Chapter 7
Gases
Pressure and Volume (Boyle’s Law)
Temperature and Volume
(Charles’ Law)
Temperature and Pressure
(Gay-Lussac’s Law)
Pressure and Volume
Experiment
Pressure Volume
(atm)
(L)
PxV
(atm x L)
1
8.0
2.0
16
2
4.0
4.0
_____
3
2.0
8.0
_____
4
1.0
16
_____
Boyle's Law
P x V = k (constant) when
T remains constant
P and V Changes
P1
V1
P2
V2
Boyle's Law
The pressure of a gas is inversely related to
the volume when T does not change
Then the PV product remains constant
P1V1
=
P2V2
P 1 V 1=
8.0 atm x 2.0 L
= 16 atm L
P 2 V 2=
4.0 atm x 4.0 L
= 16 atm L
PV Problem
Freon-12, CCl2F2, is used in refrigeration
systems. What is the new volume (L) of a
1.6 L sample of Freon gas initially at 50
mm Hg after its pressure is changed to
200 mm Hg at constant T?
PV Calculation
Prepare a data table
DATA TABLE
Initial conditions
Final conditions
P1
= 50 mm Hg
P2
= 200 mm Hg
V1
= 1.6 L
V2
= ?
?
Find New Volume (V2)
Solve for V2:
V2 =
P1V2 = P2V2
V1 x P1 /P2
V2 = 1.6 L x 50 mm Hg =
200 mm Hg
0.4 L
Learning Check GL1
A sample of nitrogen gas is 6.4 L at a
pressure of 0.70 atm. What will the new
volume be if the pressure is changed to
1.40 atm? (T constant) Explain.
1) 3.2 L
2) 6.4 L
3) 12.8 L
Solution GL1
A sample of nitrogen gas is 6.4 L at a
pressure of 0.70 atm. What will the new
volume be if the pressure is changed to
1.40 atm? (T constant)
6.4 L x 0.70 atm
=
3.2 L (1)
1.40 atm
Volume must decrease to cause an
increase in the pressure
Learning Check GL2
A sample of helium gas has a volume of
12.0 L at 600. mm Hg. What new
pressure is needed to change the
volume to 36.0 L? (T constant) Explain.
1) 200. mmHg
2) 400. mmHg
3) 1200 mmHg
Solution GL2
A sample of helium gas has a volume of
12.0 L at 600. mm Hg. What new
pressure is needed to change the
volume to 36.0 L? (T constant) Explain.
600. mm Hg x 12.0 L = 200. mmHg (1)
36.0 L
Pressure decrease when volume
increases.
Charles’ Law
V = 125 mL
V = 250 mL
T = 273 K
T = 546 K
Observe the V and T of the balloons. How
does volume change with temperature?
Charles’ Law: V and T
At constant pressure, the volume of a gas is
directly related to its absolute (K) temperature
V1 = V2
T1
T2
Learning Check GL3
Use Charles’ Law to complete the statements
below:
1. If final T is higher than initial T, final V
is (greater, or less) than the initial V.
2. If final V is less than initial V, final T is
(higher, or lower) than the initial T.
Solution GL3
V1
T1
= V2
T2
1. If final T is higher than initial T, final V
is (greater) than the initial V.
2. If final V is less than initial V, final T is
(lower) than the initial T.
V and T Problem
A balloon has a volume of
785 mL on a Fall day when
the temperature is 21°C. In
the winter, the gas cools to
0°C. What is the new volume
of the balloon?
VT Calculation
Complete the following setup:
Initial conditions
Final conditions
V1 = 785 mL
V2 = ?
T1 = 21°C = 294 K
T2 = 0°C = 273 K
V2 = _______ mL x __
V1
K = _______ mL
K
Check your answer: If temperature decreases,
V should decrease.
Learning Check GL4
A sample of oxygen gas has a volume of
420 mL at a temperature of 18°C. What
temperature (in °C) is needed to change
the volume to 640 mL?
1) 443°C
2) 170°C
3) - 82°C
Solution GL4
A sample of oxygen gas has a volume of
420 mL at a temperature of 18°C. What
temperature (in °C) is needed to change
the volume to 640 mL?
2) 170°C
T2 = 291 K x 640 mL = 443 K
420 mL
= 443 K - 273 K
= 170°C
Gay-Lussac’s Law: P and T
The pressure exerted by a confined gas
is directly related to the temperature
(Kelvin) at constant volume.
P (mm Hg)
T (°C)
936
761
691
100
25
0
Learning Check GL5
Use Gay-Lussac’s law to complete the
statements below:
1. When temperature decreases, the
pressure of a gas (decreases or increases).
2. When temperature increases, the pressure
of a gas (decreases or increases).
Solution GL5
1. When temperature decreases, the
pressure of a gas (decreases).
2. When temperature increases, the
pressure of a gas (increases).
PT Problem
A gas has a pressure at 2.0 atm at 18°C.
What will be the new pressure if the
temperature rises to 62°C? (V constant)
T = 18°C
T = 62°C
PT Calculation
P1 = 2.0 atm
P2 = ? ?
T1 = 18°C + 273 = 291 K
T2 = 62°C + 273 = 335 K
What happens to P when T increases?
P increases (directly related to T)
P 2 = P 1 x T2
T1
P2 =
2.0 atm
x
K =
K
atm
Learning Check GL6
Complete with 1) Increases 2) Decreases
3) Does not change
A. Pressure _____, when V decreases
B. When T decreases, V _____.
C. Pressure _____ when V changes from 12.0 L
to 24.0 L (constant n and T)
D. Volume _____when T changes from 15.0 °C to
45.0°C (constant P and n)
Solution GL6
A. Pressure 1) Increases, when V decreases
B. When T decreases, V 2) Decreases
C. Pressure 2) Decreases when V changes
from 12.0 L to 24.0 L (constant n and T)
D. Volume 1) Increases when T changes from 15.0
°C to 45.0°C (constant P and n)