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Ideal gases: Charles’ Law
Krisan M. Luis
Instructor
Objectives
• To describe the behavior of ideal gases
– To understand the following gas laws
• Boyle’s Law
• Charles’ Law
• Avogadro’s Law

– To understand the ideal gas equation
– To apply the above concepts on real life situations
Kinetic Theory of Gases
The particles in gases

•

Are very far apart

•

Move very fast in straight lines until they collide

•

Have no attraction (or repulsion)

•

Move faster at higher temperatures
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
Variation of gas volume with temperature
at constant pressure.

Charles’ Law

V

T

V = constant x T
V1/T1 = V2 /T2

Temperature must be
in Kelvin
T (K) = t (0C) + 273.15

5.3
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 = V2
T1
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.
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 = 398.15 K

T2 = ?

T1 = 125 (0C) + 273.15 (K) = 398.15 K
T2 =

V2 x T1
V1

=

1.54 L x 398.15 K
3.20 L

= 192 K
5.3
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 T1 = 18°C + 273 = 291 K
P2 = ? ?
T2 = 62°C + 273 = 335 K
What happens to P when T increases?
P increases (directly related to T)
P2 = P1 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)

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3 charles law

  • 1. Ideal gases: Charles’ Law Krisan M. Luis Instructor
  • 2. Objectives • To describe the behavior of ideal gases – To understand the following gas laws • Boyle’s Law • Charles’ Law • Avogadro’s Law – To understand the ideal gas equation – To apply the above concepts on real life situations
  • 3. Kinetic Theory of Gases The particles in gases • Are very far apart • Move very fast in straight lines until they collide • Have no attraction (or repulsion) • Move faster at higher temperatures
  • 4. 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?
  • 5. 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
  • 6. Variation of gas volume with temperature at constant pressure. Charles’ Law V T V = constant x T V1/T1 = V2 /T2 Temperature must be in Kelvin T (K) = t (0C) + 273.15 5.3
  • 7. 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.
  • 8. Solution GL3 V1 = V2 T1 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.
  • 9. 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 = 398.15 K T2 = ? T1 = 125 (0C) + 273.15 (K) = 398.15 K T2 = V2 x T1 V1 = 1.54 L x 398.15 K 3.20 L = 192 K 5.3
  • 10. 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?
  • 11. 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.
  • 12. 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
  • 13. 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
  • 14. 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
  • 15. 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).
  • 16. Solution GL5 1. When temperature decreases, the pressure of a gas (decreases). 2. When temperature increases, the pressure of a gas (increases).
  • 17. 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
  • 18. PT Calculation P1 = 2.0 atm T1 = 18°C + 273 = 291 K P2 = ? ? T2 = 62°C + 273 = 335 K What happens to P when T increases? P increases (directly related to T) P2 = P1 x T2 T1 P2 = 2.0 atm x K = K atm
  • 19. 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)
  • 20. 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)