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

P a
1/ V
P x V = constant
P
1 x V
1
= P
2 x V
2
Constant temperature
Constant amount of gas
2

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
P
1 x V
1
= P
2 x V
2
P
1
= 726 mmHg P
2
= ?
V
1
= 946 mL V
2
= 154 mL
P
2
=
P
1 x V
1
V
2
=
726 mmHg x 946 mL
154 mL
= 4460 mmHg
3

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

Variation of Gas Volume with Temperature at Constant Pressure
V a
T
V = constant x T
V
1
/ T
1
= V
2
/ T
2
Charles’ &
GayLussac’s
Law
Temperature must be in Kelvin
T (K) = t ( 0 C) + 273.15
5

A sample of carbon monoxide gas occupies 3.20 L at 125 0 C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?
V
1
/ T
1
= V
2
/ T
2
V
T
1
1
= 3.20 L
= 398.15 K
V
2
= 1.54 L
T
2
= ?
T
1
= 125 ( 0 C) + 273.15 (K) = 398.15 K
T
2
=
V
2 x T
1
V
1
=
1.54 L x 398.15 K
3.20 L
= 192 K
6

V a number of moles ( n )
Constant temperature
Constant pressure
V = constant x n
V
1
/ n
1
= V
2
/ n
2
7

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?
4NH
3
+ 5O
2
4NO + 6H
2
O
1 mole NH
3
1 mole NO
At constant T and P
1 volume NH
3
1 volume NO
8

Boyle’s Law
Summary of Gas Laws
9

Charles Law
10

Avogadro’s Law
11

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 nT
V = constant x = R
P nT
P
R is the gas constant
PV = nRT
12

The conditions 0 0 C and 1 atm are called standard temperature and pressure
(STP).
Experiments show that at STP, 1 mole of an ideal gas occupies
22.414 L.
PV = nRT
R =
PV nT
(1 atm)(22.414L)
=
(1 mol)(273.15 K)
R = 0.082057 L • atm / (mol • K)
13

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

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 0 C is heated to 85 0 C 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
P
1
= 1.20 atm P
2
= ?
T
1
= 291 K T
2
= 358 K
P
T
1
1
=
P
2
T
2
P
2
= P
1 x
T
2
T
1
= 1.20 atm x 358 K
291 K
= 1.48 atm
15

Sample Problem 5.2
PROBLEM:
Applying the Volume-Pressure Relationship
Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8 cm 3 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:
V
1 in cm 3
1cm 3 =1mL
V
1 in mL
10 3 mL=1L unit conversion
SOLUTION:
P
V
1
1
= 1.12 atm
= 24.8 cm
24.8 cm 3
3
1 mL
1 cm 3
V
1 in L xP
1
/P
2
V
2 in L gas law calculation
P
1
V n
1
T
1
1
=
P
2
V
2 n
2
T
2
V
2
=
P
1
V
1
= 0.0248 L
P
2
P and T are constant
10
L
3
P
V
2
2 mL
1.12 atm
2.46 atm
= 2.64 atm
= unknown
P
1
= 0.0248 L
V
1
= P
2
V
2
= 0.0105 L

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 1.00x10
3 torr. It is filled with methane at 23 0 C and 0.991 atm and placed in boiling water at exactly 100 0 C. Will the safety valve open?
PLAN:
P
1
(atm) T
1 and T
2
( 0 C)
P
1
1atm=760torr
(torr) x T
2
/T
1
K= 0 C+273.15
T
1 and T
2
(K)
P
2
(torr)
0.991 atm
P
1
V
1 n
1
T
1
SOLUTION:
P
1
= 0.991atm
T
1
= 23 0 C
=
P n
2
2
V
T
2
2
760 torr = 753 torr
1 atm
T
2
P
2
= P
1
T
1
= 753 torr
373K
296K
= 949 torr
P
2
= unknown
T
2
= 100 0 C
P
1
T
1
=
P
2
T
2

Sample Problem 5.4
Applying the Volume-Amount Relationship
PROBLEM:
PLAN:
A scale model of a blimp rises when it is filled with helium to a volume of
55 dm 3 . When 1.10 mol of He is added to the blimp, the volume is 26.2 dm 3 . How many more grams of He must be added to make it rise?
Assume constant T and P.
We are given initial n
1 and V
1 as well as the final V and convert it from moles to grams.
2
. We have to find n
2 n
1
(mol) of He x V
2
/V
1 n
2
(mol) of He subtract n
1 mol to be added x M g to be added
SOLUTION: n
1
= 1.10 mol
V
1 n
1
V
1
= 26.2 dm 3
=
V
2 n
2 n
2
= 1.10 mol
P and T are constant n
2 n
2
V
2
= n
1
55.0 dm
26.2 dm
3
3
= unknown
= 55.0 dm 3
P
1
V
1 n
1
T
1
V
2
V
1
= 2.31 mol
4.003 g He mol He
=
P n
2
2
V
T
2
2
= 9.24 g He

Sample Problem 5.5
PROBLEM:
Solving for an Unknown Gas Variable at Fixed
Conditions
A steel tank has a volume of 438 L and is filled with 0.885 kg of O
2
.
Calculate the pressure of O
2 at 21 0 C.
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 n = 0.885 kg (convert to mol)
T = 21 0 C (convert to K)
P = unknown
0.885kg
10 3 g kg mol O
2
32.00 g O
2
= 27.7 mol O
2
24.7 mol x 0.0821
atm*L mol*K
P = nRT
V
=
438 L
21 0 C + 273.15 = 294.15K
x 294.15K
= 1.53 atm

Sample Problem 5.6
PROBLEM:
Using Gas Laws to Determine a Balanced Equation
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) A
2
+ B
2
2AB (2) 2AB + B
2
2AB
2
(3) A + B
2
AB
2
(4) 2AB
2
A
2
+ 2B
2
PLAN:
SOLUTION:
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.
Looking at the relationships, the equation that shows a decrease in the number of moles of gas from 2 to 1 is equation (3).