1. The relationship between temperature and volume
Charles's Law
The relationship between temperature and volume

2. How Volume Varies With Temperature
If we place a balloon in liquid nitrogen it shrinks:
So, gases shrink if cooled.
Conversely, if we heat a gas it expands (as in a hot air balloon).
Let’s take a closer look at temperature before we try to find the exact relationship of V vs. T.

3. Temperature scales Is 20C twice as hot as 10C?
Is 20 kg twice as heavy as 10 kg?
No. 68F (20C) is not double 50F (10C)
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 0C.
We saw that doubling P yields half the V.
Yet, to investigate the effect of doubling temp-erature, we first have to know what that means.
An experiment with a fixed volume of gas in a cylinder will reveal the relationship of V vs. T…

4. Temperature vs. Volume Graph (fig.7,8 pg.430)5 10 15 20 25 30
Volume (mL)
25 mL at 22C
31.6 mL, 23.1 mL
Y=0.0847x
Temperature (C)
100
– 273

5. The Kelvin Temperature Scale
If a volume vs. temperature graph is plotted for gases, most lines can be interpolated so that when volume is 0 the temperature is -273 C.
Naturally, gases don’t really reach a 0 volume, but the spaces between molecules approach 0.
At this point all molecular movement stops.
–273C is known as “absolute zero” (no EK)
Lord Kelvin suggested that a reasonable temp-erature scale should start at a true zero value.
He kept the convenient units of C, but started at absolute zero. Thus, K = C
62C = ? K: K=C+273 = = 335 K
Notice that kelvin is represented as K not K.

6. Kelvin Practice
What is the approximate temperature for absolute zero in degrees Celsius and kelvin?
Calculate the missing temperatures
0C = _______ K 100C = _______ K
100 K = _______ C – 30C = _______ K
300 K = _______ C 403 K = _______ C
25C = _______ K 0 K = _______ C
Absolute zero is – 273C or 0 K
273
373
– 173
243 27
130
298
– 273

7. Read pages 432-3. Answer these questions:
Charles’s Law
Looking back at the temperature vs. volume graph, notice that there is a direct relationship.
It can be shown that V/T = constant
Read pages Answer these questions:
Give Charles’s law in words & as an equation.
Charles’s Law: as the temperature of a gas increases, the volume increases proportionally, provided that the pressure and amount of gas remain constant,
V1/T1 = V2/T2

8. A sample of gas occupies 3. 5 L at 300 K
A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K?
V1 = 3.5 L, T1 = 300K, V2 = ?, T2 = 200K
Using Charles’ law: V1/T1 = V2/T2
3.5 L / 300 K = V2 / 200 K
V2 = (3.5 L/300 K) x (200 K) = 2.3 L
If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be?
V1 = 1 L, T1 = 22°C = 295 K
V2 = ?, T2 = 100 °C = 373 K
V1/T1 = V2/T2, 1 L / 295 K = V2 / 373 K
V2 = (1 L/295 K) x (373 K) = 1.26 L
Do questions 16, 17, 19 on page 434