Gases
Objective
I will know the properties of gasses and be
able to calculate the quantifiable properties
of gasses
Success Criteria
I will be able to calculate each of the four
quantifiable properties using the
appropriate equation
Video Links
• Liquid Nitrogen and Balloon Demo Link
• Charles Law Balloon Demo Link
• Boyle’s Law Demo Link
Collecting Background
• 1. What are the physical properties of a gas?
2. What are the quantitative (measurable)
properties of gasses?
• 3. What are the units of pressure? How do
you convert between the units?
• 4. Describe the different ways that the
measurable properties can be related (the
laws)?
Characteristics
What are the physical properties of a gas?
• Gases are fluid
– In essence, gases can flow
• Gases have low density
• Gases are highly compressible
• Gases completely fill their container
Pressure
• Pressure is defined as the amount of force
exerted per unit area of surface
– How hard the gas is pushing on the container
• There are several units of pressure
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Atmosphere (atm)
Kilopascal (kPa)
Millimeters of Mercury (mm Hg)
Torr (torr)
Pounds per square inch (psi)
http://www.youtube.com/watch?v=XfFdNNiIAJw
http://www.youtube.com/watch?v=O9p8SRmQlPo
Pressure
• Converting Units of Pressure
• 1.00 atm=101.325 kPa=760 torr=760 mm Hg
• Practice Problems
– The vapor pressure of water at 50.0oC is 12.33kPa.
What is the value in atmospheres?
• (12.33kPa)(1.00atm/101.325kPa)=0.122atm
– In thermodynamics, the standard pressure is
100.0kPa. What is this value in mm Hg?
• (100.0kPa)(760mm Hg/101.325kPa)=750.1mm Hg
Quantifiable Properties
• There are four measurable properties of gases
– P is pressure
– T is temperature (must be measured in Kelvin)
• K = oC + 273
– V is the volume
– n is the number of moles of the gas
STP
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Standard Temperature and Pressure (STP)
Standard Temperature is 0oC
Standard Pressure is 1.00atm
ΔG° = ΔH° – T ΔS°
The superscript indicates at STP
Boyle’s Law
• Boyle’s Law discusses the relationship of pressure and
volume
• This shows that as the pressure increases, the
volume decreases.
– This is an inverse relationship
• PV = k1
• P = k1/V where k1 is the pressure-volume constant
• And if we keep Temperature and the number of
particles constant, but change
either pressure or volume we
then have:
P1V1=P2V2
Boyle’s Law
• A sample of oxygen gas has a volume of
150.0mL at a pressure of 0.947atm. What will
the volume be at a pressure of 1.000atm if
the temperature remains constant?
– P1V1=P2V2
– (0.947atm)(150.0mL)=(1.000atm)V2
– V2=142mL
– Be sure your answers are in Significant Figures!
Boyle’s Law
• A balloon has a volume of 456mL at a pressure of
761torr. It is taken under water in a submarine to
a depth where the air pressure in the submarine
is 3.3atm. What is volume of the balloon if the
temperature remains constant?
– Be sure all variables are in the same units
• (761torr)(1.00atm/760torr)=1.00atm
– P1V1=P2V2
– (1.00atm)(456mL)=(3.3atm)V2
– V2=138mL=140mL
Charles’ Law
• Charles’ Law discusses the relationship
between Temperature and Volume
• This shows that as the temperature increases,
the volume increases
– This is a direct relationship
• V/T=k2
• At a constant pressure and number of
particles, if either temperature
or volume change we have:
V1/T1=V2/T2
Charles’ Law
• A sample of neon gas has a volume of 752mL
at 25.0oC. What is the volume at 50.0oC if the
pressure remains constant?
– Change the degrees in to Kelvin
• 25.0oC + 273 = 298 K; 50.0oC + 273 = 323 K
– V1/T1=V2/T2
– (752 mL)/(298 K)=V2/(323K)
– V2 = 815 mL
Gay-Lussac’s Law
• Gay-Lussac’s Law discusses the relationship
between temperature and pressure
• This shows that as the temperature increases, the
pressure increases.
• P = k3T where k3 is the temperature-pressure
constant
• P/T = k3
• At a constant volume and number of particles, if
either the temperature or
pressure changes, we use this
equation:
P1/T1=P2/T2
Gay-Lussac’s Law
• At 122oC, the pressure of a sample of nitrogen
is 1.07atm. What will be the pressure at
205oC, assuming constant volume?
– First change temperature to Kelvin
• 122oC + 273 = 395 K; 205oC + 273 = 478 K
– P1/T1=P2/T2
– (1.07 atm)/(395 K)=P2/(478 K)
– P2= 1.29 atm
Combined Gas Law
• When the three different laws just discussed
are analyzed, what would they be if they are
all combined?
• PV = k1
• V/T=k2
• P/T = k3
Combined Gas Law
• PV/T = k4
• If the number of moles remains constant, we
can solve for any of the other variables
• P1V1/T1= P2V2/T2
• Also used is Standard Temperature and
Pressure (STP)
Combined Gas Law
• At a pressure of 780.0 torr and 24.2oC, Xenon
has a volume of 0.350L. What will the volume
be at STP?
• First change temperature to Kelvin
– 24.2oC + 273 = 297.2 K
• Change the pressure to atmospheres
– (780.0 torr)(1.00 atm/760 torr) = 1.03 atm
– P1V1/T1= P2V2/T2
– (1.03atm)(0.350 L)/(297.2K) = (1.00atm)V2/(273K)
– V2 = 0.330 L
Diffusion and Effusion
• Diffusion is the movement of particles from
high density to low density
– Popping a balloon
• Effusion is the passage of gas molecules under
pressure through a tiny opening
– Nail in a tire
Graham’s Law of Diffusion
• In Graham’s Law of Diffusion, we will analyze
the kinetic energy of two different gases
– Kinetic energy formula is 1/2mv2
• m is the mass (in this case molar mass)
• v is the velocity of the molecule
– Graham’s Law states that the kinetic energy of the
first gas will equal the kinetic energy of the second
gas
m1v12= m2v22
Graham’s Law of Diffusion
• Oxygen molecules have an average speed of
480 m/s at room temperature. At the same
temperature, what is the average speed of
molecules of sulfur hexafluoride?
– Molar mass of Oxygen is 32.00 g/mol (O2)
– Molar mass of SF6 is 146.07 g/mol
– m1v12= m2v22
– (32.00 g/mol)(480 m/s)2= (146.07 g/mol) v2
– v = 220 m/s
Dalton’s Law of Partial Pressures
• Dalton’s law of partial pressures states that the total
pressure of a mixture of gases is equal to the sum of
the partial pressures of each gas
• PT = P1 + P2 + P3 + ……
• A 42.0 mL sample of hydrogen is collected over water
at 30.0oC. What is the total pressure if the pressure of
the dry gas 845.0 torr and the vapor pressure of water
at 30.0oC is 18.6 torr?
• PT = P1 + P2 + P3 + ……
• PT = 845.0 torr + 18.6 torr
• PT = 863.6 torr
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Essential
Essential Information
1.00 atm=101.325 kPa=760 torr=760 mm Hg
STP: 273 K and 1.00 atm
Dalton’s Law: PT = P1 + P2 + P3 + ……
Combined Gas Law: P1V1/T1= P2V2/T2
Gay-Lussac’s Law: P1/T1=P2/T2
Charles’ Law: V1/T1=V2/T2
Gramham’s Law: m1v12= m2v22
Boyle’s Law: P1V1=P2V2
Ideal Gas Law: PV=nRT where R is the gas constant