1. Gas Laws
Videodisk Unit 5
Demo
Imploding Can
Eggs-in-the-Bottle

2. Real Gases
An ideal gas adheres to the Kinetic Theory exactly in all situations.
Real gases deviate from ideal behavior at high pressures and low temperatures.
When the pressure is high, it becomes more difficult to compress a gas because the particles actually have a volume of their own.
When the temperature is low, gas particles slow down and attractions between them become significant as they clump together and form liquids.

3. We must define some terms:
n = moles of gas particles
V = volume (of the container)
T = temperature (must be in Kelvin)
P = pressure
(You will see these variables in a variety of gas laws)

4. All temperatures in gas problems
Kelvin is the only temperature scale that measures absolute speed of particles.
K = oC + 273
Absolute Zero is the lowest possible temperature, no movement in the molecules.
All temperatures in gas problems
must be in Kelvin.

5. Pressure 1 atm = 101.325 kPa = 760 torr = 760 mmHg
1 atm is the normal atmospheric pressure at sea level.
Pressure changes with altitude.
Air pressure is measured with a barometer.

6. Standard Temperature and Pressure (STP)
Standard Temperature = 0oC = 273 K
Standard Pressure = 1 atm or equivalent

7. Pressure, Volume, & Temperature
Boyle’s Law
Pressure and volume are inversely proportional if the temperature remains constant
P1 V1 = P2 V2
Robert Boyle

8. Temperature must be in Kelvin
Jacques Charles
Charles’ Law
Volume and temperature are directly proportional if pressure remains constant
V1 = V2
T1 T2
Temperature must be in Kelvin

9. Temperature must be in Kelvin
Joseph Louis Gay-Lussac
Gay-Lussac’s Law
Pressure and temperature are directly proportional if volume remains constant
P1 = P2
T1 T2
Eggs-in-the-Bottle Demo
*see attached
Temperature must be in Kelvin

10. AP Chemistry - Unit 2 - Chapter 5
12/25/2018
CB p. 8
Avogadro’s law
For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures).
V = volume of the gas
n = number of moles of gas
n1/V1 = n2/V2

11. Avogadro’s Principle
At STP, 1 mole of gas is equal to 22.4 L

12. Temperature must be in Kelvin
Combined Gas Law
P1 V1 = P2 V2
T T2
You can combine the above laws, then only one law is needed.
Temperature must be in Kelvin
Cross out any constants

13. Sample Problem
At conditions of 785 torr of pressure and 15.0 oC temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 745 torr and 30.0oC?
T1 = 15.0 oC = 288 K
T2 = 30.0 oC = 303 K

14. P1V1 = P2V2
T T2
(785 torr)(45.5 mL) = (745 torr) V2
288 K K
V2 = 50.4 mL

15. Example
On a cold morning (10.0 oC) a group of hot-air balloonists start filling their balloon with air, using a large fan. After the balloon is three-fourths filled, they turn on the propane burner to heat the air. At what Celsius temperature will the air completely fill the envelope to its maximum capacity of m3?

16. P1V1 = P2V2
T T2
Pressure is constant
V1 = ¾ x 1700 m3 V2 = m3
T1 = 283 K T2 = ?
1275 m3 = m3
283 K T2
T2 = 377 K
T2 = 377 K – 273 = 104 oC

17. Ideal Gas Law P V = n R T (R = 0.0821 atm L/mol K)
(R = kPa L/mol K)
(R = 62.4 mmHg L/mol K)
We get the R by subbing in one mole of gas.
Be careful of your units.
R is the universal gas constant. An “R” value is picked based upon the unit being used to measure pressure.

18. (1.50 atm)(1.00 L) = (n)(0.0821 Latm/molK)(373 K)
Example
How many moles of a gas at 100.oC does it take to fill a 1.00-L flask to a pressure of 1.50 atm?
PV = nRT
(1.50 atm)(1.00 L) = (n)( Latm/molK)(373 K)
n = mol

19. Example
What is the volume occupied by 9.45 g of C2H2 at STP?
PV = nRT
9.45 g C2H2 x 1 mol C2H2 = mol C2H2
g C2H2
(1.00 atm) V = ( mol)( Latm/molK)(273 K)
V = 8.14 L

20. Gas Stoichiometry
Only gas volumes at STP (Avogadro’s Principle 1 mol = 22.4 L) can be entered into a stoichiometry equation
If gas is at a different temperature & pressure, use PV=nRT to convert liters to moles and then continue with stoichiometry

21. Sample Problem
3 H2 + N2  2 NH3
A chemist might commonly perform this reaction (Haber process) in a chamber at 327oC under a pressure of 900. mm Hg. How many grams of ammonia would be produced from liters of hydrogen at the above conditions?

22. (900. mmHg)(166.3 L) = n (62.4 L mmHg/mol K)(600.K)
3 H2 + N2  2 NH3
PV = nRT
(900. mmHg)(166.3 L) = n (62.4 L mmHg/mol K)(600.K)
n = 4.00 mol H2
4.00 mol H2 x 2 mol NH3 x 17g NH3 = g NH3
3 mol H mol NH3