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The Chapter Behavior of Gases
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Section 14.3 Ideal Gases l\
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STP Standard temperature and pressure 1atmosphere 0 degrees Celsius
STP is also a motor oil. That’s cool, if irrelevant.
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The Gas Laws are mathematical.
The gas laws will describe HOW gases behave.
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Four Variables Describe a Gas
pressure (P) in atm 2. volume (V) in Liters 3. temperature (T) in Kelvin 4. amount (n) in moles
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Four Variables Describe a Gas
pressure (P) in atm 2. volume (V) in Liters 3. temperature (T) in Kelvin 4. amount (n) in moles Not Held constant in Section 14.3
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Ideal Gas Law
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5. The Ideal Gas Law #1 R = 0.08206 L x atm) / (mol x K)
Equation: P x V = n x R x T R = L x atm) / (mol x K) The other units must match the value of the constant, in order to cancel out.
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Sample Problem What is the pressure in atmospheres exerted by a mol sample of nitrogen gas in a 10.0 L container at 298 K? P = 1.22 atm
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A 1. 50 mole sample of Helium is placed in a 15. 0 L container at 400K
A 1.50 mole sample of Helium is placed in a 15.0 L container at 400K. What is the pressure in the container? Answer: 3.28 atm
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Evan has 23. 7 grams of nitrogen gas in a 30
Evan has 23.7 grams of nitrogen gas in a 30.8 L container with a pressure of 1.5 atm. What is the temperature of the gas? Answer: 670 K
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Susan has 2. 03 moles of Helium in a 400
Susan has 2.03 moles of Helium in a mL container at 282 degrees Celsius. What is the pressure of this gas in kPa? Answer: kPa
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Ideal Gases We are going to assume the gases behave “ideally”- in other words, they obey the Gas Laws under all conditions of temperature and pressure
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Ideal Gases An ideal gas does not really exist, but it makes the math easier and is a close approximation. Particles have no volume? Wrong! No attractive forces? Wrong!
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Ideal Gases There are no gases which are truly “ideal”; however,
Real gases behave this way at a) high temperature, and b) low pressure.
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#6. Ideal Gas Law 2 Equation: P x V = m x R x T M
Allows LOTS of calculations, and some new items are: m = mass, in grams M = molar mass, in g/mol Molar mass = m R T P V
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Sample Problem At 28°C and atm, 1.00 L of gas has a mass of 5.16 g. What is the molar mass of this gas? Molar Mass = 131 g/mol
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Density Density is mass divided by volume m V so, m M P V R T D = D =
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Practice Problems What is the molar mass of a gas if g of the gas occupies a volume of 125 mL at 20.0°C and atm? 83.8 g/mol What is the density of a sample of ammonia gas, NH3, if the pressure is atm and the temperature is 63.0°C? 0.572 g/L NH3
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The density of a gas was found to be 2. 0 g/L at 1. 50 atm and 27°C
The density of a gas was found to be 2.0 g/L at 1.50 atm and 27°C. What is the molar mass of the gas? 33 g/mol What is the density of argon gas,Ar, at a pressure of 551 torr and a temperature of 25°C? 1.18 g/L Ar
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Real Gases and Ideal Gases
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Ideal Gases don’t exist, because:
Molecules do take up space There are attractive forces between particles - otherwise there would be no liquids formed
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Real Gases behave like Ideal Gases...
When the molecules are far apart. The molecules do not take up as big a percentage of the space We can ignore the particle volume. This is at low pressure
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Real Gases behave like Ideal Gases…
When molecules are moving fast This is at high temperature Collisions are harder and faster. Molecules are not next to each other very long. Attractive forces can’t play a role.
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Section 14.4 Gases: Mixtures and Movements
OBJECTIVES: Relate the total pressure of a mixture of gases to the partial pressures of the component gases.
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Section 14.4 Gases: Mixtures and Movements
OBJECTIVES: Explain how the molar mass of a gas affects the rate at which the gas diffuses and effuses.
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#7 Dalton’s Law of Partial Pressures
For a mixture of gases in a container, PTotal = P1 + P2 + P P1 represents the “partial pressure”, or the contribution by that gas. Dalton’s Law is particularly useful in calculating the pressure of gases collected over water.
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Collecting a gas over water
Connected to gas generator Collecting a gas over water
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If the first three containers are all put into the fourth, we can find the pressure in that container by adding up the pressure in the first 3: 2 atm + 1 atm + 3 atm = 6 atm 1 2 3 4
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Diffusion is: Molecules moving from areas of high concentration to low concentration. Example: perfume molecules spreading across the room. Effusion: Gas escaping through a tiny hole in a container. Both of these depend on the molar mass of the particle, which determines the speed.
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Diffusion: describes the mixing of gases
Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing. Molecules move from areas of high concentration to low concentration.
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Effusion: a gas escapes through a tiny hole in its container
-Think of a nail in your car tire… Diffusion and effusion are explained by the next gas law: Graham’s
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8. Graham’s Law RateA MassB RateB MassA =
The rate of effusion and diffusion is inversely proportional to the square root of the molar mass of the molecules. Derived from: Kinetic energy = 1/2 mv2 m = the molar mass, and v = the velocity.
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Graham’s Law With effusion and diffusion, the type of particle is important: Gases of lower molar mass diffuse and effuse faster than gases of higher molar mass. Helium effuses and diffuses faster than nitrogen – thus, helium escapes from a balloon quicker than many other gases
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End of Chapter 14
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