Transcript Slide 1

Units
mm, torr, atm, Pa,
kPa, N/cm2
Pressure:
Force
Area
Devices
Boyle’s
Law
PV=k
Gases
Charles’
Law
V/T=k
Barometer
Manometer
The way we
choose to view it
A.P. Chem. Ch. 5
Avog.
Law
V/n=k
CaPtot= Pa
Dalton’s
Law
Ptot=
Pa+ Pb…
Graham’s Law
Ra/Rb =
(MMb/MMa)1/2
Stoich.
At STP
0.0821 L atm/mol K
8.31 J/ mol K
PV= nRT
Dimensionless Pts.
In constant motion
Colliding 100% elast.
Creating pressure
w/o influence
In such a way that
Temp is dir. Prop. To
average KE.
The Way It Really Is.
KMT
Derivation
P = dRT/MM
P = 2 nNA1/2mu2
3V
P = 2 nKEper mol
3V
KEper mol = (3/2)RT
Non- STP
22.4 L =
1 mole
Due to Avogadro’s
Hypothesis.`
urms= (3RT/MM)1/2
Particle interactions
P= nRT – a(n/V)2
V-nb
Volumes of particles
Format of the Test
•Option 1 – The marathon problem (1
system; many parts) 70 points
•Option 2 – The regular test (single
parts; many systems) 70 points
13 multiple choice [~20 minutes]
2 free response [~25 minutes]
Problems to emphasize in your HW:
43, 61, 67, 83, 85
Container
(3.0 L)
A
B
C
Gas
Methane
Ethane
Butane
Formula
CH4
C2H6
C4H10
Molar Mass
(g/mol)
16
30.
58
Temp. (ºC)
27
27
27
Pressure
(atm)
2.0
4.0
2.0
The average kinetic energy of the gas molecules is
(A) greatest in container A
(B) greatest in container B
How would you
calculate the average
KE per mole of gas?
(C) greatest in container C Avg. KEmol = 1.5 (8.31 J/mol K) 300.K
(D) the same for all three containers
Avg. KEmol = 3740 J/mol
Container
(3.0 L)
A
B
C
Gas
Methane
Ethane
Butane
Formula
CH4
C2H6
C4H10
Molar Mass
(g/mol)
16
30.
58
Temp. (ºC)
27
27
27
Pressure
(atm)
2.0
4.0
2.0
The average velocity of the gas molecules is
(A) greatest in container A
(B) greatest in container B
(C) greatest in container C
How would you
calculate the velocity of
the methane gas?
urms = [3(8.31J/mol K)(300. K)/0.016 kg]½
(D) the same for all three containers urms = 680 m/s
Container
(3.0 L)
A
B
C
Gas
Methane
Ethane
Butane
Formula
CH4
C2H6
C4H10
Molar Mass
(g/mol)
16
30.
58
Temp. (ºC)
27
27
27
Pressure
(atm)
2.0
4.0
2.0
The number of gas molecules is
(A) greatest in container A
(B) greatest in container B
(C) greatest in container C
How would you
calculate the number of
particles of ethane?
n = (4.0 atm)(3.0L)
(0.0821 L atm/mol K)(300. K)
(D) the same for all three containers
n = 0.49 mol
Container
(3.0 L)
A
B
C
Gas
Methane
Ethane
Butane
Formula
CH4
C2H6
C4H10
Molar Mass
(g/mol)
16
30.
58
Temp. (ºC)
27
27
27
Pressure
(atm)
2.0
4.0
2.0
The density of the gas is
(A) greatest in container A
(B) greatest in container B
(C) greatest in container C
How would you
calculate the density of
the ethane gas?
d = (4.0 atm)(30. g/mol)
(0.0821 L atm/mol K)(300. K)
(D) the same for all three containers
d = 4.9 g/L
Container
(3.0 L)
A
B
C
Gas
Methane
Ethane
Butane
Formula
CH4
C2H6
C4H10
Molar Mass
(g/mol)
16
30.
58
Temp. (ºC)
27
27
27
Pressure
(atm)
2.0
4.0
2.0
If the containers were opened simultaneously the
diffusion rates of the gas molecules out of the
containers through air would be Calculate how much
faster the methane
(A) greatest in container A
diffuses compared to
(B) greatest in container B
the ethane.
½
(C) greatest in container C (ratemeth/rateeth) = (30.g/mol/16g/mol)
(ratemeth/rateeth) = 1.4 (times faster)
(D) the same for all three containers
Consider the production of water at 450. K and
1.00 atm:
2 H2(g) + O2(g) --- > 2 H2O(g)
If 300.0 mL of hydrogen is mixed with 550.0 mL
of oxygen, determine the volume of the reaction
mixture when the reaction is complete.
You Have:
nH2= 0.00812 mol
nO2= 0.0149 mol
Since H2 is the L.R. then 300.0 mL
of it is used and only 150 mL of the
O2 is used. (2:1 mole ratio can be
You Need:
Twice as many moles of H2 than O2
(from balanced reaction equation)
of H2O will be produced.
Consequence:
H2 is the L.R.
viewed as 2:1 volume ration as well due to
Avogadro’s Hypothesis) Also, 300.0 mL
The reaction mixture remaining will
be 600 mL in volume from the
300.0mL of water made and the
300.0 mL of O2 unused.
UF6= 352 g
Urms= [3(8.31Jmol-1K-1)(330.0K)/0.352kg]½
Urms= 153 m/s
1.03 mg O2 = 3.22 x 10-5 mol O2
P O2 = 0.00380 atm
0.41 mg He = 1.0 x 10-4 mol He
P He = 0.012 atm
P tot = 0.016 atm
XO2 (Ptot) = PO2
XO2 = PO2 /(Ptot) = 0.24
XHe = 0.76
P = dRT/MM
MMair = 29 g/mol
dair@25°C, 740 t = 1.2 g/L
UH2
UI2
√
=
52 s
timeH2
MMI2
MMH2
√
=
amount H2/timeH2
=
RateH2
RateI2
=
amount I2/timeI2
253.8g
2.02 g
timeI2
timeH2
timeH2= 4.6 s