1. GAS LAWS!

2. COMPRESSIBILITY
Compressibility is a measure of how much the VOLUME of matter can DECREASE under pressure.

3. COMPRESSIBILITY OF GASES
According to the KMT gas particles are so SMALL in relation to the distances between them that their individual volumes are virtually insignificant. This means that there is a lot of EMPTY space between the individual gas particles. This space between the particles explains why gases can be COMPRESSED.

4. VARIABLES THAT DESCRIBE GASES
Temperature (T) (KELVIN)
PRESSURE (P) (atmosphere)
Volume (V) (mL)
Number of MOLES (N)

5. TEMPERATURE Gas laws use temperature in the SI unit of KELVIN.
TEMPERATURE= measure of
the AVERAGE kinetic
energy of Particles
Converting CELSIUS to Kelvin:
Kelvin= Celsius Temperature +273
EX: What is 16 degrees Celsius in Kelvin?

6. TEMPERATURE (T) Gas laws use temperature in the SI unit of KELVIN.
Converting Celsius Temperatures to Kelvin:
Kelvin= Celsius Temperature +273
EX: What is 16 degrees Celsius in Kelvin?
Kelvin= = 289

7. PRESSURE (P)
Pressure is caused by MOVING molecules hitting container walls
PRESSURE can be measured in atmospheres, torr’s, mmHG, kPa, or psi
The SI unit for pressure is ATMOSPHERES.
Conversion of Pressure:
1 atm= 760 torr= 760mmHG= kPa= 14.7psi

8. PRESSURE CONVERSION EXAMPLES
How many torrs are in 2 atm?
How many HGmm are in 2 atm?
How many atm are in 100 kPa?

9. PRESSURE CONVERSION ANSWERS
How many torrs are in 2 atm?
1520 torr
How many HGmm are in 2 atm?
1520 Hgmm
How many atm are in 50 kPa?
.494 atm

10. VOLUME (V) Volume is measured in MILLILITERS.
Volume is the amount of SPACE occupied by a substance.

11. NUMBER OF MOLES (n)
The number of MOLES represents the number of PARTICLES contained in a system.
6.02X particles = 1 mole
22.4 L= 1 mole

12. IDEAL GAS LAW!
The ideal gas law is the mathematical relationship among pressure (P), volume (V), temperature(T) , and the number of moles(n) of a gas:
PV = nRT

13. WHAT IS ‘R’?
“R” is the IDEAL GAS CONSTANT. The UNIT of pressure need to match up!
R = L • atm
mol • K
R = 8.31 L • kPa
R = 62.4 L • mmHg

14. Sample Problem
 What volume will 2.50 mol of hydrogen (H2) occupy at °C and 1.5 atm?

15. Solve…
Use the Ideal Gas Law: PV = nRT
Solve for V:
V = nRT P

16. Substitute: V = 2.50 mol X 0.0821 L atm X 253 K 1.5 atm mol K
Don’t forget to convert temperature to Kelvin!!!
V = 2.50 mol X L atm X 253 K
1.5 atm mol K

17. Final Answer
V = 34.6 L

18. PART 2!

19. BOYLES LAW PRESSURE vs. VOLUME
As one increase the other decreases. This is an INVERSE relationship.

20. BOYLES LAW GRAPH
As the Volume decreases the pressure increases.

21. BOYLES LAW EQUATION k = VxP
k is a CONSTANT for a certain sample of gas that depends on the mass of the gas and the temperature.

22. BOYLES LAW EQUATION RATIO
If we have a set of new conditions for the same sample of gas, they will have the same k so we can set up a RATIO.

23. BOYLES LAW EXAMPLE

24. BOYLES LAW EXAMPLE
Consider a 1.53-L sample of gaseous SO2 at a pressure of 5.6 x 103 Pa. If the pressure is changed to 1.5 x 104 Pa at constant temperature, what will be the new volume of the gas?

25. CHARLES LAW TEMPERATURE vs. VOLUME
Charles Law states that the volume of a fixed mass of gas varies directly with temperature at a constant pressure.

26. Charles’ Law
DIRECT Relationship: As temperature increases the volume increases.

27. CHARLES LAW EQUATION V= kT or k=V/T
K is a CONSTANT for certain sample of gas that depends on the mass of gas and its pressure.

28. CHARLES LAW EQUATION- RATIO
If we have a set of new conditions for the same sample of gas they will have the same k so we can use a ratio to solve for our variable.
V1/T1 = k = V2/ T2
V1/T1 = V2/ T2

29. CHARLES LAW EXAMPLE
A gas with a volume of 600 mL has a temperature of 30 0C. At constant pressure the gas is heated until the gas expands to 1,200 mL. What is the new temperature of the gas if the pressure remains constant?

30. CHARLES LAW ANSWER
The first step is to convert the temperature from Celsius to Kelvin: = 303 K
Rearrange Charles’ law to solve for the new temperature: T2= V2 T1 / V1
T2 = 1200mL X 303 K = 2 X 303K = mL
T2 = 606K

31. GAY-LUSSAC’s GAS LAW
Lussac’s Gas Law states that if the TEMPERATURE of a gas is increased, and the volume is held constant, the pressure of the gas will also INCREASE.

32. Direct Relationship
Gay-Lussac's Law is a DIRECT relationship when one increases the other also increases.
Pressure
TEMPERATURE

33. GAY LUSSACS EQUATION P ÷ T = k
If we have two sets of data for the same gas we can set the equations EQUAL to each other.
We know this: P1 ÷ T1 = k
And we know this: P2 ÷ T2 = k
Since k = k, we can conclude that:
P1 ÷ T1 = P2 ÷ T2

34. GAY LUSSACS EXAMPLE
10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What would be the required temperature (in Celsius) to change the pressure to standard pressure( kPa)?

35. GAY LUSSACS STEPS
STEP 1: Change 25.0°C to K and remember that standard pressure in kPa is
STEP 2: Insert values into the equation.
STEP 3: Solve for x

36. GAY LUSSACS ANSWER
The answer is K, but the question asks for Celsius, so you subtract 273 to get the final answer of 38.3°C.
BONUS: Can you convert the temperature to KELVIN?

37. COMBINED GAS LAW
BOYLES LAW, Charles law, and Gay Lussacs Law can be combined to create a COMBINED GAS LAW.
The Combined gas Law equation is:
P1V1 / T1 = P2V2 / T2
P= Pressure
V= VOLUME
T= Temperature.

38. ……..Now It’s time for you to practice the different kinds of Gas Laws in your Gas Laws Packet. If you don’t finish in class- It’s HOMEWORK!