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Pressure units summary:
1.000 atm
= 760.0 mmHg
= 760.0 torrs
= 101.3 kPa
= 101,300 Pa (N/m2)
= 14.7 psi (lbs./in2)
Boyle’s Law:
P∝
•
𝟏
𝑽
PV = Constant
P1V1 = P2V2
•
A sample of a gas has an initial
volume of 3.95 L at a pressure of
705 mm Hg. If the volume of the
gas is increased to 5.38 L (without
the temperature changing), what
is the final pressure?
A snorkeler with a lung capacity of 6.3 L inhales a lungful of
air at the surface, where the pressure is 1.0 atm. The
snorkeler then descends to a depth of 25 m, where the
pressure increases to 3.5 atm. Assuming a constant
temperature, what is the volumetric capacity (in L) of the
snorkeler’s lungs at this depth?
Charles Law:
At constant pressure: temperature and
volume are directly proportional
As the temperature goes up, the volume gets larger.
As the temperature goes down, the volume goes down.
Temperture ∝ Volume
𝑽
= Constant
𝑻
𝑽𝟏 𝑽𝟐
=
𝑻𝟏 𝑻𝟐
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As the temperature goes down, the volume goes down.
As the temperature goes up, the volume gets larger.
Charles Law Example Problems:
•
A helium balloon is cooled from 25 oC to liquid nitrogen
temperatures (77 K). If the initial volume of the balloon is 5.8
L, what is its volume after it cools? Assume the pressure is
constant and the helium does NOT liquefy.
When working with gases: Always use
the absolute temperature scale (Kelvin):
Kelvin = oC + 273.15
•
A balloon with an initial volume of 3.2 L at a temperature of
299 K is warmed to 376 K. What is its volume at this final
temperature?
Gay-Lussac Law:
At constant volume: temperature and
pressure are directly proportional
As the temperature goes up, the pressure goes up.
As the temperature goes down, the pressure goes down.
Temperture ∝ Pressure
𝑷
𝑻
= Constant
𝑷𝟏
𝑻𝟏
𝑷
= 𝑻𝟐
𝟐
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Gay-Lussac Law:
As the temperature goes up, the pressure goes up.
Gay-Lussac Practice Problem:
• A closed metal cylinder of gas having a pressure
of 3.14 atm at 22.5 oC is heated to 311.4 oC. If
the metal cylinder can withstand a maximum
pressure of 9.95 atm before structurally failing,
predict whether the cylinder will explode after
being heated to 311.4 oC.
When working with gases: Always use
the absolute temperature scale (Kelvin):
Kelvin = oC + 273.15
Combine Gas Law:
• When the total quantity of gas is constant, we can
combine the three gas laws just described into one
relationship
𝑷𝟏 𝑽 𝟏
𝑻𝟏
= Constant
𝑷𝟏 𝑽 𝟏
𝑻𝟏
=
𝑷𝟐 𝑽 𝟐
𝑻𝟐
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Combine Gas Law Practice Problems:
• A sample of gas occupies 175 mL at 22 C and
218 mm Hg. What is the pressure of the gas if
the volume is decreased to 122 mL and the
temperature increased to 75 C?
iClicker Participation Question:
Understanding the Combined Gas Law
A cylinder of gas starts out with a volume of 5.0 L, a
pressure of 1.5 atm and a temperature of 50 oC. If the gas
is compressed to 2.5 L, while the temperature is
reduced, what will happen to the pressure?
A. The pressure will go INCREASE
B. The pressure will go DECREASE
C. The pressure stay the SAME
D. Cannot determine: Not enough
information provided
Avogadro’s Law:
When all else is constant: the volume of a gas is directly
proportional to the number of moles of the gas
Volume ∝ 𝒏 (# of moles of gas)
The more gas present,
the larger the volume.
𝑽
= Constant
𝒏
𝑽𝟏
𝒏𝟏
𝑽
= 𝒏𝟐
𝟐
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Avogadro’s Law
• The volume of gases are related to the number of moles
present.
• For convenience, a Standard Temperature and Pressure
(STP) was invented to provide a useful set of reference
conditions.
• Standard Temperature = 0˚C = 273.15 K
• Standard Pressure = 1.000 atm
At STP: 1.00 mol of gas = 22.4 Liters
iClicker Participation Question:
Stoichiometric Analysis in terms of Avogadro’s Law
How much oxygen gas would be required to react with
5 L of propane gas?
A. 1 L
B. 5 L
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
C. 10 L
D. 25 L
E. Not enough information provided
Ideal Gas Law:
𝑷𝑽
= 𝑼𝒏𝒊𝒗𝒆𝒓𝒔𝒂𝒍 𝑮𝒂𝒔 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 (𝑹)
𝒏𝑹𝑻
PV = nRT
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Ideal Gas Law:
• R is a constant for all gases. The value of R depends on the units
we use for the various properties.
»
»
»
»
R  0.08206
 8.314
Volume = Liter (L)
Temperature = Kelvin (K)
Number of moles = moles (mol)
Pressure units vary. We commonly use: atm, Pa, kPa, mmHg
L  atm
K mol
L kPa
K mol
 62.37
 8.314
L mmHg
K mol
 8314
L Pa
K mol
J
K mol

Example Problems
• The pressure in a 10.0 L flask containing only
propane is 0.912 atm at 78 C. How many
moles of propane are in the flask?
• A 50.0 L cylinder of He has a pressure of 151
atm at 298 K. What mass of He (in grams) is
inside this cylinder?
iClicker Participation Question:
Understanding Gas Density in terms of Avogadro’s Law
Which gas would be the most dense?
A. CH4
B. O2
C. N2
D. CO2
E. Not enough information provided
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Example Problems
• What is the density of butane at STP? What
about when the gas is at 20.1 oC & 758.5
mmHg?
• What volume of O2(g), measured at 18.9 C and
700. torr, is needed for the complete
combustion of 12.1 grams of propene, C3H6?
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