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## GAS LAWS

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### 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. – PowerPoint PPT presentation

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Title: GAS LAWS

1
CHAPTER 14
• GAS LAWS

2
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
• P1V1P2V2

3
• P1V1P2V2
• where P1 and V1 are the pressure volume before
the gas expands
• and P2 and V2 are the pressure volume after the
gas expands.

4
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5
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)
• 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.

6
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

7
• V1 V2
• 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.

8
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9
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 V2
• 293 K 302 K and V2
2.1L

10
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

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12
Combined Gas Law
• -combined the three gas laws into a single
expression.
• V1P1 V2P2
• T1 T2
• The Vice President is higher than the
• Treasurer

CHARLES LAW
GAY-LUSSACS LAW
13
COMBINED GAS LAW DOESNT
• ACCOUNT FOR THE AMOUNT OF GAS
• IT JUST LOOKS AT PRESSURE, VOLUME, TEMPERATURE

14
Ideal Gas Law
• -allows you to solve for the number of moles of a
gas(n).
• PV nRT
• Ppressure in kPa or atm
• Vvolume in L
• Ttemperature in K
• nof moles
• RIdeal gas constant (R) is 8.31 (L x kPa)
• (K x mol)
• 0.0821
(L x atm)
• (K x
mol)

15
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.

16
Do Now
• If 4.50 g of methane gas (CH4) is in a 2.00-L
container at 35C, what is the pressure inside
the container?
• Solution PVnRT - Ideal Gas Law Equation.
• P ?
• V 2.00 L
• R 8.31 L kPa
• T 308 K

mol K
17
Do Now
• If 4.50 g of methane gas (CH4) is in a 2.00-L
container at 35C, what is the pressure inside
the container?
• Solution PVnRT - Ideal Gas Law Equation.
• P 3.60 x 102 kPa.
• V 2.00 L
• R 8.31 L kPa
• T 308 K

mol K
18
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?
19
Mathematical Relationship
Each lego block has dimensions 4 cm x 2 cm x 1 cm
How tall is the stack of lego blocks?
20
History
• 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
Daltons Law.

21
Observe and Find a Pattern
Partial Pressures of Compressed Air Partial Pressures of Compressed Air Partial Pressures of Compressed Air
(Assuming air is 80 Nitrogen and 20 Oxygen) (Assuming air is 80 Nitrogen and 20 Oxygen) (Assuming air is 80 Nitrogen and 20 Oxygen) (Assuming air is 80 Nitrogen and 20 Oxygen)
Depth (meters) Absolute Pressure P O2 P O2 P O2 P N2
0 14.7 2.94 2.94 2.94 11.76
33 29.4 5.88 5.88 5.88 23.52
66 44.1 8.82 8.82 8.82 35.28
99 58.8 11.76 11.76 11.76 47.04
132 73.5 14.70 14.70 14.70 58.80
165 88.2 17.64 17.64 17.64 70.56
198 102.9 20.58 20.58 20.58 82.32
231 117.6 23.53 23.53 23.53 94.08
264 132.3 26.46 26.46 26.46 105.8
297 147.0 29.40 29.40 29.40 117.6
22
Explain
• 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.
• Explain why we can use a gas mixture, such as
air, to study the general behavior of an ideal
gas under ordinary conditions.

23
Predict and Test
• 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.

24
Predict and Test
• 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.

25
Derive
• 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
Daltons Law of Partial Pressures. How can we
qualitatively explain this mathematical
relationship?

26
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 wont 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 27C 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?

27
Observe and Explain
KClO3
• 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.

28
Daltons 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.
• PtotalP1 P2 P3 ..

29
Daltons 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?
• -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

30
Ideal vs. Real Gases
• An ideal gas is a hypothetical concept. No gas
exactly follows the ideal gas law.

31
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!!!!!!!!

32
Real Gases
• 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

33
Grahams Law of Effusion
• the rate of effusion and diffusion of a gas is
inversely proportional to the square root of its
molar mass .
• 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.

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35
Grahams Law of Effusion
• Practice
• Does He effuse faster or slower than O2? What is
the relative rate of diffusion of He compared to
O2?

Answer 2.8, He is 2.8 times much faster