Topic 4- Gases - Kenmore Town of Tonawanda UFSD / …€™s Law Temperature vs. Pressure (Constant...
Transcript of Topic 4- Gases - Kenmore Town of Tonawanda UFSD / …€™s Law Temperature vs. Pressure (Constant...
I. Kinetic Molecular Theory – (under ideal circumstances) - explains how an “ideal” gas acts Let’s watch The Kinetic Molecular Theory: Properties of Gases (6:09)…http://education-
portal.com/academy/lesson/the-kinetic-molecular-theory-properties-of-gases.html#lesson
A. Particles in an ideal gas…
• have no volume.
• have elastic collisions.
• are in constant, random, straight-line
motion.
• don’t attract or repel each other.
• have an avg. KE directly related to
Kelvin temperature (KE = ½ MV2) S.Panzarella
B. Gas behavior is most
ideal…
at low pressures
at high temperatures
in nonpolar atoms/molecules of low
molecular mass ie. H2 and He
think about the conditions for a beach vacation
C. Real Gases Particles….
Real gases deviate from
an ideal gas at low
temperatures and high
pressures.
• Have their own volume
(due to increased pressure)
• Attract each other (due to
increased temps
S.Panzarella
Guide pg 4-5 #1-11
1. C
2. C
3. D
4. C
5. B
6. C
7. B
8. Real gases have a small,
but significant volume or
they have weak IMF’s
9. A
10. A
11. A
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CHEM DO review – You try 5. Under the same conditions of temperature and pressure, which of the following gases would behave most like an ideal gas?
(1) He(g) (2) NH3(g)
(3) Cl2(g) (4) CO2(g)
6. Which gas has properties that are most similar to those of an ideal gas?
(1) N2 (2) O2 (3) He (4) Xe
7. One reason that a real gas deviates from an ideal gas is that the molecules of the real gas have
(1) a straight-line motion
(2) no net loss of energy on collision
(3) a negligible volume
(4) forces of attraction for each other
1. Which gas will most closely resemble an ideal gas at STP? 1) SO2 2) NH3 3) Cl2 4) H2 2. At STP, which gas would most likely behave as an ideal gas? (1) H2 (2) CO2 (3) Cl2 (4) SO2 3. Which gas has properties that are most similar to those of an ideal gas? (1) O2 (2) H2 (3) NH3 (4) HCl 4. Under which conditions does a real gas behave most like an ideal gas? (1) at high temperatures and low pressures (2) at high temperatures and high pressures (3) at low temperatures and low pressures (4) at low temperatures and high pressures
III. The Gas Laws Shows the relationships
between TEMPERATURE, PRESSURE, VOLUME and number of moles of a gas
Used to determine what effect changing one of those variables will have on any of the others.
S.Panzarella
Normal conditions…… • Standard Temperature &
Pressure - *STP
• 273 K (or 0◦C) and 101.3 kPa or 1 ATM
• Temperature MUST be (KELVIN) when working with gases.
K = ºC + 273
S.Panzarella
*Found in
Reference Table A
1 Atm = 760 torr =760 mmHg = 101.3 kPa = 14.7 lb/in2
1. Combined Gas law
Combines 3 gas laws
# moles are held constant
P, T and V change Let’s watch (1:05): http://www.youtube.com/watch?v=t-Iz414g
ro&feature=player_embedded
Formula: found on Table T
P1V1
T1 =
P2V2
T2 S.Panzarella
GIVEN:
V1 = 2.00 L
P1 = 80.0 kPa
T1 = 300 K
V2 = 1.00 L
P2 = 240. kPa
T2 = ?
WORK:
P1V1T2 = P2V2T1
(80.0 kPa)(2.00 L)(T2)
= (240. kPa) (1.00 L) (300 K)
T2 = 450 K
COMBINED Gas Law Problem #1
2.00 L sample of gas at 300. K and a pressure of
80.0 kPa is placed into a 1.00 L container at a
pressure of 240. kPa. What is the new
temperature of the gas?
S.Panzarella
GIVEN:
V1 = 7.84 cm3
P1 = 71.8 kPa
T1 = 25°C = 298 K
V2 = ?
P2 = 101.3 kPa
T2 = 273 K
WORK:
P1V1T2 = P2V2T1
(71.8 kPa)(7.84 cm3)(273 K)
= (101.3 kPa) V2 (298 K)
V2 = 5.09 cm3
COMBINED Gas Law Problem #2
A gas occupies 7.84 cm3 at 71.8 kPa &
25°C. Find its volume at STP.
S.Panzarella
2. Boyle’s Law
Video: Let’s watch (1:01):
http://www.youtube.com/watch?feature=player
_embedded&v=DcnuQoEy6wA
Pressure vs. Volume
(Constant Temperature):
(think: squeezing a balloon)
As pressure is INCREASED,
volume is DECREASED
S.Panzarella
GIVEN:
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
P1V1T2 = P2V2T1
BOYLE’S Law Problems
A gas occupies 100. mL at 150.
kPa. Find its volume at 200. kPa.
P V
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
S.Panzarella
3. Charles’ Law Let’s Watch Video (33 sec) : http://www.youtube.com/watch?v=XHiYKfAmTMc&feature=player_embedded
Volume vs. Temperature (Constant
Pressure) (think: hot air balloon)
As temperature is INCREASED, volume is
INCREASED
• As the temperature of the water increases, the volume of the
balloon increases.
S.Panzarella
GIVEN:
V1 = 5.00 L
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
P1V1T2 = P2V2T1
CHARLES’ Law Problems
A gas occupies 5.00L at 36°C. Find
its volume at 94°C.
T V
(5.00 L)(367 K)=V2(309 K)
V2 = 5.94 L
S.Panzarella
4. Gay-Lussac’s Law
Temperature vs. Pressure (Constant
Volume)- (think: car tires or pressure
cooker)
As temperature is INCREASED,
pressure is INCREASED
S.Panzarella
GIVEN:
P1 = 1.00 atm
T1 = 200 K
P2 = ?
T2 = 800 K
WORK:
P1V1T2 = P2V2T1
GAY-LUSSAC’S Law Problem
A 10.0 L sample of gas in a rigid container at 1.00 atm and 200. K is heated to 800. K. Assuming that the volume remains constant, what is the new pressure of the gas?
P T
(1.00 atm) (800 K) = P2(200 K)
P2 = 4.00 atm
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Topic 4 RB ANSWERS
47) 1
48) 3
49) 1
50) 4
51) 1
52) 3
53) 3
54) 1
55) 2
56) 2
57) 3
58) 3
59) 4
60) 4
61) 3
62) 2
63) 3
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5. Graham’s Law
States that the rate of effusion
(diffusion) of a gas is inversely
proportional to the square root of the gas’s molar mass.
Helium effuses (and diffuses) nearly three
times faster than nitrogen at the same
temperature.
S.Barry S.Panzarella
Graham’s Law cont.
Gases of SMALL MASS (or DENSITY) diffuse faster than gases of higher molar mass.
ex. At STP, which gas will diffuse more rapidly? Use Table S
a) He b) Ar c) Kr d) Xe
S.Barry S.Panzarella
6. Dalton’s Law
At constant volume and temperature,
the total pressure exerted by a mixture of gases
is equal to the sum of the partial pressures of
the component gases.
Formula
Three gases are combined in container T
Ptotal = P1 + P2 + P3 ...
S.Panzarella
GIVEN:
PH2 = ?
Ptotal = 94.4 kPa
PH2O = 2.72 kPa
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH2 + 2.72 kPa
PH2 = 91.7 kPa
Dalton’s Law example #1
Hydrogen gas is collected over water at 22.5°C and 2.72 kPa. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.
S.Panzarella
GIVEN:
PO2 = ?
PN2 = 79 kPa
PCO2 = 0.034 kPa
Pothers = 0.95 kPa
Ptotal = 101.3 kPa
WORK:
Ptotal = PO2 + PN2 + PCO2 + Poth
101.3 kPa = PO2 + 79 kPa + 0.034 kPa + 0.95 kPa
101.3 kPa = PO2 + 79.984 kPa
21.316 kPa = PO2
Ex. Air contains oxygen, nitrogen, and carbon dioxide
and other gases. What is the partial pressure of O2
at 101.3 kPa of pressure if the PN2 = 79 kPa , the
PCO2 = 0.034 kPa and Pothers = 0.95 kPa?
Dalton’s Law example #2
S.Panzarella
III. Avogadro’s Hypothesis Let’s watch http://education-portal.com/academy/lesson/molar-volume-using-
avogadros-law-to-calculate-the-quantity-or-volume-of-a-gas.html#lesson
Definition: Equal volume of gases at the same
temperature and pressure contain equal
numbers of particles (molecules) regardless of
their mass
Let’s watch (2:05)
http://www.youtube.com/watch?v=fexEvn0ZOpo&feature=player_embedded S.Panzarella
1) Which rigid cylinder contains the same number of gas molecules at
STP as a 2.0-liter rigid cylinder containing H2(g) at STP?
(a) 1.0-L cylinder of O2(g)
(b) 2.0-L cylinder of CH4(g)
(c) 1.5-L cylinder of NH3(g)
(d) 4.0-L cylinder of He(g)
2) Which two samples of gas at STP contain the same total number of
molecules?
(a) 1 L of CO(g) and 0.5 L of N2(g)
(b) 2 L of CO(g) and 0.5 L of NH3(g)
(c) 1 L of H2(g) and 2 L of Cl2(g)
(d) 2 L of H2(g) and 2 L of Cl2(g)
3) At the same temperature and pressure, which sample contains the
same number of moles of particles as 1 liter of O2(g)?
(a) 1 L Ne(g) (c) 0.5 L SO2(g)
(b) (b) 2 L N2(g) (d) 1 L H2O(ℓ)
B. Molar volume:
The number of molecules in 22.4 L of any gas at STP has been chosen as a standard unit called 1 mole
1 mole = 22.4 L of any gas at STP contains 6.02x1023 particles
22.4 L of any gas at STP is = to its molecular mass
S.Panzarella
Molar volume cont.
Density = molecular mass of gas volume (22.4 L)
ex. What is the density of 1 mole of oxygen gas?
ex. Which gas has a density of 1.70 g/L at STP?
a) F2 b) He c) N2 d) SO2
S.Panzarella
IV. Ideal Gas Law
PV=nRT
Recall - Avogadro’s law, which is
derived from this basic idea, says
that the volume of a gas maintained
at constant temperature and
pressure is directly proportional to
the number of moles of the gas
Video: http://education-portal.com/academy/lesson/the-ideal-gas-law-and-the-
gas-constant.html#lesson
S.Panzarella
PV
T
V
n
PV
nT
Ideal Gas Law video http://education-portal.com/academy/lesson/using-the-ideal-gas-law-to-predict-the-effect-of-
changes-to-a-gas.html#lesson
= k
UNIVERSAL GAS
CONSTANT
R=0.0821 Latm/molK
R=8.315 dm3kPa/molK
= R
You don’t need to memorize these values!
• Merge the Combined Gas Law with Avogadro’s Principle:
n is the
number of
moles of the
gas
S.Panzarella
GIVEN:
P = ? atm
n = 0.412 mol
T = 16°C = 289 K
V = 3.25 L
R = 0.0821Latm/molK
WORK:
PV = nRT
P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K
P = 3.01 atm
Ideal Gas Law Problem #1
Calculate the pressure in atmospheres
of 0.412 mol of He at 16°C & occupying
3.25 L.
S.Panzarella