The relationship between temperature and

volume

If we place a balloon in liquid

nitrogen it shrinks:

How Volume Varies With Temperature

So, gases shrink if cooled.Conversely, if we heat a

gas it expands (as in a hot air balloon).

No. 68F (20C) is not double 50F (10C)

Yes. 44 lb (20 kg) is double 22 lb (10 kg)What’s the difference?• Weights (kg or lb) have a minimum value of 0.• But the smallest temperature is not 0C.• We saw that doubling P yields half the V.• An experiment with a fixed volume of gas in a

cylinder will reveal the relationship of V vs. T…

Temperature scalesIs 20C twice as hot as 10C?

Is 20 kg twice as heavy as 10 kg?

CHARLES' LAW

If a gas has a constant mass and is held at a constant pressure then

the volume divided by the temperature (in kelvins) is a

constant value.

V = kT or = k

VT

Mathematically,

Graphically

Using a glass syringe, a scientist draws exactly 25.5 cm3 of dry oxygen at 20°C

from a metal cylinder. She heats the syringe to 65 °C. What volume will the

oxygen occupy?1. What are you trying to

determine?

What volume will the oxygen occupy?

Vf or V2

Step 2: Identify what is given:Vi = 25.5 cm3

Ti = 20.0 °C

Tf = 65.0 °C

Step 3: Convert temperatures from °C to KTi = 20.0 °C Tf = 65.0 °C

= 20.0 °C + 273 = 65.0 °C + 273

= 293 K = 338 K

Step 4: Use equation to determine final volume:

=

=

x 338 K = Vf

Vi

Ti

Vf

Tf

25.5293

Vf

33825.5

293

Vf = 29.42 cm3

A balloon is filled with 2.50 L of dry helium at 23.5°C. the balloon is placed in a

freezer overnight. The resulting volume is found to be 2.15L. What was the

temperature (in °C) in the freezer.

1. What are you trying to determine?

What was the temperature (in °C) in the freezer.

Tf

Step 2: Identify what is given:Vi = 2.50 L

Ti = 23.5 °C

Vf = 2.15 L

Step 3: Convert temperature from °C to KTi = 23.5 °C + 273

= 297 K

Step 4: Use equation to determine final temperature:

=

= 2.50 L (Tf) = 297K (2.15L)

Tf = 255.42K

Vi

Ti

Vf

Tf

2.50297

2.15 Tf

Cross multiply

Divide by 2.50L to isolate Tf

Step 5: Convert the temperature back to °C

Tf = 255.42K – 273

= -17.6 °C

3. A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K?

4. If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be?

V1 = 3.5 L, T1 = 300K, V2 = ?, T2 = 200KUsing Charles’ law: V1/T1 = V2/T2

3.5 L / 300 K = V2 / 200 KV2 = (3.5 L/300 K) x (200 K) = 2.3 L

V1 = 1 L, T1 = 22°C = 295 KV2 = ?, T2 = 100 °C = 373 KV1/T1 = V2/T2, 1 L / 295 K = V2 / 373 KV2 = (1 L/295 K) x (373 K) = 1.26 L