Lecture 22: Gases 1 - Pennsylvania State...

Sheets Lecture 22

Lecture 22: Gases 1 Read: BLB 10.1–10.5 HW: BLB 10.2,19a,b,23,26,30,39,41,45,49 Sup 10:1–6 Know: • gases & gas laws • PV=nRT(!!!!) • partial pressures • density & molecular mass Exam #2: “NO SCORE”? Not a problem. Please talk with Mike Joyce in 210 Whitmore Final Exam: Wednesday, December 17 @ 8 am; MUST register on elion for a final exam conflict or overload by Oct 26. See http://www.registrar.psu.edu/exams/exam_overload.cfm http://www.psu.edu/dus/handbook/exam.html#conflict Need help?? Get help!! TAs in CRC (211 Whitmore) and SI—hours on Chem 110 website; my office hours (Mon 12:30-2 & Tues 10:30-12 in 324 Chem Bldg [or 326 Chem])

Sheets Lecture 22

Itʼs a gas, gas, gas gas molecules: • are constantly • are far apart (10 times as far as they are big;

occupy < 0.1% of the total volume) • move in lines from collision to collision • move at higher temperatures • move at lower temperatures as a result: • expand in whatever volume is available • have pressure (collisions with walls) • mix completely with one another • can be easily compressed • when cooled, will eventually condense

(compression may be necessary) • gases are described in terms of (P),

(T), (V), & # (n) • all gases behave similarly at low • gases will mix in all proportions with other gases to form mixtures

Sheets Lecture 22

Pressure • pressure: force/unit area

force: kg•m•s–2, Newton (N) unit area: m2

• SI unit for pressure: 1 N•m–2 = 1 Pa (Pascal) • standard atmospheric pressure**: 1 atm ≡ 1.013 × 105 Pa 1 atm = 760 torr (or mm Hg) [1 atm = 14.7 lb/in2] **real atmospheric pressure varies with altitude, temperature and weather • how big is one atm? 10 tons / m2! or 600 lbs/head

Sheets Lecture 22

Measuring pressure • barometer used to measure Patm by balancing forces; (Torricelli, early 1600ʼs)

• measure P in terms of height of Hg 1 atm ≡ 760 torr “=” 760 mm Hg

[know this conversion!] [factoid: 760 mm Hg = 10 m H2O]

Sheets Lecture 22

The ideal gas law • state of gas described by: P, V, T in K (degrees Kelvin), & n • STP: standard temperature (273.15 K) & pressure (1 atm) NOTE: STP for gases is not the same as standard state conditions used for ΔH!! (more in Chap 5) • absolute temperature in Kelvin (K):

°C + 273.15 = K • Avagadro: V ∝ n

V/n = constant* (P,T fixed) *proportionality constant is the same for ALL gases

a

V

n

Sheets Lecture 22

The ideal gas law (cont.) • Boyle: V ∝ 1/P

PV = constant (T,n fixed)

• Charles: V ∝ T

V/T = constant (P,n fixed) note: T in absolute T (K)! • All of these laws can be described by one equation: PV = nRT • R = gas constant; units of R are very, very important!

R =0.08206Latm

mol K

R =8.314J

mol K

R =1.987cal

mol K

a

V

T (°C)

-273.15°C = 0 K

a

1 atm 2 atm 4 atm

a

V

P

a

V

1/P

Sheets Lecture 22

Example: • for problems: given known quantities, solve for the unknown What is the volume (V) occupied by 1.00 mole of gas at exactly 0°C and 1 atm (STP)?

V = ?

P = 1atm

n = 1mol

T = 0° + 273 = 273K

R =0.08206L atm

mol K

PV = nRT

Sheets Lecture 22

Example: The gas in a 750 mL vessel at 105 atm and 27°C is expanded into a vessel of 54.5 L and –10°C. What is the final pressure?

V1= 750mL

1L

1000mL

!"#

$%&= 0.75L

P1= 105atm

T1= 27° + 273 = 300K

V2= 54.5L

P2= ?

T2= !10° + 273 = 263K

PV = nRT so P1V1= nRT

1

n is not changing & R is a constant, so...

nR =P1V1

T1

& nR =P2V2

T2

Sheets Lecture 22

Density and molecular weight Example: can derive d = PM/RT from PV=nRT where d = density M = molar mass You should derive this on your own! (see BLB pp 406–407) Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21°C and 740.0 torr.

d = 1.17g / L

P = 740.0torr( )1atm

760torr

!"#

$%&= 0.974atm

T = 21° + 273 = 294K

M = ?d = PM / RT

Sheets Lecture 22

Example: What is the density of oxygen, in grams per liter, at 25°C and 0.850 atm?

d = ? need g / L

P = 0.850atm

T = 25° + 273 = 298K

M = 32g /mol

Sheets Lecture 22

Daltonʼs law of partial pressures • partial pressure: the pressure a gas would have if it was the only gas in the container • Daltonʼs law of partial pressures: total pressure is equal to the sum of partial pressures • mole fraction: ratio of na/ntot; dimensionless Xa = na/ntot note: mole fractions must sum to 1

Ptot

= P1+P

2+P

3+ ...

P =nRT

V

Ptot

=n1RT

V+n2RT

V+n3RT

V+ ...

Ptot

= n1+ n

2+ n

3+ ...( )

RT

V

ntot

= n1+ n

2+ n

3+ ...

Ptot

=ntotRT

V X1= n

1/ n

tot

P1=n1RT

V

P1=n1

ntot

Ptot

P1=n1

ntot

ntotRT

V

!"#

$%&

P1= X

1Ptot

Sheets Lecture 22

Example: Mix 5 moles of CO2, 2 moles of N2 and 1 mole of Cl2 in a 40 L container at 0°C. (a) What is the pressure in the container? (b) What is the partial pressure of each gas in the container? (c) What volume does each gas occupy in the container?

nCO2

= 5mol

nN 2

= 2mol

nCl 2

= 1mol

ntot

= 8mol

XCO2

=

XN 2

=

XCl 2

=

check: X! = 1.000[ ]

V = 40L

T2= 0° + 273 = 273K

Ptot

= ?

Ptot

=ntotRT

V

Ptot

=

therefore,

PCO2

=

PN 2

=

PCl 2

=

check: P! = Ptot

= 4.48atm[ ]

Sheets Lecture 22

Example: 3.0 L of He at 5.0 atm and 25°C are combined with 4.5 L of Ne at 2.0 atm and 25°C in a 10 L vessel at constant temperature. What is the partial pressure of the He in the 10 L vessel?

on to next page…

He (initial):

VHe

= 3L

PHe

= 5.0atm

T = 25° + 273 = 298K

nHe

=PHeVHe

RT

nHe

=

Ne (initial):

VNe

= 4.5L

PNe

= 2.0atm

T = 25° + 273 = 298K

nNe

=PNeVNe

RT

nNe

=

ntot

= nHe

+ nNe

ntot

=

therefore,

XHe

=

XNe

=

note: do not need XNe

for problem

check: X! = 1.000[ ]

Sheets Lecture 22

Ptot

in 10L vessel: (final)

Vtot

= 10L

Ptot

= ??

ntot

= 0.981mol

T = 25° + 273 = 298K

Ptot

=ntotRT

Vtot

Ptot

=

Now we need to find PHe

(final)

PHe

=

Sheets Lecture 22

Before next class: Read: BLB 10.6–9 HW: BLB 10:5,8,59,61,71,75,77,82,83,84 Sup 10:7–15 Know: • kinetic-molecular theory • effusion & diffusion • real gases (van der Waals) Answers: p. 7: 22.4 L p. 8: Pfinal = 1.27 atm p. 9: 29.0 g/mol p. 10: 1.11 g/L p. 12: (a) Ptot = 4.48 atm; (b) PCO2 = 2.80 atm, PN2 = 1.12 atm,

PCl2 = 0.56 atm; (c) 40 L p. 13: 1.5 atm

Sheets Page 1 Lecture 23

Lecture 23: Gases 2 Read: BLB 10.6–9 Read: BLB 10.6–9 HW: BLB 10:5,8,59,61,71,75,77,82,83,84 Sup 10:7–15 Know: • kinetic-molecular theory • effusion & diffusion • real gases (van der Waals) Exam #2: “NO SCORE”? Not a problem. Please talk with Mike Joyce in 210 Whitmore Final Exam: Wednesday, December 17 @ 8 am; MUST register on elion for a final exam conflict or overload by Oct 26. See http://www.registrar.psu.edu/exams/exam_overload.cfm http://www.psu.edu/dus/handbook/exam.html#conflict Need help?? Get help!! TAs in CRC (211 Whitmore) and SI—hours on Chem 110 website; my office hours (Mon 12:30-2 & Tues 10:30-12 in 324 Chem Bldg [or 326 Chem]) Bonus deadline for BST #8: Intermolecular forces, Oct. 30


Collecting a gas over water Example: During a reaction, N2 is collected over H2O. How many milligrams of nitrogen were collected under these conditions? Pbar = 742 torr (=Ptot); V = 55.7 mL; T = 23°C (296 K) NOTE: Ptot = Pgas + PH2O vapor vapor pressure at 23°C = 21.07 torr (BLB Appendix B)

PN 2

=

PN 2

=

nN 2

=PN 2V

RT

nN 2

=


Kinetic-molecular theory (KMT) • PV = nRT explains how gases behave • KMT explains why gases behave they way they do; gives a view of gases on a molecular level The 5 key postulates of KMT 1. -line motion in random directions 2. molecules are —volume they occupy is small compared to the total volume 3. intermolecular forces—molecules are not “sticky” 4. collisions 5. mean kinetic energy ε ∝ T (in K)

ε = average kinetic energy of a molecule m = mass of molecule (in kg) u = root mean squared (rms) speed of a

molecule (in m/s)

! = 12mu

2


Kinetic-molecular theory (cont.) • KMT provides molecular level explanation for why gases behave as they do • T ↑ at constant (V, n) ⇒ P

T ↑ , ε ↑, u ↑; thus: more collisions per unit time & and harder collisions, so P

• V ↑ at constant (T, n) ⇒ P

constant T means constant ε & u; thus, longer distances between collisions & fewer collisions per unit time with walls, so P


Temperature & molecular speeds • some molecules move slowly & some molecules move quickly: mean speed is in middle of of molecular speeds • entire distribution (& mean) shift ↑ for higher temperature; that is, when T ↑, fewer molecules move slowly, more molecules move quickly, so average speed ↑

Temperature & kinetic energy

T ! = 1

2mu

2

ε = average kinetic energy of a molecule

m = mass of molecule (in kg) u = root mean squared (rms) speed of a

molecule (in m/s)

! =1

2mu

2=3RT

2N

N = Avagadroʼs number

• at a given T, all gases have same average KE, independent of m & u; that is as m↑ , u


Diffusion & effusion

• from KMT:

u =3RT

M u = rms speed M = molar mass (this is not m from previous page!) • thus, gases with lower molar mass have higher rms speed • Grahamʼs law of effusion: r1

r2

=M

2

M1

• effusion: escape of gas through a pinhole • diffusion: spread of one gas throughout a space or second substance; even though diffusion is more complicated (due to gas molecular collisions), it still obeys Grahamʼs law


Collisions & diffusion • diffusion: movement of molecules interrupted by collisions • rate of diffusion follows Grahamʼs Law: depends on molecular speed & therefore, [1/M]0.5 • at STP, molecules collide ~1010 times/s for each molecule! • this means that although molecular speeds are high at STP, molecules donʼt go very far net distance traveled << (speed × time) • mean free path (mfp): distance traveled between collisions (~50 nm at 1 atm); mfp↓ as density (& P) ↑ factoid: if there were no collisions, O2 would traverse this room (~20 m) in ~0.04 s. BUT, in the absence of drafts, it takes many, many days!!?!


Before next class: Read: BLB 10.6–9 HW: BLB 10:5,8,59,61,71,75,77,82,83,84 Sup 10:7–15 Know: • kinetic-molecular theory of real gases • real gases (van der Waals) Answers: p. 2: 60.9 mg

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