The Chapter 14 Behavior of Gases

Section 14.3 Ideal Gases l\

STP Standard temperature and pressure 1atmosphere 0 degrees Celsius STP is also a motor oil. That’s cool, if irrelevant.

The Gas Laws are mathematical. The gas laws will describe HOW gases behave.

Four Variables Describe a Gas pressure (P) in atm 2. volume (V) in Liters 3. temperature (T) in Kelvin 4. amount (n) in moles

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

Ideal Gas Law

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 = 0.08206 L x atm) / (mol x K) The other units must match the value of the constant, in order to cancel out.

Sample Problem What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K? P = 1.22 atm

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

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

Susan has 2. 03 moles of Helium in a 400 Susan has 2.03 moles of Helium in a 400.1 mL container at 282 degrees Celsius. What is the pressure of this gas in kPa? Answer: 23400 kPa

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

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!

Ideal Gases There are no gases which are truly “ideal”; however, Real gases behave this way at a) high temperature, and b) low pressure.

#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

Sample Problem At 28°C and 0.974 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

Density Density is mass divided by volume m V so, m M P V R T D = D =

Practice Problems What is the molar mass of a gas if 0.427 g of the gas occupies a volume of 125 mL at 20.0°C and 0.980 atm? 83.8 g/mol What is the density of a sample of ammonia gas, NH3, if the pressure is 0.928 atm and the temperature is 63.0°C? 0.572 g/L NH3

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

Real Gases and Ideal Gases

Ideal Gases don’t exist, because: Molecules do take up space There are attractive forces between particles - otherwise there would be no liquids formed

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

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.

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.

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.

#7 Dalton’s Law of Partial Pressures For a mixture of gases in a container, PTotal = P1 + P2 + P3 + . . . 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.

Collecting a gas over water Connected to gas generator Collecting a gas over water

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

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.

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.

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

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.

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

End of Chapter 14