Ideal gases: Charles’ Law

Krisan M. LuisInstructor

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 mLT = 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 isdirectly related to its absolute (K) temperature

V1 = V2

T1 T2

Variation of gas volume with temperatureat constant pressure.

5.3

V a T

V = constant x T

V1/T1 = V2 /T2T (K) = t (0C) + 273.15

Charles’ Law

Temperature must bein Kelvin

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 = 3.20 L

T1 = 398.15 K

V2 = 1.54 L

T2 = ?

T2 = V2 x T1

V1

1.54 L x 398.15 K3.20 L= = 192 K

5.3

V1 /T1 = V2 /T2

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

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 conditionsV1 = 785 mL V2 = ?

T1 = 21°C = 294 K T2 = 0°C = 273 K

V2 = _______ mL x __ K = _______ mL

V1 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°CT2 = 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 100761 25691 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 = atm

K

?

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)