1. Apparatus for Studying the Relationship Between
Pressure and Volume of a Gas
As P (h) increases
V decreases

2. Boyle’s Law P a 1/V Constant temperature P x V = constant
Constant amount of gas
P x V = constant
P1 x V1 = P2 x V2

3. P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL
A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?
P x V = constant
P1 x V1 = P2 x V2
P1 = 726 mmHg
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1 V2
726 mmHg x 946 mL
154 mL =
P2 =
= 4460 mmHg

4. As T increases V increases
Variation in Gas Volume with Temperature at Constant Pressure
As T increases
V increases

5. Variation of Gas Volume with Temperature at Constant Pressure
Charles’ & Gay-Lussac’s Law
V a T
Temperature must be
in Kelvin
V = constant x T
V1/T1 = V2 /T2
T (K) = t (0C)

6. A sample of carbon monoxide gas occupies 3. 20 L at 125 0C
A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?
V1 /T1 = V2 /T2
V1 = 3.20 L
V2 = 1.54 L
T1 = K
T2 = ?
T1 = 125 (0C) (K) = K
V2 x T1 V1
1.54 L x K
3.20 L =
T2 =
= 192 K

7. Avogadro’s Law V a number of moles (n) V = constant x n
Constant temperature
Constant pressure
V a number of moles (n)
V = constant x n
V1 / n1 = V2 / n2

8. 4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P
Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?
4NH3 + 5O NO + 6H2O
1 mole NH mole NO
At constant T and P
1 volume NH volume NO

9. Summary of Gas Laws
Boyle’s Law

10. Charles Law

11. Avogadro’s Law

12. Ideal Gas Equation 1 Boyle’s law: P a (at constant n and T) V
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
V a nT P
V = constant x = R nT P
R is the gas constant
PV = nRT

13. PV = nRT PV (1 atm)(22.414L) R = = nT (1 mol)(273.15 K)
The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).
Experiments show that at STP, 1 mole of an ideal gas occupies L.
PV = nRT
R = PV nT =
(1 atm)(22.414L)
(1 mol)( K)
R = L • atm / (mol • K)

14. PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.7 L
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
T = 0 0C = K
P = 1 atm
PV = nRT
n = 49.8 g x
1 mol HCl
36.45 g HCl
= 1.37 mol
V =
nRT P
V =
1 atm
1.37 mol x x K
L•atm
mol•K
V = 30.7 L

15. PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 =
Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?
PV = nRT
n, V and R are constant nR V = P T
= constant
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K P1 T1 P2 T2 =
P2 = P1 x T2 T1
= 1.20 atm x
358 K
291 K
= 1.48 atm

16. Sample Problem 5.2
Applying the Volume-Pressure Relationship
PROBLEM:
Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8 cm3 at 1.12 atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64 atm. Assuming constant temperature, what is the new volume of air (in L)?
PLAN:
SOLUTION:
P and T are constant
V1 in cm3
P1 = 1.12 atm
P2 = 2.64 atm
unit conversion
1cm3=1mL
V1 = 24.8 cm3
V2 = unknown
V1 in mL
1 mL
1 cm3 L
103 mL
103 mL=1L
24.8 cm3
= L
V1 in L
gas law calculation
xP1/P2
P1V1
n1T1
P2V2
n2T2
P1V1 = P2V2 =
V2 in L
P1V1 P2
V2 =
1.12 atm
2.46 atm = L = L

17. Sample Problem 5.3
Applying the Pressure-Temperature Relationship
PROBLEM:
A steel tank used for fuel delivery is fitted with a safety valve that opens when the internal pressure exceeds 1.00x103 torr. It is filled with methane at 230C and atm and placed in boiling water at exactly 1000C. Will the safety valve open?
PLAN:
SOLUTION:
P1(atm)
T1 and T2(0C)
P1 = 0.991atm
P2 = unknown
1atm=760torr
K=0C
T1 = 230C
T2 = 1000C
P1(torr)
T1 and T2(K)
P1V1
n1T1
P2V2
n2T2 = P1 T1 P2 T2 =
x T2/T1
P2(torr)
0.991 atm
1 atm
760 torr
= 753 torr
P2 = P1 T2 T1
= 753 torr
373K
296K
= 949 torr

18. Sample Problem 5.4
Applying the Volume-Amount Relationship
PROBLEM:
A scale model of a blimp rises when it is filled with helium to a volume of 55 dm3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm3. How many more grams of He must be added to make it rise? Assume constant T and P.
PLAN:
We are given initial n1 and V1 as well as the final V2. We have to find n2 and convert it from moles to grams.
n1(mol) of He
SOLUTION:
P and T are constant
x V2/V1
n1 = 1.10 mol
n2 = unknown
P1V1
n1T1
P2V2
n2T2 =
n2(mol) of He
V1 = 26.2 dm3
V2 = 55.0 dm3
subtract n1 V1 n1 V2 n2 =
n2 = n1 V2 V1
mol to be added
x M
n2 = 1.10 mol
55.0 dm3
26.2 dm3
4.003 g He
mol He
g to be added
= 9.24 g He
= 2.31 mol

19. Sample Problem 5.5
Solving for an Unknown Gas Variable at Fixed Conditions
PROBLEM:
A steel tank has a volume of 438 L and is filled with kg of O2. Calculate the pressure of O2 at 210C.
PLAN:
V, T and mass, which can be converted to moles (n), are given. We use the ideal gas law to find P.
SOLUTION:
V = 438 L
T = 210C (convert to K)
n = kg (convert to mol)
P = unknown
0.885kg
103 g kg
mol O2
32.00 g O2
= 27.7 mol O2
210C = K
24.7 mol
294.15K
atm*L
mol*K
0.0821 x
438 L
P =
nRT V
= 1.53 atm =

20. Sample Problem 5.6
Using Gas Laws to Determine a Balanced Equation
PROBLEM:
The piston-cylinders below depict a gaseous reaction carried out at constant pressure. Before the reaction, the temperature is 150K; when it is complete, the temperature is 300K.
New figures go here.
Which of the following balanced equations describes the reaction?
(1) A2 + B AB
(2) 2AB + B AB2
(3) A + B AB2
(4) 2AB A2 + 2B2
PLAN:
We know P, T, and V, initial and final, from the pictures. Note that the volume doesn’t change even though the temperature is doubled. With a doubling of T then, the number of moles of gas must have been halved in order to maintain the volume.
SOLUTION:
Looking at the relationships, the equation that shows a decrease in the number of moles of gas from 2 to 1 is equation (3).