10/21/2019
1
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)
1
2
3
10/21/2019
2
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 modelwhich 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 constantrandom motion.
5 Parts of Kinetic-Molecular Theory
Collisions are elastic(no KE lost).
(ideally)
4
5
6
10/21/2019
3
Ideally: Vgas = Vcontainer
5 Parts of Kinetic-Molecular Theory
4) Volume of 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.)
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 ↓
7
8
9
10/21/2019
4
(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)
10
11
12
10/21/2019
5
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)
PV = nRT
= Rconstant
given on exam
=P1V1
n1T1
P2V2
n2T2(initial) (final)
1. The pressure on a 411 mL sample of gas is decreased from 812 torr to 790 torr. How will volume of the gas be affected?
P1
P2
V2
V1
PV = nRT
= ?
(812)(0.411) = nRT
Ideal-Gas Changes PV = nRT
(790) (?) = nRT V increased
13
14
15
10/21/2019
6
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. How will the temperature in be affected?
V1
T1
V2 T2 = ?
Ideal-Gas Changes PV = nRT
PV = nRT
(1.20)(10.0)
(1.20)(8.83) = nR(?) T decreased
= nR(298)
=
HW p. 432 #1, 23, 26
Ideal-Gas Calc’s 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 Calc’s
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
(760 mmHg)(0.0192 L) = n (62.36)(623 K)
n = 0.000376 mol
PV = nRT R = 62.36 L∙torrmol∙K
gramsmole
M = __0.0500 g_0.000376 mol
=AlCl3 =
n = mM
M = mn
so…
(given on exam)
mTPVM = ?
PV = nRT
16
17
18
10/21/2019
7
TPVV = ?@STP
Ideal-Gas Calc’s
3. A gas at 25.0oC and 1.20 atm occupies a volume of 4.28 L.What volume will it occupy at STP?
(1.20 atm)(4.28 L) = n (0.08206)(298 K)
n = 0.210 mol
PV = nRTR = 0.08206 L∙atm
mol∙K
PV = nRT
HW p. 438 #92, 29,
46, 35, 38
(1.00)V = (0.210)(0.08206)(273)
V = 4.70 L
• The volume of 1 mole of any gas at STPis:
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…ONLYat STP!!!
or22.4 L/mol22.4 L∙mol–1
Gas Stoich with Molar Volume1. Calculate the mass of NH4CI reacted with
Ca(OH)2 to produce 11.6 L of NH3(g) at STP.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 at STP.
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 x2 mol NH3
53.49 g NH4Cl =1 mol NH4Cl
9.85 g BaCO3 x 1 mol BaCO3 x197.34 g BaCO3
1 mol CO2 x1 mol BaCO3
22.4 L CO2 =1 mol CO2
19
20
21
10/21/2019
8
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)
V = ____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)(would be 89.6 L
if at STP)
Mole Fraction (XA)
PA = Ptotal x XA
XA =moles of A
total moles
• The mole fraction (XA) is like the % 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
#636466
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 levelsinside & outside =
22
23
24
10/21/2019
9
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?
25
26
27
10/21/2019
10
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)
KEAr = KEHe
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.433 #8, 74, 76a
28
29
30
10/21/2019
11
(usually)
“Real” (non-Ideal) GasesIn the real world, the behavior of gases only conforms to the ideal-gas equation under “ideal” conditions.
Ideal: (High T) (Low P)
WHY?(weak IMAFs)
(negligible)
(high KE) (high Vtotal)
(ONLY under ideal conditions)
PV = nRT
(tiny Vparticles)(negligible)
Ideal Gas vs Non-Ideal Gas
) (V ) = nRT(P
“observed” P too lowb/c at low Temp
IMAFs not negligible,collisions less frequent
and of less force
“observed” V too highb/c at high P,
particle volumenot negligible
compared to total volume
(ideal P) (ideal V)
NON
T: ↓P: ↑
IDEAL
T: ↑P: ↓
more KEovercome IMAFsmore total volume
less KEsignificant IMAFsless total volume
IMAFs
Vgas = Vcontainer
n2aV2 − nb+
HW p.437 #81,82,83
– Vparticles
Slides #35-37 demonstrate calculation of gas changes
using the combined gas law.
This is no longer required on the new AP Exam
31
32
33
10/21/2019
12
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
34
35
36
10/21/2019
13
PV = nRT
0.500 mol O2 x 2 mol O3 = 3 mol O2
3. A 12.2 L sample of 0.500 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?
n1
n2 = ?
V2
V1
O2 O33 20.333mol O3
n2n1
Ideal-Gas Changes PV = nRT
(1.00)(12.2) = (0.500) R (298)(1.00) (?) = (?) R (298)(1.00) V =
V = 8.19 L(0.333) R (298)
37