)/(1.09) = 72.6 g/mol

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AP ExamplesAP Examples Density of ammonia at 23ºC & 735 torr? Density of ammonia at 23ºC & 735 torr? Ans.Ans.

D = MP/RT = 0.68 g/LD = MP/RT = 0.68 g/L

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AP ExamplesAP Examples A compound has empirical formula CHCl. A A compound has empirical formula CHCl. A

256 mL flask at 100.ºC & 750 torr contains 256 mL flask at 100.ºC & 750 torr contains 0.80 g of the 0.80 g of the gaseousgaseous compound. What is compound. What is the molecular formula? Steps. . .the molecular formula? Steps. . .

What do you need to get the molecular What do you need to get the molecular formula? . . .formula? . . .

The molecular mass, M.The molecular mass, M. How do you get this?How do you get this? M = g/mol. You are given the grams, so find M = g/mol. You are given the grams, so find

the number of moles using . . .the number of moles using . . . PV = nRT and solve for n . . . Try it! . . .PV = nRT and solve for n . . . Try it! . . .

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AP Examples cont.AP Examples cont. EF of CHCl. 0.80 g are in 256 mL flask at EF of CHCl. 0.80 g are in 256 mL flask at

100.ºC & 750 torr. What is the MF?100.ºC & 750 torr. What is the MF? n = 0.0083 from n = PV/RTn = 0.0083 from n = PV/RT What’s M? . . .What’s M? . . . M M 97 from 0.80 g/0.0083 moles = g/mol 97 from 0.80 g/0.0083 moles = g/mol How do you find molecular formula from EF How do you find molecular formula from EF

and MF?and MF? Find how many Efs are in the MF (CHCl) Find how many Efs are in the MF (CHCl)

then multiply the subscripts of the EF by that then multiply the subscripts of the EF by that number. Try it! . . .number. Try it! . . .

97/49 97/49 2, so formula is C 2, so formula is C22HH22ClCl22

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5.5 Dalton’s Law of Partial Pressures5.5 Dalton’s Law of Partial Pressures The total pressure in a container is The total pressure in a container is

the sum of the pressure each gas the sum of the pressure each gas would exert if it were alone in the would exert if it were alone in the container.container.

The total pressure is the sum of the The total pressure is the sum of the partial pressures.partial pressures.

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Figure 5.12 p. 206Figure 5.12 p. 206Partial Pressure of Each Partial Pressure of Each Gas in a MixtureGas in a Mixture

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We can find out the pressure in the fourth container by adding up the pressure in the first 3.

2 atm 1 atm 3 atm 6 atm

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Dalton’s LawDalton’s Law PPTotalTotal = P = P11 + P + P22 + P + P33 + P + P44 + P + P55 ... ...

For each, P still = nRT/VFor each, P still = nRT/V

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Dalton's LawDalton's Law PPTotalTotal = n = n11RT + nRT + n22RT + nRT + n33RT +...RT +...

V V V V V V In the same container R, T and V are In the same container R, T and V are

the same.the same.

PPTotalTotal = (n = (n11+ n+ n22 + n + n33+...)RT+...)RT

V V

PPTotalTotal = (n = (nTotalTotal)RT)RT

V V

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Examples pp

In a balloon the total pressure is 1.3 atm. What is the pressure of oxygen if the pressure of nitrogen is 720 mm Hg? . . .

PTotal = P1 + P2 + P3 . . .

1.3 atm = Po2 + 720 mm Hg of N2

Convert to same units (we’ll use mm Hg) 1.3 atm(760 mm Hg/1 atm) = Po2 + 720 mm Hg

988 mm Hg = Po2 + 720 mm Hg so, . . . Po2 = 988 mm Hg - 720 mm Hg = 268 mm Hg Can also express as 0.35 atm.

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The mole fractionThe mole fraction Ratio of moles of the substance to Ratio of moles of the substance to

the the totaltotal moles. moles.

symbol is Greek letter chi symbol is Greek letter chi

= n= n11 = P= P1 1

n nTotal Total PPTotalTotal

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The partial pressure of nitrogen in air is 592 The partial pressure of nitrogen in air is 592 torr. The partial pressure of oxygen in air is torr. The partial pressure of oxygen in air is 160 torr. What is the mole fraction of 160 torr. What is the mole fraction of oxygen? Ans. . .oxygen? Ans. . .

0.213 Why isn’t it 0.270? . . .0.213 Why isn’t it 0.270? . . . Because you have to divide by the Because you have to divide by the totaltotal

pressure, which is 752 torr.pressure, which is 752 torr.

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ExamplesExamples The partial pressure of nitrogen in air is 592 The partial pressure of nitrogen in air is 592

torr. Air pressure is 752 torr, what is the torr. Air pressure is 752 torr, what is the mole fraction of nitrogen? Steps. . .mole fraction of nitrogen? Steps. . .

= 592 torr/752 torr = 0.787= 592 torr/752 torr = 0.787 What is the partial pressure of nitrogen if What is the partial pressure of nitrogen if

the container holding the air is compressed the container holding the air is compressed to 5.25 atm? . . .to 5.25 atm? . . .

0.787 x 5.25 atm = 4.13 atm0.787 x 5.25 atm = 4.13 atm Note: can use mole fraction regardless of Note: can use mole fraction regardless of

units (don’t have to convert).units (don’t have to convert).

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3.50 L

O2

1.50 L

N2

2.70 atm When these valves are opened, what is each partial pressure and the total pressure? When these valves are opened, what is each partial pressure and the total pressure?

Hint.Hint. Use mole fraction to get the new partial pressures (or Boyle’s law), then add to get Use mole fraction to get the new partial pressures (or Boyle’s law), then add to get

total pressure. Ans.total pressure. Ans. CHCH44 = 1.2 atm = 1.2 atm NN22 = 0.763 = 0.763 OO22 = 0.294 = 0.294 Sum = 2.127Sum = 2.127

4.00 L

CH4

4.58 atm 0.752 atm

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Vapor PressureVapor Pressure Water evaporates!Water evaporates! When that water evaporates, the When that water evaporates, the

vapor has a pressure.vapor has a pressure. Gases are often collected over water Gases are often collected over water

so the vapor. pressure of water must so the vapor. pressure of water must be subtracted from the total be subtracted from the total pressure.pressure.

It must be given.It must be given.

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Figure 5.13Figure 5.13The Production of Oxygen by Thermal The Production of Oxygen by Thermal

Decomposition of KCIODecomposition of KCIO33

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NN22O can be produced by the following O can be produced by the following

reactionreaction

what volume of Nwhat volume of N22O collected O collected

over water at a total pressure of 94 kPa over water at a total pressure of 94 kPa and 22ºC can be produced from 2.6 g of and 22ºC can be produced from 2.6 g of NHNH44NONO33? (the vapor pressure of water ? (the vapor pressure of water

at 22ºC is 21 torr). Steps . . .at 22ºC is 21 torr). Steps . . .

NH4NO3(s) heat N2O(g) + 2H2O(l)

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ExampleExample What vol. NWhat vol. N22O over water at total pressure of 94 O over water at total pressure of 94 kPakPa & &

22ºC can be produced from 2.6 g NH22ºC can be produced from 2.6 g NH44NONO33? (Vp? (Vpwaterwater at at 22ºC is 21 22ºC is 21 torrtorr))

Make sure reaction is balanced (it is)Make sure reaction is balanced (it is) M of ammonium nitrate = 80. gM of ammonium nitrate = 80. g Use spider to get moles of NUse spider to get moles of N220 = .0325, then use PV = 0 = .0325, then use PV =

nRT, but first have to:nRT, but first have to: Convert 21 Convert 21 torrtorr to 2.8 to 2.8 kPakPa (or (or vice versavice versa)) 94.0 - 2.8 = 91.2 kPa N94.0 - 2.8 = 91.2 kPa N22OO convert P to atm d/t R ( or use 8.31 for kPa)!!!convert P to atm d/t R ( or use 8.31 for kPa)!!! Convert C to KConvert C to K Ans. 6.6 LAns. 6.6 L

NH4NO3(s) heat N2O(g) + 2H2O(l)

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5.6 Kinetic Molecular Theory5.6 Kinetic Molecular Theory Theory tells Theory tells whywhy the things happen. the things happen. explains why ideal gases behave the explains why ideal gases behave the

way they do.way they do. AssumptionsAssumptions that simplify the theory, that simplify the theory,

but don’t work in real gases.but don’t work in real gases. The particles are The particles are so smallso small we can we can

ignore their volumeignore their volume.. The particles are in The particles are in constant motionconstant motion and and

their their collisions cause pressurecollisions cause pressure. .

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The Average speed of an oxygen molecule is 1656 km/hr at 20ºC

The molecules don’t travel very far without hitting each other so they move in random directions.

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Kinetic Molecular TheoryKinetic Molecular Theory The particles do not affect each other, The particles do not affect each other,

neither attracting or repellingneither attracting or repelling.. The The averageaverage kinetic energy is kinetic energy is

proportionalproportional to the to the KelvinKelvin temperature. temperature. Appendix 2 shows the derivation of the Appendix 2 shows the derivation of the

ideal gas law and the definition of ideal gas law and the definition of temperature.temperature.

We need the formula KE = 1/2 mvWe need the formula KE = 1/2 mv22

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What it tells usWhat it tells us (KE)(KE)avgavg = 3/2 RT = 3/2 RT This the meaning of temperature.This the meaning of temperature. u is the particle velocity.u is the particle velocity. u is the average particle velocity.u is the average particle velocity. u u 22 is the average particle velocity is the average particle velocity

squared.squared. the root mean square velocity is the root mean square velocity is

u u 2 = 2 = uurmsrms

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Combine these two equationsCombine these two equations (KE)(KE)avgavg = n = nAA(1/2 mu (1/2 mu 22 ) ) (KE)(KE)avgavg = =

3/2 RT3/2 RT

70

Combine these two equationsCombine these two equations (KE)(KE)avgavg = n = nAA(1/2 mu (1/2 mu 22 ) ) (KE)(KE)avgavg = 3/2 RT = 3/2 RT

Where M is the molar mass Where M is the molar mass in in kgkg/mole /mole ((watch for test questionwatch for test question!!), and ), and R has the units 8.3145 R has the units 8.3145 JJ/K mol./K mol.

The velocity will be in The velocity will be in m/sm/s

u = 3RT

Mrms

71

ExampleExample pp pp Calculate the root mean square velocity of Calculate the root mean square velocity of

carbon dioxide at 25ºC.carbon dioxide at 25ºC.

Must express R using joules & M = Kg/mol (not g, Must express R using joules & M = Kg/mol (not g, d/t R)d/t R)

R = 8.3145 J/K•Mol, M = R = 8.3145 J/K•Mol, M = 0.00.044 44 KgKg/mol ans. . ./mol ans. . . 411 m/sec411 m/sec If time, do for hydrogen, for chlorine. Ans. . .If time, do for hydrogen, for chlorine. Ans. . . Hydrogen is 1928 m/s; Chorine is 324 m/sHydrogen is 1928 m/s; Chorine is 324 m/s Notice that the less massive hydrogen has a Notice that the less massive hydrogen has a

higher velocity.higher velocity.

urms = 3RT

M

72

Range of velocitiesRange of velocities The average distance a molecule The average distance a molecule

travels before colliding with another travels before colliding with another is called the mean free path and is is called the mean free path and is small (around 10small (around 10-7-7m)m)

73

Figure 5.18Figure 5.18Path of One Path of One Particle in a Particle in a Gas.Gas.Any given Any given particle will particle will continuously continuously change its course change its course as a result of as a result of collisions with collisions with other particles, as other particles, as well as with the well as with the walls of its walls of its container.container.

74

Range of velocitiesRange of velocities Temperature is an average. There are Temperature is an average. There are

molecules of molecules of manymany speeds in the speeds in the average.average.

This is shown on a graph called a This is shown on a graph called a velocity distribution . . .velocity distribution . . .

75

num

ber

of p

arti

cles

Molecular Velocity

273 K

76

num

ber

of p

arti

cles

Molecular Velocity

273 K

1273 K

77

num

ber

of p

arti

cles

Molecular Velocity

273 K

1273 K

2273 K

78

VelocityVelocity Average increases as temperature Average increases as temperature

increases.increases. Spread increases as temperature Spread increases as temperature

increases.increases.

79

5.7 Effusion & Diffusion5.7 Effusion & Diffusion EffusionEffusion: Passage of gas through a : Passage of gas through a

small hole, into a vacuum.small hole, into a vacuum. The effusion rate measures how fast The effusion rate measures how fast

this happens.this happens. Graham’s LawGraham’s Law: The rate of effusion is : The rate of effusion is

inversely proportional to the square inversely proportional to the square root of the mass of its particles.root of the mass of its particles.

Rate effusion for gas 1

Rate effusion for gas 2

M2

M1

80

Figure 5.21 p. 219The Effusion of a Gas into an Evacuated ChamberFigure 5.21 p. 219The Effusion of a Gas into an Evacuated Chamber

Rate of effusion is inversely proportional to the square root of the mass of the gas molecules.

81

DerivingDeriving The rate of effusion should be The rate of effusion should be

proportional to uproportional to urmsrms

Effusion Rate 1 = uEffusion Rate 1 = urms rms 11

Effusion Rate 2 = u Effusion Rate 2 = urms rms 22

82

Deriving Graham’s LawDeriving Graham’s Law The rate of effusion should be The rate of effusion should be

proportional to uproportional to urmsrms

Effusion Rate 1 = uEffusion Rate 1 = urms rms 11

Effusion Rate 2 = u Effusion Rate 2 = urms rms 22

effusion rate 1

effusion rate 2

urms 1

urms 2

3RT

M1

3RT

M2

M2

M1

83

DiffusionDiffusion The spreading of a gas through a room.The spreading of a gas through a room. Slow, considering molecules move at 100’s Slow, considering molecules move at 100’s

of meters per second.of meters per second. Collisions with other molecules slow down Collisions with other molecules slow down

diffusions.diffusions. Best estimate of diffusion is Graham’s Law.Best estimate of diffusion is Graham’s Law. Do not confuse rate with time. Do not confuse rate with time. Time is in seconds (or min). Rate is in 1/s Time is in seconds (or min). Rate is in 1/s

(or 1/min), etc. (or 1/min), etc. The following questions test your The following questions test your

understanding of this.understanding of this.

84

Figure 5.23Figure 5.23HCI(HCI(gg) and NH) and NH33((gg) Meet in a ) Meet in a

TubeTube

85

Whiteboard ExamplesWhiteboard Examples pp pp

An element effuses through a porous cylinder 3.20 An element effuses through a porous cylinder 3.20 times times slowerslower than helium. What is it’s molar mass? than helium. What is it’s molar mass? Steps on whiteboard to get . . .Steps on whiteboard to get . . .

40.96 (Probably Argon = 39.948 g/mol)40.96 (Probably Argon = 39.948 g/mol) 0.00251 mol NH0.00251 mol NH33 effuses via a hole in 2.47 min, how effuses via a hole in 2.47 min, how

much HCl would effuse in the same time? Your ans.?much HCl would effuse in the same time? Your ans.? 0.0017 mol HCl0.0017 mol HCl Some NSome N22 effuses via a hole in 38 seconds. What is effuses via a hole in 38 seconds. What is

molecular weight of gas that effuses in 55 seconds molecular weight of gas that effuses in 55 seconds under identical conditions? Your answer? . . . (hint: under identical conditions? Your answer? . . . (hint: think if Nthink if N22 is faster or slower. If faster then the rate is faster or slower. If faster then the rate must be > 1, if slower, then the rate is a fraction.must be > 1, if slower, then the rate is a fraction.

59 g/mol59 g/mol

86

5.8 Real Gases5.8 Real Gases Real molecules Real molecules dodo take up space and take up space and

they they dodo interact with each other interact with each other (especially polar molecules).(especially polar molecules).

Need to add Need to add correction factorscorrection factors to the to the ideal gas law to account for these.ideal gas law to account for these.

87

VolumeVolume Correction Correction The actual volume free to move in is less The actual volume free to move in is less

because of particle size.because of particle size. More molecules (More molecules (nn) will have more effect.) will have more effect. Corrected volume V’ = V - nb. Substitute Corrected volume V’ = V - nb. Substitute

this for V in PV = nRTthis for V in PV = nRT bb is a constant that differs for each gas. is a constant that differs for each gas.

PP’’ = = nRTnRT

(V- (V-nnbb))

88

PressurePressure correction correction Because the molecules are attracted to Because the molecules are attracted to

each other, the pressure on the container each other, the pressure on the container will be will be lessless than ideal than ideal

Depends on the number of molecules per Depends on the number of molecules per liter (liter (n ÷ Vn ÷ V). “a” is a unique constant.). “a” is a unique constant.

Since Since twotwo molecules interact, the effect molecules interact, the effect must be must be squaredsquared..

89

Pressure correctionPressure correction

Because the molecules are attracted to Because the molecules are attracted to each other, the pressure on the container each other, the pressure on the container will be less than idealwill be less than ideal

Depends on the number of molecules per Depends on the number of molecules per liter (liter (n ÷ Vn ÷ V). “a” is a unique constant). “a” is a unique constant

Since two molecules interact, the effect Since two molecules interact, the effect must be squared.must be squared.

Pobserved = P’ - a

2

( )Vn

90

AltogetherAltogether PPobsobs= nRT - a n = nRT - a n 22 V-V-

nb Vnb V

Called the Van der Waal’s equation if Called the Van der Waal’s equation if

rearrangedrearranged Corrected Corrected Corrected Corrected

Pressure Pressure Volume Volume “P” “P” x x “V”“V” == nRTnRT

( )

P + an

V x V - nb nRTobs

2

91

Where does it come fromWhere does it come from a and b are determined by a and b are determined by

experiment.experiment. Different for each gas.Different for each gas. Bigger molecules have larger b.Bigger molecules have larger b. a depends on both size and polarity.a depends on both size and polarity. once given, plug and chug.once given, plug and chug.

92

Figure 5.24 p. 222Figure 5.24 p. 222Plots of Plots of PV/nRTPV/nRT Versus Versus PP for Several Gases for Several Gases (200 K)(200 K)

Behavior is close to Behavior is close to ideal ONLY at low ideal ONLY at low pressures (less than pressures (less than 1 atm)1 atm)

93

Figure 5.25Figure 5.25Plots of Plots of PV/nRTPV/nRT Versus Versus PP for for Nitrogen Gas Nitrogen Gas at Three at Three TemperaturesTemperaturesDeviations Deviations are smaller at are smaller at higher higher temperatures.temperatures.

94

Real Real vsvs Ideal Behavior (test question) Ideal Behavior (test question) Gases act most ideally if at:Gases act most ideally if at: Low pressureLow pressure High temperature, and have . . . High temperature, and have . . . Low mass and . . .Low mass and . . . Low polarityLow polarity This provides least deviation from the This provides least deviation from the

expected collisions derived from the expected collisions derived from the kinetic molecular theory.kinetic molecular theory.

Be able to explain Be able to explain whywhy this is so. this is so.

95

ExampleExample Calculate the pressure exerted by 0.5000 Calculate the pressure exerted by 0.5000

mol Clmol Cl22 in a 1.000 L container at 25.0ºC in a 1.000 L container at 25.0ºC Using the Using the idealideal gas law. . . gas law. . . P = nRT/V = P = nRT/V = 12.2 atm12.2 atm Using Using Van der Waal’sVan der Waal’s equation, where equation, where

– a = 6.49 atm La = 6.49 atm L22 /mol /mol22

– b = 0.0562 L/mol . . .b = 0.0562 L/mol . . . P = nRT/V - nb - a (n/v)P = nRT/V - nb - a (n/v)22 = = 11.0 atm11.0 atm Note the difference.Note the difference.

96

5.9 Chemistry in the Atmosphere5.9 Chemistry in the Atmosphere Acid RainAcid Rain There will be AP questions on this. . .There will be AP questions on this. . .

97

Sources of Acid Rain

l Power stations

l Oil refineries

l Coal with high S content

l Car and truck emissions

l Bacterial decomposition, and lightning

hitting N2 in the atmosphere.

98

Acid Rain

l Unpolluted rain has a pH of 5.6

l Rain with a pH below 5.6 is “acid rain“

l CO2 in the air forms carbonic acid

CO2 + H2O H2CO3

l Adds to H+ of rain

H2CO3 H+ (aq) + HCO3-(aq)

99

SO2 26 million tons in 1980

NO and NO2 22 million tons in 1980

Mt. St Helens (1980) 400,000 tons SO2

l Reactions with oxygen in air form SO3

2SO2 + O2 2 SO3

l Reactions with water in air form acids

SO3 + H2O H2SO4 sulfuric acid

NO + H2O HNO2 nitrous acid

HNO2 + H2O HNO3 nitric acid

Sources of Acid Rain

100

Effects of Acid Rain

l Leaches Al from soil, which kills fish

l Fish kills in spring from runoff due to accumulation

of large amounts of acid in snow

l Dissolves waxy coatings that protect leaves from

bacteria

l Corrodes metals, textiles, paper and leather

l Dissolves minerals Mg, Ca, and K from the soil and

waxy coatings that protect leaves from bacteria.

l Read 5.9 in book for the rest of the story

Education

Ch5 z5e gases