Lecture 22: Gases 1
Read:
HW:
BLB 10.1–10.5
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
Fina l Exa m: 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])
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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
(T),
(V), & #
(P),
(n)
• all gases behave similarly at low
• gases will mix in all proportions with other
gases to form
mixtures
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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
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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]
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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
V
n
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Lecture 22
The ideal gas law (cont.)
• Boyle:
V ∝ 1/P
PV = constant (T,n fixed)
V
P
V
1 atm
2 atm
• Charles: V ∝ T
V/T = constant (P,n fixed)
note: T in absolute T (K)!
1/P
4 atm
V
-273.15°C = 0 K
T (°C)
• All of these laws can be described by one
equation: PV = nRT
• R = gas constant; units of R are very, very
important!
L atm
J
R = 0.08206
R = 8.314
mol K
mol K
cal
R =1.987
mol K
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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
0.08206L atm
R=
mol K
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PV = nRT
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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?
1L $
!
V1 = 750mL #
= 0.75L
" 1000mL &%
V2 = 54.5L
P1 = 105atm
T2 = !10° + 273 = 263K
T1 = 27° + 273 = 300K
P2 = ?
PV = nRT so P1V1 = nRT1
n is not changing & R is a constant, so...
nR =
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P1V1
T1
Page 8
& nR =
P2V2
T2
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
d = 1.17g / L
torr.
! 1atm $
P = ( 740.0torr ) #
= 0.974atm
" 760torr &%
d = PM / RT
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T = 21° + 273 = 294K
M =?
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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
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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 Ptot = P1 + P2 + P3 + ...
P=
Ptot
Ptot
ntot
nRT
V
n RT n2RT n3RT
= 1
+
+
+ ...
V
V
V
RT
= ( n1 + n2 + n3 + ...)
V
= n1 + n2 + n3 + ...
Ptot =
ntot RT
V
• mole fraction: ratio of na/ntot;
dimensionless
Xa = na/ntot
X 1 = n1 / ntot
n1RT
P1 =
V
n
P1 = 1 Ptot
ntot
note: mole fractions must sum
n1 ! ntot RT $
P
=
to 1
1
n #" V &%
tot
P1 = X 1Ptot
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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?
nCO 2 = 5mol
nN 2 = 2mol
nCl 2 = 1mol
Ptot =
ntot RT
V
Ptot =
ntot = 8mol
X CO 2 =
XN2 =
X Cl 2 =
[ check: ! X = 1.000 ]
V = 40L
T2 = 0° + 273 = 273K
Ptot = ?
therefore,
PCO 2 =
PN 2 =
PCl 2 =
[ check: ! P = P
tot
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Page 12
= 4.48atm
]
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?
He (initial):
VHe = 3L
PHe = 5.0atm
T = 25° + 273 = 298K
nHe =
PHeVHe
RT
nHe =
ntot = nHe + nNe
ntot =
therefore,
X He =
X Ne =
note: do not need X Ne for problem
Ne (initial):
VNe = 4.5L
[ check: ! X
= 1.000
]
PNe = 2.0atm
T = 25° + 273 = 298K
nNe =
PNeVNe
RT
nNe =
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Lecture 22
Ptot in 10L vessel: (final)
Vtot = 10L
Ptot = ??
ntot = 0.981mol
T = 25° + 273 = 298K
Ptot =
ntot RT
Vtot
Ptot =
Now we need to find PHe (final)
PHe =
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Lecture 22
Before next class:
Read:
HW:
BLB 10.6–9
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
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Lecture 22
Lecture 23: Gases 2
Read:
Read:
HW:
BLB 10.6–9
BLB 10.6–9
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
Fina l Exa m: 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
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Lecture 23
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 =
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Lecture 23
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)
!=
1
mu 2
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)
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Lecture 23
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
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Lecture 23
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
mu 2
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
3RT
mu 2 =
2
2N
N = Avagadroʼs number
• at a given T, all gases have same average
KE, independent of m & u; that is as m↑, u
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Lecture 23
Diffusion & effusion
• from KMT:
3RT
u=
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
M2
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
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Lecture 23
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!!?!
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Lecture 23
Before next class:
Read:
HW:
BLB 10.6–9
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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Lecture 23