Transcript Chapter 14

CHAPTER 14
GAS LAWS
Boyles Law:
Pressure (kPa)
Volume (L)
-for a given mass of gas at constant temperature,
the volume of the gas varies inversely with
pressure.
– Pressure increases, Volume decreases
– Inversely proportional
– Smaller volume- more pressure b/c particles collide
more often
EX: syringe
P1V1=P2V2
P1V1=P2V2

where P1 and V1 are the pressure & volume
before the gas expands
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and P2 and V2 are the pressure & volume after
the gas expands.
Boyles Law:
Pressure (kPa)
Volume (L)
Practice: If you had a gas that exerted 202 kPa of
pressure and took up a space of 3.00 liters, & you
decided to expand the space to 7.00 liters, what would
be the new pressure? (temperature remains constant)
Answer: P1V1 = P2V2
So, P1 = 202 kPa, V1 = 3.00L, V2 = 7.00L, and you
need to solve for P2, the new pressure. Plug the
numbers into the equation, & you have:
(P1V1)/V2 = P2
(202 kPa) x (3.00 L) = (P2) x (7.00 L). P2 = 86.6 kPa.
Charles’ Law:
Temp. (K)
Volume (L)
-the volume of a fixed mass of gas is directly
proportional to its Kelvin temperature if the
pressure is kept constant.
– Temperature increases, Volume increases
– Directly proportional
– Higher temperature, gas particles speed up and
move farther away from one another
EX: heating a sealed container
V1 = V2
T1
T2
V1 = V 2
T1
T2

where V1 and T1 are the volume &
temperature before the gas expands

and V2 and T2 are the volume & temperature
after the gas expands.
Charles’ Law:
Temp. (K)
Volume (L)
Practice: If you took a balloon outside that was at 20oC
at 2L in volume, & it heated up to 29oC, what would
its volume be? Assume constant pressure.
Answer: V1 / T1 = V2 / T2
–
V1 = 2.0L, T1 = 20oC, T2 = 29oC, you must solve for V2.
Wait!! You have to convert the temperatures to Kelvin
So:
T1 = 20oC = 20oC + 273 = 293 K
T2 = 29oC = 29oC + 273 = 302 K
–
Now, you can plug in the numbers and solve for V2.
2.0L
293 K
=
V2
302 K
and V2 = 2.1L
Gay-Lussac's Law :
Temp. (K)
Pressure (kPa)
-the pressure of a fixed mass of gas is directly
proportional to its temperature if the volume
is kept constant.
– Temperature increases, Pressure increases
– Directly proportional
P1 = P2
T1
T2
Combined Gas Law:
-combined the three gas laws into a single
expression.
V
P
=
V
P
1
1
2
2 CHARLES’ LAW
BOYLE’S LAW
T1
T2
GAY-LUSSAC’S LAW
The Vice President is higher than the
Treasurer
COMBINED GAS LAW DOESN‘T:
ACCOUNT FOR THE AMOUNT OF GAS
IT JUST LOOKS AT PRESSURE, VOLUME, &
TEMPERATURE
Ideal Gas Law:
-allows you to solve for the number of moles of a
gas(n).
PV = nRT
P=pressure in kPa or atm
V=volume in L
T=temperature in K
n=#of moles
R=Ideal gas constant (R) is 8.31 (L x kPa)
(K x mol)
0.0821 (L x atm)
(K x mol)
Ideal Gas Law:
PV = nRT
Practice:
A sample of O2 gas has a volume of 4.52L at a temp. of 10oC and
a pressure of 110.5 kPa. Calculate the number of moles of O2
gas present in this sample.
Answer: Rearrange to solve for n (number of moles):
n = PV/RT.
P = 110.5 kPa,
V = 4.52L,
T = 10oC + 273 = 283K, and
R = 8.31 L x kPa/K x mol.
n= (110.5 kPa)(4.52L)/(8.31L kPa/K mol)(283 K) = .212 moles.
Do Now
If 4.50 g of methane gas (CH4) is in a 2.00-L
container at 35°C, what is the pressure inside
the container?
 Solution: PV=nRT - Ideal Gas Law Equation.
P=?
V = 2.00 L

n = 4.50 g CH4 x 1 mol / 16.0 g CH4
R = 8.31 L * kPa
mol * K
T = 308 K
= 0.281 mol
Do Now
If 4.50 g of methane gas (CH4) is in a 2.00-L
container at 35°C, what is the pressure inside
the container?
 Solution: PV=nRT - Ideal Gas Law Equation.
P = 3.60 x 102 kPa.
V = 2.00 L

n = 4.50 g CH4 x 1 mol / 16.0 g CH4
R = 8.31 L * kPa
mol * K
T = 308 K
= 0.281 mol
Mathematical Relationship
Each slice weighs 0.5 lbs.
Eight slices make up a pie.
How much does this pie weigh?
What can I ignore in my calculation?
Mathematical Relationship
Each lego block has dimensions
4 cm x 2 cm x 1 cm
How tall is the stack of
lego blocks?
History
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John Dalton (1766-1844) was
an English chemist and
physicist born in Cumberland,
England.
Early in life, influenced by
meteorology.
Researched color-blindness
a.k.a. “Daltonism.”
Most notably known for
compiling fundamental ideas
into a universal atomic theory.
His interest in gases and gas
mixtures lead him to
investigate humidity. This
ultimately lead to Dalton’s Law.
Observe and Find a Pattern
Partial Pressures of Compressed Air
(Assuming air is 80% Nitrogen and 20% Oxygen)
Depth (meters)
Absolute Pressure
P O2
P N2
0
14.7
2.94
11.76
33
29.4
5.88
23.52
66
44.1
8.82
35.28
99
58.8
11.76
47.04
132
73.5
14.70
58.80
165
88.2
17.64
70.56
198
102.9
20.58
82.32
231
117.6
23.53
94.08
264
132.3
26.46
105.8
297
147.0
29.40
117.6
Explain
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The Ideal Gas Law holds for virtually any
gas, whether pure or a mixture, at
ordinary conditions for two reasons:
– Gases mix homogeneously in any proportions.
– Each gas in a mixture behaves as if it were
the only gas present.
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Explain why we can use a gas mixture,
such as air, to study the general behavior
of an ideal gas under ordinary conditions.
Predict and Test
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What is the relationship between the number of particles
in containers A and C and the partial pressures of A and
C.
Predict what the reading will be for container T.
Predict and Test
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What is the relationship between the number of particles
in containers A and C and the partial pressures of A and
C.
Predict what the reading will be for container T.
Derive
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Based on the outcome of the experiment
relative to the prediction, make a judgment
about whether the experiment disproved the
hypothesis or not.
Write a mathematical expression that relates
absolute (total) pressure to the individual gas
pressures.
–
Ptotal = P1 + P2 + P3 + …this is known as Dalton’s
Law of Partial Pressures. How can we qualitatively
explain this mathematical relationship?
Mathematical Application

It was a dark and stormy night and your “friend”
has locked you in the lab without a key. You
remembered that the oldest window in the lab is
always cracked open on a diagonal that won’t
ever seem to budge completely. Someone
calculated that you need a total pressure of 7.10
x 106 kPa to exert a strong enough force to bust
open the window. Coincidentally, you have 2.00
moles of Ne, 4.00 moles of Xe, and 6.00 moles
of Ar in a 5.00-L vessel at 27°C standing next to
your lab station. Is the total pressure of the
mixture enough for you to escape and make it to
the club in time or will you be left alone in the
lab synthesizing aspirin to relieve you of your
torment?
Observe and Explain
KClO3
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Gas produced is less dense than water, so it replaces the
water in the test tube.
Gas collected is not pure because it contains vapor from
the water.
Dalton’s Law of Partial Pressure:
-for a mixture of gases in a container, the total
pressure exerted is the sum of the pressures
that each gas would exert if it were alone.
Ptotal=P1 + P2 + P3 +…..
Dalton’s Law of Partial Pressure:
Practice:
Air contains O2, N2, CO2, and trace amounts of other
gases. What is the partial pressure of O2 (PO2) at
101.3 kPa of pressure if PN2 =79.10 kPa, PCO2 =0.040
kPa, and Pothers =0.94 kPa?
Answer: Ptotal = 101.3 kPa
-Ptotal = PO2 + PN2 + PCO2 + Pothers
-PO2 = Ptotal - (PN2 + PCO2 + Pothers)
-PO2 = 101.3 kPa - (79.10 kPa + 0.040kPa + 0.94kPa)
-PO2 = 21.22 kPa
Ideal vs. Real Gases
An ideal gas is a hypothetical concept. No
gas exactly follows the ideal gas law.
Ideal Gases
Always a gas
 Not attracted to one another
 Have no volume
 Follow the gas laws for all conditions of
pressure and temperature
 Ideal gases DO NOT exist!!!!!!!!
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Real Gases
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Can be liquified and sometimes solidified
by cooling and applying pressure
 (ideal gases can not)
Have a finite volume
 Are attracted to one another, especially at
low temperatures
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Graham’s Law of Effusion:
the rate of effusion and diffusion of a gas is
inversely proportional to the square root of its
molar mass .
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–
–
Lighter gas goes faster.
Heavier gas goes slower.
To figure out how much faster take the square root
of the gfm of the heavier gas and divide by the
square root of the gfm of the lighter gas.
Rate(lighter gas )
Rate( heavier gas )

gfm( heavier gas )
gfm(lighter gas )
Graham’s Law of Effusion:
Practice:
Does He effuse faster or slower than O2? What is
the relative rate of diffusion of He compared to
O2?
Rate( He )
Rate(O2 )

32g ( heavier gas )
4 g (lighter gas )
Answer: = 2.8, He is 2.8 times much faster