1. Section 2 – The Gas Laws
Scientists have been studying physical properties of gases for hundreds of years. In 1662, Robert Boyle discovered that gas pressure and volume are related mathematically. The observations of Boyle and others led to the development of the gas laws. The gas laws are simple mathematical relationships between the volume, temperature, pressure, and amount of a gas.

2. The gas laws
Since gases are highly compressible and will expand when heated, these properties have been studied extensively.
The relationships between volume, pressure, temperature and moles are referred to as the gas laws.
To understand the relationships, we must look a few concepts.

3. Units we will be using
Volume liters, although other units could be used.
Temperature Must use an absolute scale.
K - Kelvin is most often used.
Pressure Atm, torr, mmHg, lb/in2.
- use what is appropriate.
Moles We specify the amounts in
molar quantities.

4. Gas laws
Laws that show the relationship between volume and various properties of gases
Boyle’s Law
Charles’ Law
Gay-Lussac’s Law
Avogadro’s Law
The Ideal Gas Equation combines several of these laws into a single relationship.

5. Temperature and number of moles must be held constant!
Boyle’s law
The volume of a gas is inversely proportional to its pressure.
PV = k or
P1 V1 = P2 V2
Temperature and number of moles must be held constant!

7. Boyle’s law can be used to compare changing conditions for a gas
Using P1 and V1 to stand for initial conditions and P2 and V2 to stand for new conditions results in the following equations
P1V1 =k P2V2 = k
P1V1 = P2V2

8. Practice Problem
A sample of oxygen gas has a volume of 150. mL when its pressure is atm. What will the volume of the gas be at a pressure of atm if the temperature remains constant?
Given: V1 of O2 = 150. mL; P1 of O2 = atm; P2 of O2 = atm
Unknown: V2 of O2 in mL

9. Rearrange the equation for Boyle’s law (P1V1 = P2V2) to obtain V2
Given: V1 of O2 = 150. mL; P1 of O2 = atm; P2 of O2 = atm Unknown: V2 of O2 in mL
Rearrange the equation for Boyle’s law (P1V1 = P2V2) to obtain V2

10. Practice Problems
A balloon filled with helium gas has a volume of 500 mL at a pressure of 1 atm.The balloon is released and reaches an altitude of 6.5 km, where the pressure is 0.5 atm. Assuming that the temperature has remained the same, what volume does the gas occupy at this height?
Answer 1000 mL He

11. A gas has a pressure of 1. 26 atm and occupies a volume of 7. 40 L
A gas has a pressure of 1.26 atm and occupies a volume of 7.40 L. If the gas is compressed to a volume of 2.93 L, what will its pressure be, assuming constant temperature?
Answer 3.18 atm

12. Divers know that the pressure exerted by the water increases about 100 kPa with every 10.2 m of depth. This means that at 10.2 m below the surface, the pressure is 201 kPa; at 20.4 m, the pressure is 301 kPa; and so forth. Given that the volume of a balloon is 3.5 L at STP and that the temperature of the water remains the same, what is the volume 51 m below the water’s surface?
Answer 0.59 L

13. Charles’s Law: Volume-Temperature
Charles’s law  the volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature

14. Charles’ law When you heat a sample of a gas, its volume increases.
The pressure and number
of moles must be held constant.

15. Charles’ Law Placing an air filled balloon near liquid nitrogen (77 K)
will cause the volume to be reduced. Pressure and
the number of moles are constant.

16. The temperature −273.15°C is referred to as absolute zero and is given a value of zero in the Kelvin scale
K = °C
V = kT or V/T = k

17. Practice Problems
A sample of neon gas occupies a volume of 752 mL at 25°C.What volume will the gas occupy at 50°C if the pressure remains constant?
Given: V1 of Ne = 752 mL; T1 of Ne = 25°C = 298 K; T2 of Ne = 50°C = 323 K
Note that Celsius temperatures have been converted to kelvins.This is a very important step for working the problems in this chapter
Unknown: V2 of Ne in mL

18. Given: V1 of Ne = 752 mL; T1 of Ne = 25°C = 298 K; T2 of Ne = 50°C = 323 K Unknown: V2 of Ne in mL

19. A helium-filled balloon has a volume of 2. 75 L at 20. °C
A helium-filled balloon has a volume of 2.75 L at 20.°C.The volume of the balloon decreases to L after it is placed outside on a cold day. What is the outside temperature in K? in °C?
Answer 262 K, or −11°C

20. A gas at 65°C occupies 4.22 L. At what Celsius temperature will the volume be 3.87 L, assuming the same pressure?
Answer 37°C

21. Gay-Lussac’s Law: Pressure-Temperature
Gay-Lussac’s law: The pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature
P = kT or P/T = k

22. Gay-Lussac’s Law Law of of Combining Volumes.
At constant temperature and pressure, the volumes of gases involved in a chemical reaction are in the ratios of small whole numbers.
Studies by Joseph Gay-Lussac led to a better understanding of molecules and their reactions.

23. Gay-Lussac’s Law Example. Reaction of hydrogen and oxygen gases.
Two ‘volumes’ of hydrogen will combine with one ‘volume’ of oxygen to produce two volumes of water.
We now know that the equation is:
2 H2 (g) + O2 (g) H2O (g) + H2 O2
H2O

24. The gas in an aerosol can is at a pressure of 3. 00 atm at 25°C
The gas in an aerosol can is at a pressure of 3.00 atm at 25°C. Directions on the can warn the user not to keep the can in a place where the temperature exceeds 52°C. What would the gas pressure in the can be at 52°C?
Given: P1 of gas = 3.00 atm; T1 of gas = 25°C = 298 K; T2 of gas = 52°C = 325 K
Unknown: P2 of gas in atm

25. Given: P1 of gas = 3.00 atm; T1 of gas = 25°C = 298 K; T2 of gas = 52°C = 325 K Unknown: P2 of gas in atm

26. Before a trip from New York to Boston, the pressure in an automobile tire is 1.8 atm at 20.°C. At the end of the trip, the pressure gauge reads 1.9 atm. What is the new Celsius temperature of the air inside the tire? (Assume tires with constant volume.)
Answer 36°C

27. At 120. °C, the pressure of a sample of nitrogen is 1. 07 atm
At 120.°C, the pressure of a sample of nitrogen is atm. What will the pressure be at 205°C, assuming constant volume?
Answer 1.30 atm

28. A sample of helium gas has a pressure of 1. 20 atm at 22°C
A sample of helium gas has a pressure of 1.20 atm at 22°C. At what Celsius temperature will the helium reach a pressure of 2.00 atm?
Answer 219°C

29. Avogadro’s law
Equal volumes of gas at the same temperature and pressure contain equal numbers of molecules.
V = k n
V V2
n n2 =

30. Avogadro’s law If you have more moles of a gas, it takes up more
space at the same temperature and pressure.

31. Standard conditions (STP)
Remember the following standard conditions.
Standard temperature = K
Standard pressure = 1 atm
At these conditions:
One mole of a gas has
a volume of 22.4 liters.

32. The ideal gas law
A combination of Boyle’s, Charles’ and Avogadro’s Laws
PV = nRT
P = pressure, atm
V = volume, L
n = moles
T = temperature, K
R = L atm/K mol
(gas law constant)

33. Example What is the volume of 2.00 moles of gas at
3.50 atm and K?
PV = nRT
V = nRT / P
= (2.00 mol)( L atm K-1mol-1)(310.0 K)
(3.50 atm)
= 14.5 L

34. Ideal gas law PV R = nT R = = 0.08206 atm L mol-1 K-1
R can be determined from standard conditions. PV nT
R =
( 1 atm ) ( 22.4 L )
( 1 mol ) ( K)
R =
= atm L mol-1 K-1

35. Ideal gas law
When you only allow volume and one other factor to vary, you end up with one of the other gas laws.
Just remember
Boyle Pressure
Charles Temperature
Avogadro Moles

36. Ideal gas law P1V1 P2V2 n1T1 n2T2 = R = This one equation says it all.
Anything held constant will
“cancels out” of the equation

37. Ideal gas law P1V1 P2V2 n1T1 n2T2 = P1V1 = P2V2
Example - if n and T are held constant
P1V1
n1T1
P2V2
n2T2 =
This leaves us
P1V1 = P2V2
Boyle’s Law

38. Example If a gas has a volume of 3.0 liters at 250 K,
what volume will it have at 450 K ?
Cancel P and n
They don’t change
P1V1
n1T1
P2V2
n2T2 =
We end up with
Charles’ Law V1 T1 V2 T2 =

39. Example If a gas has a volume of 3.0 liters at 250 K,
what volume will it have at 450 K ?
P1V1
n1T1
P2V2
n2T2 V1 T1 = V2 T2 =
V2 = (3.0 l) (450 K)
(250 K)
= 5.4 L

40. Dalton’s law of partial pressures
The total pressure of a gaseous mixture is the sum of the partial pressure of all the gases.
PT = P1 + P2 + P
Air is a mixture of gases - each adds it own pressure to the total.
Pair = PN2 + PO2 + PAr + PCO2 + PH2O

41. Partial pressure example
Mixtures of helium and oxygen are used in scuba diving tanks to help prevent “the bends.”
For a particular dive, 46 liters of O2 and 12 liters of He were pumped in to a 5 liter tank. Both gases were added at 1.0 atm pressure at 25oC.
Determine the partial pressure for both gases in the scuba tank at 25oC.

42. Partial pressure example
First calculate the number of moles of each gas using PV = nRT.
nO2 = = 1.9 mol
nHe = = 0.49 mol
(1.0 atm) (46 l)
( l atm K-1 mol-1)(298.15K)
(1.0 atm) (12 l)
( l atm K-1 mol-1)(298.15K)

43. Partial pressure example
Now calculate the partial pressures of each.
PO2 = = 9.3 atm
PHe = = 2.4 atm
Total pressure in the tank is 11.7 atm.
(1.9 mol) ( K) ( l atm K-1 mol-1)
(5.0 l)
(0.49 mol) ( K) ( l atm K-1 mol-1)
(5.0 l)

44. Diffusion

45. Diffusion and effusion
The random and spontaneous mixing of molecules.
Effusion
The escape of
molecules through
small holes in a
barrier.

46. Practice Problems
A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm.What volume will it have at atm and 10.°C? Given: V1 of He = 50.0 L; T1 of He = 25°C = 298 K; T2 of He = 10°C = 283 K; P1 of He = 1.08 atm; P2 of He = atm Unknown: V2 of He in L

47. Given: V1 of He = 50.0 L; T1 of He = 25°C = 298 K; T2 of He = 10°C = 283 K; P1 of He = 1.08 atm; P2 of He = atm Unknown: V2 of He in L

48. The volume of a gas is 27. 5 mL at 22. 0°C and 0. 974 atm
The volume of a gas is 27.5 mL at 22.0°C and atm. What will the volume be at 15.0°C and atm?
Answer 26.3 mL

49. A 700. mL gas sample at STP is compressed to a volume of 200
A 700. mL gas sample at STP is compressed to a volume of 200. mL, and the temperature is increased to 30.0°C. What is the new pressure of the gas in Pa?
Answer 3.94 × 105 Pa, or 394 kPa

50. Given: PT = Patm = 731. 0 torr; PH2O = 17
Given: PT = Patm = torr; PH2O = 17.5 torr (vapor pressure of water at 20.0°C); Patm = PO2 + PH2O Unknown: PO2 in torr
PO2 = Patm − PH2O
PO2 = torr − 17.5 torr = torr

51. Some hydrogen gas is collected over water at 20. 0°C
Some hydrogen gas is collected over water at 20.0°C.The levels of water inside and outside the gas-collection bottle are the same. The partial pressure of hydrogen is torr. What is the barometric pressure at the time the gas is collected?
Answer torr

52. Helium gas is collected over water at 25°C
Helium gas is collected over water at 25°C.What is the partial pressure of the helium, given that the barometric pressure is mm Hg?
Answer mm Hg