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Unit 6 (Chp 10):
Gases
John Bookstaver
St. Charles Community College
St. Peters, MO
2006, Prentice Hall, Inc.
Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Characteristics of Gases
• Unlike liquids and solids, they… expand to fill their containers.
(indefinite volume) are highly compressible. have extremely low densities.
Atmospheric Pressure:weight of air per area
Pressure• Pressure is the amount
of force applied per area.
P =FA
22,000 lbs!!!(per sq. meter)
Atmospheric Pressure
(weight of air)
empty space(a vacuum)
h760 mm
1 atm = 760 mmHg= 760 torr= 101.3 kPa
1 N1 m2
Units (at sea level)
Pressure
STP(standard T & P)273 K 1 atm
1) 657 mmHg to atm2) 830 torr to atm3) 0.59 atm to torr
657760
830760
0.59 x 760
Kinetic-Molecular Theory
KMT is a model which explains the
Properties(P, V, T, n)
and
Behavior(motion, energy, speed, collisions)
of gases.
2) Gas pressure is caused by collisions with the container walls.
P = FA
P = FA
P = FA
1) Gas particles are in constant random motion.
5 Parts of Kinetic-Molecular Theory
Collisions are elastic (no KE lost). (ideally)
Ideally: Vgas = Vcontainer
5 Parts of Kinetic-Molecular Theory
4) Volume of gas particles is negligible, compared to total volume of container.
Vgas = Vcontainer – Vparticles
In a 1.000 L container,
gas only expands into about 0.999 L of
volume
(negligible)1.000 L container
has 1.000 L of gas
3) Attractive forces (IMAFs) are negligible.
IMAFs
5) Average KE of gas particles is……directly proportional to Kelvin Temp.
(K not oC)no negative temp’s, no negative energies, no negative volumes, etc.)
video clip
KEavg α T
5 Parts of Kinetic-Molecular Theory
Boyle’s Law (P & V)
1 atm 2 atm
10 L
5 L
(inversely proportional)
P ↑ , V ↓
(directly proportional)
Charles’ Law (V & T)
how absolute zero was estimated
150 K
60 L30 L
300 K
T ↑ , V ↑
(directly proportional)
Lussac’s Law (P & T)
300 K
100 kPa 500 kPa
600 K
200 kPa
T ↑ , P ↑
Avogadro's Hypothesis• At the same ___ & ___, equal _________ of
gas must contain equal _________.
CO2
HeO2
1 mol He
2 mol Hen ↑ , V ↑
Avogadro's Law (V & n)
All at:P = 1 atmT = 25oCV = 1.0 L
moles (n)
add gas
volumesT P(particles)
(directly proportional)
V 1/P (Boyle’s law)V T (Charles’s law)P T (Lussac’s law)V n (Avogadro’s law)
• So far we’ve seen that
PVnT
= R
ideal gas constant: R = 0.08206 L∙atm/mol∙K
constant
PV = nRTIdeal Gas Law
(all gases same ratio)
NO Units of : mL , mmHg , kPa , grams , °C
given on exam
(changes in P,V,T,n)
P ↑ , V ↓ T ↑ , V ↑
T ↑ , P ↑
n ↑ , V ↑
(directly proportional)(inversely proportional)
NOTgiven on examPV = nRT
= Rconstant
given on exam
P2V2
n2T2
P1V1
n1T1
==P1V1
n1T1
P2V2
n2T2(initial) (final)
1. The pressure on a 411 mL sample of gas is decreased from 812 torr to 790 torr. What will be the new volume of the gas?
= 422 mLP1
P2
V2
V1
P2V2
n2T2
P1V1
n1T1
=
= ?
(812)(411) = (790) V2
(812)(411)(790)
= V2
Ideal-Gas Changes PV = nRT
PVnT
= R
2. A 10.0 L sample of a gas is collected at 25oC and then cooled to a new volume of 8.83 L while the pressure remains at 1.20 atm. What is the final temperature in oC?
V1
T1
V2 T2P2V2
n2T2
P1V1
n1T1
== ?
(10.0)(298)
= (8.83)T2
T2 (10.0) = (8.83)(298)
(8.83)(298)(10.0)
=T2
Ideal-Gas Changes
= 263 K = –10 oC
PV = nRT
PVnT
= R
0.502 mol O2 x 2 mol O3 =
3 mol O2
3. A 13.1 L sample of 0.502 moles of O2 is held under conditions of 1.00 atm and 25.0oC. If all of the O2 is converted to Ozone (O3), what will be the volume of O3? = 8.73 L
n1
n2 = ?
V2
V1
O2 O33 2
0.335mol O3
n2
n1
P2V2
n2T2
P1V1
n1T1
= (13.1)(0.502)
= V2 .
(0.335)
Ideal-Gas Changes PV = nRTHW p. 432 # 1, 23, 89, 26
PVnT
= R
Ideal-Gas Equation PV = nRT1. A 5.00 L He balloon has 1.20 atm at 0.00oC. How many moles of He gas are in the balloon?
(1.20 atm)(5.00 L)
= n (0.08206)(273 K)
(1.20)(5.00)(0.08206)(273)
= n
n = 0.268 mol He
PV = nRT R = 0.08206L∙atm∙mol–1∙K–1
How many atoms
of He?
PT
V
n
Ideal-Gas Equation
Molar Mass
2. A sample of aluminum chloride gas weighing 0.0500 g at 350.oC and 760 mmHg of pressure occupies a volume of 19.2 mL.Calculate the Molar Mass of the gas.
133 g/mol
(1.00 atm)(0.0192 L)
= n (0.08206)(623 K)
n = 0.000376 mol
PV = nRT R = 0.08206 L∙atmmol∙K
gramsmole
M = __0.0500 g_0.000376 mol
=AlCl3 =
HW p. 438 #92, 29, 46, 38, 35
n = mM
M = mn
so…
(given on exam)
mTPVM = ?
PV = nRT
• The volume of 1 mole of any gas at STP will be:
Vm = _____
PVm = nRT
the ______ of ______of any gas at ____.
volume 1 mole
STP
(1.00 atm) Vm
= (1 mol )(0.08206)(273 K)
Molar Volume:
22.4 L1 mol
but…ONLY at STP!!!
Gas Stoich with Molar VolumeThe reactions below occurred at STP.1. Calculate the mass of NH4CI reacted with
Ca(OH)2 to produce 11.6 L of NH3(g) .
2 NH4Cl + Ca(OH)2 2 NH3 + CaCI2 + 2 H2O
2. Calculate the volume of CO2 gas produced when 9.85 g of BaCO3 is decomposed.
BaCO3(s) BaO(s) + CO2(g)
27.7 g
1.12 L
11.6 L NH3 x 1 mol NH3 x22.4 L NH3
2 mol NH4Cl x 2 mol NH3
53.49 g NH4Cl = 1 mol NH4Cl
9.85 g BaCO3 x 1 mol BaCO3 x197.34 g BaCO3
1 mol CO2 x 1 mol BaCO3
22.4 L CO2 = 1 mol CO2
3. What volume of O2 gas is produced from
490 g KClO3 at 298 K and 1.06 atm?
KClO3(s) KClO(s) + O2(g)
= ____L O2
490 g KClO3 xg KClO3 mol KClO3
mol KClO3 x mol O2 mol O2122.55
=1
1 1
92.3 L O2
NOT at STP
Molar Mass of KClO3 is 122.55 g/mol
PV = nRTuse…
4.00
(1.06 atm) V
= (4.00 mol )(0.08206)(298 K)
Mole Fraction (XA)
PA = Ptotal x XA
XA =moles of A
total moles
• The mole fraction (XA) is like a % of total moles that is A, but without the % or x 100.
WS 6b#1-4
Ptotal = PA + PB + PC + …
Dalton’s Law of Partial Pressures
HWp. 436
#63 64 66
Ptotal = PH2O + Pgas
• When one collects a gas over water, there is water vapor mixed in with the gas.
• To find only the pressure of the gas, one must subtract the water vapor pressure from the total pressure. Pgas = Ptotal – PH2O
equalize water level inside & outside Ptot Patm=
1. Calculate the mass of 0.641 L of H2 gas collected over water at 21.0oC with a total pressure of 750. torr.The vapor pressure of water at 21.0oC is 20.0 torr.
Ptotal = PH2O + Pgas PV = nRT
750. = 20.0 + PH2
PH2 = 730. torr
PH2 = 730/760 = 0.961 atm
(0.961)(0.641) = nH2 (0.08206)(294)
nH2 = 0.0255 mol
mH2 = 0.0514 g
↑ T , ↑ v↑ M , ↓ v
Study the models below. What can be said quantitatively about the molecular speed (v) of a gas in relation to its molar mass (M) and its temperature (T)?
Compare the molecular speed (v) of these gases:
1) at 25.0 oC (i) Helium (ii) Oxygen (O2)
2) at 50.0 oC (i) Helium (ii) Oxygen (O2)
3) Does the data support your conclusions from the models on the previous slide about effects of T and M on v ? ↑ T , ↑ v↑ M , ↓ v
1360 m/s 482 m/s
1420 m/s 502 m/s
KE = ½ mv2
(given on exam)
WHY?
Distributions of Molecular Speed
average molecular speed (v)
KE = ½ mv2
(KMT)
(given on exam)
Therefore:
T & v are __________proportional
directly
Temp (K) & KEavg
are directly proportional T α KEavg
↑ T , ↑ v
Gases at the same Temp, have the same _____.
Speed vs. Molar MassKEavg
½ mv2 = ½ mv2
KE1 = ½ m1v12
Ar: M = 40 g/mol He: M = 4.0 g/mol
KE = ½ mv2
(at same T)
KEHe = KEAr
KE2 = ½ m2v22
↑ M , ↓ v ↓ M , ↑ v
M & v are __________proportional
inversely
Effusion
escape of gas particles
through a tiny hole
spread of gas particles throughout
a space
Diffusion↑T, ↑v
↑M,↓v
KE = ½ mv2
64 g/mol (SO2)16 g/mol (CH4)
_____ is __ times faster than _____.
CH4
SO2
2
KE = ½ mv2
½ mv2 = ½ mv2
HW p.437 #8,74,76a
(usually)
Real (non-Ideal) GasesIn the real world, the behavior of gases only conforms to the ideal-gas equation under “ideal” conditions.
Non-Ideal: (Low T) (High P)
Ideal: (High T) (Low P)
WHY?
(weaker IMAFs)
(stronger IMAFs)
(negligible)
(not negligible)
(high KE)(high Vtotal)
(low KE) (low Vtotal)
(ONLY under ideal conditions)
PV = nRT
Ideal Gas vs Non-Ideal Gas
) (V ) = nRT(P“observed” P too low b/c attractive forces
not negligible,collisions less
frequent and of less force
“observed” V too high b/csize of particlesnot negligible
compared to total volume
(ideal P) (ideal V)
NON
T: ↓P: ↑
IDEAL
T: ↑P: ↓
more KE/speedweaker IMAFsmore avg. dist.
less KE/speedstronger IMAFsless avg. dist.
IMAFs
Vgas = Vcontainer
n2aV2 − nb+
HW p.437 #81,82,83
– Vparticles