> rtq7 bjbjUU "~77>lzzzzzzz###8$$>$TF=%%"%%<&'''<<<<<<<$B? bA<z'''''<k*zz%<&=k*k*k*'z%z<&<k*'<k*k*,r6Tzz7%%`)RT #'77=0F=+7A)A7k*zzzzGayLussacs Law and Avogadros Law
CSCOPE Unit 09 Lesson 01 Day 3
Vocabulary
Avogadros LawAt constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas.Boyles LawFor a given mass of gas at constant temperature, the volume of a gas varies inversely with the pressure.Charles LawFor a given mass of gas at constant pressure, the volume of a gas is directly proportional to the Kelvin temperature. The temperature must be in Kelvin.Combined Gas LawA gas law that combines Boyle's Law, Charles' Law, and GayLussac's Law, it states the ratio of the product of pressure and volume to the absolute temperature of a gas is equal to a constant, this gas law is used when pressure, volume, and temperature are all changing, the temperature must be in KelvinDaltons Law of Partial PressuresAt constant temperature and volume, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures. GayLussacs LawFor a given mass of gas at constant volume, the pressure is directly proportional to the Kelvin temperature.Ideal Gas LawA gas law that describes the relationships between measurable properties of an ideal gas, it describes the physical behavior of an ideal gas in terms of the temperature, volume, pressure, and number of moles of a gas that are present, this gas law is used when no variables P, V, or T are changing, the temperature must be in Kelvin
GayLussacs Law the TemperaturePressure Relationship
1. Verbal definition
For a given mass of gas at constant volume, the pressure is directly proportional to the Kelvin temperature.
The temperature must be in Kelvin.
2. Equation
P1=P2T1T2Model
A sample of a gas has a pressure of 98.9 kPa at 253 K. What is its pressure
at 298 K?
GivenFindP1 = 98.9 kPa
T1 = 253 K
T2 = 298 K P2 = ?
P1=P2T1T2
98.9 kPa=P2253 K298 K
(253 K) P2 = (98.9 kPa)(298 K)
P2 =(98.9 kPa)(298 K)253 K
P2 = 116.49 kPa
P2 = 116 kPa
Examples
How many times larger or smaller would the pressure of a sample of a gas in a container be if the Kelvin temperature of the gas were quadrupled?
A sample of a gas has a pressure of 1.005 atm at 30.0 (C. What is its pressure
at a temperature of 60.0 (C?
Avogadros Law the QuantityVolume relationship
1. Verbal definition
At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas.
2. Mathematical definition
V1=V2n1n2
Model
If 3.50 mol of a gas have a volume of 78.45 L at a certain temperature and pressure what would the volume be if 1.50 mol of the same gas were added while keeping pressure and temperature constant?
GivenFindn1 = 3.50 mol
V1 = 78.45 Ln2 = ?
V2 = ?
n2 = 3.50 mol + 1.50 mol = 5.00 mol
V1=V2n1n2
78.45 L=V23.50 mol5.00 mol
(3.50 mol) V2 = (78.45 L)( 5.00 mol)
V2 =(78.45 L)(5.00 mol)3.50 mol
V2 = 112 L
V2 = 112 L
Examples
How many times larger or smaller would the volume of a sample of a gas in a container be if the number of moles of gas in that container doubled?
A mixture of hydrogen and oxygen with a 2:1 ratio has a volume of 67.2 L. What is the volume of the water vapor when the H2 and the O2 react according to the equation:
2 H2 (g) + O2 (g) ( 2 H2O (g)
while keeping pressure and temperature constant?
Conceptual Exercises
01. How many times larger or smaller would the pressure of a sample of a gas
in a container be if the temperature of the gas went from 200. K to 400. K?
02. How many times larger or smaller would the temperature of a sample of
a gas in a container be if the pressure of the gas went from 2.0 atm to
8.0 atm?
03. How many times larger or smaller would the volume of a sample of a gas
in a container be if the number of moles of gas went from 1.00 mol to
3.00 mol?
04. How many times larger or smaller would the volume of a sample of a gas
in a container be if the number of moles of gas went from 8.0 mol to
2.0 mol?
Exercises
Procedure
1. Complete a Given and Find on your own.
2. Dont forget to convert Celsius to Kelvin as necessary.
3. Write the correct formula and plug in the numbers.
4. Show all of your math.
5. Box the final answer and dont forget to include the units.
GayLussacs Law Problems
05. A sample of a gas has a pressure of 777 mm Hg at a temperature of 99 K.
What is its pressure at 205 K?
06. A sample of a gas has a temperature of 668 K and a pressure of 0.884 atm.
If the pressure is increased to 12.75 atm, what is the new temperature?
Avogadros Law Problems
07. If 4.5 mol of a gas have a volume of 98.7 L at a certain temperature and
pressure what would the volume be if 2.5 mol of the same gas were added
while keeping pressure and temperature constant?
08. A mixture of nitrogen and oxygen in a 1:3 ratio has a volume of 4.00 L.
What is the volume of the nitrogen trioxide when the nitrogen and oxygen
react according to the equation:
N2 (g) + 3 O2 (g) ( 2 NO3 (g)
while keeping pressure and temperature constant?
Mixed Boyles Law, Charles Law, GayLussacs Law, and Avogadro Law Problems
09. At 25.0 (C a 2.50 L sample of gas has a pressure of 82.2 kPa. If the
temperature is decreased to 0.00 (C what will the new pressure be?
(NOTE: T is in (C NOT in K!!)
10. If 2.25 mol of a gas have a volume of 51.0 L at 0.00 (C and a pressure of
3.75 atm what would the volume be if 1.20 mol of the same gas were added
while keeping pressure and temperature constant?
11. At 278.2 K a sample of oxygen gas has a volume of 1125 L and a pressure
of 1229 mm Hg. If the pressure is changed to 1012 mm Hg while keeping the
temperature constant what will the new volume be?
12. At 246 (C a sample of oxygen gas has a volume of 478 mL and a pressure
of 3247 kPa. If the volume is changed to 569 mL what will the new
temperature be?
PAGE
PAGE 1
Unit 09 Lesson 01 Day 3
DOl
+
,
\
d
h
p
q
}
JKPQXYcdopz{5OJQJ\^JH*OJQJ^JOJQJ\^J>*OJQJ^J5>*OJQJ\^JOJPJQJ^J56OJQJ\]^JOJQJ^JH$CDOP_`f$$IflF`L
# 064
la<$If $$Ifa$$a$Qtk $$Ifa$$$IflpF`L
# 064
la<$IfQRSTUvvv$If$$IflF`L
# 064
la<UVcdzqkk$If $$Ifa$$$IflpF`L
# 064
la<vvv$If$$IflF`L
# 064
la<Dzqkk$If $$Ifa$$$IflpF`L
# 064
la<DEFGHvvv$If$$IflF`L
# 064
la<HIklzqkk$If $$Ifa$$$IflpF`L
# 064
la<vvv$If$$IflF`L
# 064
la<tzqkk$If $$Ifa$$$IflpF`L
# 064
la<tuvwxvvv$If$$IflF`L
# 064
la<xy zqkk$If $$Ifa$$$IflpF`L
# 064
la<
,
zzzztzjz` ^` ^`p^p$$IflF`L
# 064
la<
p $$Ifl4Fp@\
,064
la $$Ifa$^
1<=CH}}}wnn $$Ifa$^$$Ifl4Fp@\
,064
laHIWbmu$Ifp$$Ifl0pPP064
lauvy~ $$Ifa$^p$$Ifl0pPP064
la vvv $$Ifa$$$Ifl4Fp@\
,064
layppp $$Ifa$^$$Ifl4Fp@\
,064
laz8qqq $$Ifa$$$Ifl4Fp\
x
064
lazpppja $$Ifa$$If ^`$$Ifl4Fp\
x
064
la ~ $$Ifa$$Ifq$$Ifl40
$ 064
la{$If ^`q$$Ifl40
$ 064
la
` ^`^]$$Ifl064
la
ST"#./34>?IJSTXY\]`afgopst}~5OJQJ^JH*OJQJ^J jOJQJ^J5OJQJ\^JB*OJQJ^JphOJQJ^JP
!
"
#
$
%
&
'
(
)
*
+
,

.
/
0
1
2
3
e
z
^`
!f $$Ifl4Fp@\
,064
la$If $$Ifa$ ^`!"#*}}}wnn $$Ifa$^$$Ifl4Fp@\
,064
la
!($Ifp$$Ifl0@PP064
la()+QRUWZ $$Ifa$^p$$Ifl0@PP064
laZ[^_b vvv $$Ifa$$$Ifl4Fp@\
,064
labcdlnqyppp $$Ifa$^$$Ifl4Fp@\
,064
laqr{zPqqq $$Ifa$$$Ifl4Fp\
064
lazpppja $$Ifa$$If ^`$$Ifl4Fp\
064
la,~ $$Ifa$$Ifq$$Ifl40@*~ 064
la{$If ^`q$$Ifl40@*~ 064
la 9;BDHIMNGHQR;<MNaeFGLMEQRTUZ jOJQJ^J5OJQJ\^JOJPJQJ^JB*OJQJ^Jph jOJQJ^J5OJQJ^JH*OJQJ^JOJQJ^JH6T` ^` ^`^]$$Ifl
064
la(V^VWXOFUVak
$cde ^`p^p^^R`$a$ ^`oWEF ^` ^``Z[%&IJRSVWtu
źCJCJOJQJ^J0JmHnHu0J
j0JU jUB*OJQJ^Jph jOJQJ^J5OJQJ\^JOJQJ^J5OJQJ^J4lM ^``x$a$h]h&`#$` 1h/ =!"#$%
i8@8NormalCJ_HaJmH sH tH D@D Heading 1$$@&a$5OJQJ\^JV"V Heading 2dd@&[$\$5CJ$OJPJQJ\^JaJ$<A@<Default Paragraph Font0 @0Footer
H$PJ,@,Header
!&)@&Page NumberFO!FHeading 1 Char5CJOJQJ\^JaJ.U@1. Hyperlink>*B*phJ^BJNormal (Web)dd[$\$OJPJQJ^J"W@Q"Strong5\&X@a&Emphasis6]^r^highlightexampledd[$\$B*
OJPJQJ^Jph>V@>FollowedHyperlink>*B*ph~$CDOP_`QRSTUVcdDEFGHIkltuvwxy,1<=CHIWbmuvy~
! " # $ % & ' ( ) * + ,  . / 0 1 2 3 e z
!
"
#
*
!()+QRUWZ[^_bcdlnqr{6
T
VWXOFUVak
$cdeRoWEFlMx00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000@0@0@0@00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000@0@0@0@0@0
0 $$===@Z*6<QUDHtx
Hu
!(ZbqV !"#$%&'()+,./012345789:;=> @!!8@0(
B
S ?
$*iqx{#'/24qt
$'/[]_a9@,
5
T
Y
IRUWLT&.v{joTV463333333333333333333333333333333333
/ 1 abLloyd CrosbyC:\Users\User\Desktop\0 CSCOPE 0130215\CSCOPE 1213 0130215\UNIT 09 Gases\Unit 09 LS 01 Day 3 GayLussac and Avogadro STU.docLloyd CrosbyIC:\Users\User\Desktop\Unit 09 LS 01 Day 3 GayLussac and Avogadro STU.docLloyd CrosbyIC:\Users\User\Desktop\Unit 09 LS 01 Day 3 GayLussac and Avogadro STU.doci6$>.
VVIuKc
VuKc OP_`QRSTUVcdDEFGHIkltuvwxy1<=CHImuvy~
!
"
*
()QRUWZ[^_bcdlnqr{@50@UnknownGz Times New Roman5Symbol3&z ArialG5
jMS Mincho3 fgI&??Arial Unicode MS7&@Calibri?5 z Courier New;Wingdings"1hc'c'A
+
,0YD3qHXThe Electromagnetic SpectrumLloyd CrosbyLloyd CrosbyOh+'0X
,8@HPhe
Lloyd CrosbygneloyNormal.doty
Lloyd Crosbygne3oyMicrosoft Word 9.0 @@iR@iRA,Unit 09 LS 01 Day 3 GayLussac and Avogadro՜.+,0hp
l+
The Electromagnetic SpectrumTitle
!"#$%&'()*+,./0123456789:;<=>?ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`jklmnopsvRoot Entry FPsgRu@1Table@AWordDocument"~SummaryInformation(DocumentSummaryInformation8iCompObjjObjectPoolK)RK)R
FMicrosoft Word Document
MSWordDocWord.Document.89qOh+'0X
,8@HPhe
Lloyd CrosbygneloyNormal.doty
Lloyd Crosbygne3oyMicrosoft Word 9.0 @@iR@iRA,Unit 09 LS 01 Day 3 GayLussac and Avogadro