1. Chapter 11 Gases 11.6 The Combined Gas Law
Under water, the pressure on a diver is greater than the atmospheric pressure.

2. Summary of Gas Laws

3. Combined Gas Law
The combined gas law uses Boyle’s law, Charles’s law, and Gay-Lussac’s law (n is constant).
P1 V1 = P2 V2
T1 T2

4. Example of Using the Combined Gas Law
A sample of helium gas has a volume of L, a pressure of atm, and a temperature of 29 °C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n constant)?
STEP 1 Organize the data in a table of initial and final conditions.
Conditions 1 Conditions 2
P1 = atm P2 = atm
V1 = L (180. mL) V2 = 90.0 mL
T1 = 29 °C = 302 K T2 = ??

5. Example of Using the Combined Gas Law (continued)
STEP 2 Rearrange the gas law for the unknown.
Solve the combined gas law for T2.
P1 V1 = P2 V2
T1 T2
T = T1 P2V2
P1V1
STEP 3 Substitute values into the gas law to solve for the unknown.
T2 = 302 K x atm x mL = K
0.800 atm mL
= 604 K = °C

6. Learning Check
A gas has a volume of 675 mL at 35 °C and atm pressure. What is the volume (mL) of the gas at -95 °C and a pressure of 802 mmHg (n constant)?

7. Solution
STEP 1 Organize the data in a table of initial and final conditions.
Conditions 1 Conditions 2
T1 = 35 °C = 178K T2 = -95 °C = 178K
V1 = 675 mL V2 = ???
P1 = 646 mmHg P2 = 802 mmHg
STEP 2 Rearrange the gas law for the unknown.
P1 V1 = P2V2
T1 T2
V2 = V1 P1T2
P2T1

8. Solution (continued)
STEP 3 Substitute values into the gas law to solve for the unknown.
V2 = mL x 646 mmHg x 178K = mL
802 mmHg x 308K

9. Chapter 11 Gases 11.7 Volume and Moles (Avogadro’s Law)
Balloons rise in the air because helium is less dense than air.

10. Avogadro's Law: Volume and Moles
Avogadro’s law states that
the volume of a gas is directly related to the number of moles(n) of gas
T and P are constant
V1 = V2
n n2

11. Learning Check
If 0.75 mol of helium gas occupies a volume of 1.5 L, what volume will 1.2 mol of helium occupy at the same temperature and pressure?
1) L
2) 1.8 L
3) 2.4 L
Balloons rise in the air because helium is less dense than air.

12. Solution
STEP 1 Organize the data in a table of initial and final conditions.
Conditions Conditions 2 Know Predict
V1 = 1.5 L V2 = ? V increases
n1 = mole n2 = 1.2 moles n increases

13. Solution (continued) STEP 2 Rearrange the gas law for the unknown.
V1 = V2
n n2
V2 = V1n2 n1
STEP 3 Substitute values into the gas law to solve for the unknown.
V2 = 1.5 L x mol He = 2.4 L
0.75 mol He

14. STP
The volumes of gases can be compared at STP (standard temperature and pressure) when they have
the same temperature
Standard temperature (T) = 0 °C or 273 K
the same pressure
Standard pressure (P) = 1 atm (760 mmHg)

15. Molar Volume The molar volume of a gas
is measured at STP (standard temperature and pressure)
is 22.4 L for exactly1 mol of a gas

16. Molar Volume as a Conversion Factor
The molar volume of a gas is about the same as the volume of three basketballs.
The molar volume at STP
has about the same volume as three basketballs
can be used to write conversion factors
22.4 L and 1 mol
1 mol L

17. Using Molar Volume

18. Example of Using Molar Volume
What is the volume occupied by 2.75 mol of N2 gas
at STP?
STEP 1 Identify given and needed.
Given mol of N Need liters of N2
STEP 2 Write a plan.
moles of N2 liters of N2

19. Example of Using Molar Volume (continued)
STEP 3 Write conversion factors.
1 mol of gas = L
1 mol gas and L
22.4 L mol gas
STEP 4 Set up problem with factors to cancel units.
2.75 mol N2 x L = L of N2
1 mol N2

20. Learning Check
How many grams of He gas are present in 8.00 L of He at STP?
1) g
2) g
3) g

21. Solution STEP 1 Identify given and needed.
Given L of He Need grams of He
STEP 2 Write a plan.
liters of He moles of He grams of He

22. Solution (continued) STEP 3 Write conversion factors.
1 mol of gas = L
1 mol gas and L
22.4 L mol gas
1 mol of He = g
4.003 g He and 1 mol He
1 mol He g He
STEP 4 Set up problem with factors to cancel units.
8.00 L x 1 mol He x g He = 1.43 g of He (3)
22.4 L mol He

23. Chapter 11 Gases 11.8 The Ideal Gas Law
Dinitrogen oxide is used as an anesthetic in dentistry.

24. Ideal Gas Law
The relationship between the four properties (P, V, n, and T) of a gas can be written equal to a constant R.
PV = R nT
Rearranging this relationship gives the ideal gas law.
PV = nRT

25. Universal Gas Constant, R
The universal gas constant, R, can be calculated at STP using a temperature of 273 K, a pressure of 1.00 atm, a quantity of
1 mol of a gas, and a molar volume of 22.4 L.
P V
R = PV = (1.00 atm)(22.4 L)
nT (1 mol) (273K)
n T
= L • atm
mol • K

26. Learning Check
Another value for the universal gas constant is obtained using mmHg for the pressure at STP.
What is the value of R when a pressure of
760 mmHg is placed in the R value expression?

27. Solution
What is the value of R when a pressure of 760 mmHg (at STP) is placed in the R value expression?
R = PV = (760 mmHg) (22.4 L)
nT (1 mol) (273 K)
= L • mmHg mol • K

28. Summary of Units for the Universal Gas Constant

29. Using the Ideal Gas Law

30. Learning Check
Dinitrogen oxide (N2O), laughing gas, is used by dentists as an anesthetic. If a 20.0-L tank of laughing gas contains 2.86 mol of N2O at 23 °C, what is the pressure (mmHg) in the tank?
Dinitrogen oxide is used as an anesthetic in dentistry.

31. Solution STEP 1 Organize the data given for the gas. R = 62.4 L • mmHg
mol • K
V = L
T = 23°C = 296 K
n = mol
P = ?

32. Solution (continued)
STEP 2 Solve the ideal gas law for the unknown. Rearrange the ideal gas law for P.
P = nRT V
STEP 3 Substitute gas data and calculate the unknown quantity.
P = (2.86 mol)(62.4 L • mmHg)(296 K)
(20.0 L) (mol • K)
= x 103 mmHg

33. Learning Check
A cylinder contains 5.0 L of O2 at 20 °C and 0.85 atm. How many grams of oxygen are in the cylinder?

34. Solution STEP 1 Organize the data given for the gas.
P = 0.85 atm, V = 5.0 L, T = 293 K,
R = L • atm , n ( or g =?)
mol • K
STEP 2 Solve the ideal gas law for the unknown. Rearrange the ideal gas law for n (moles).
n = PV RT

35. Solution (continued)
STEP 3 Substitute gas data and calculate the unknown quantity.
= (0.85 atm)(5.0 L)(mol • K) = mol of O2
(0.0821atm • L)(293 K)
Convert the moles of gas to the grams of gas using its molar mass.
= mol O2 x g O2 = 5.8 g of O2
1 mol O2

36. Calculating the Molar Mass of a Gas
What is the molar mass of a gas if g of the gas
occupies 215 mL at atm and 30.0 °C?
STEP 1 Organize the data given for the gas.
R = L atm/mol K
P = atm
V = L
n = ? mol
T = °C = 303 K

37. Calculating the Molar Mass of a Gas (continued)
STEP 2 Solve the ideal gas law for the unknown. Solve for the moles (n) of gas.
n = PV = (0.813 atm) (0.215 L) mol•K = mol
RT ( L • atm)(303 K)
STEP 3 Substitute gas data and calculate the unknown quantity.
Set up the molar mass relationship.
Molar mass = g = g = g/mol
mol mol