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The relationship between temperature andvolume
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If we placea balloonin liquid
nitrogen itshrinks:
How Volume Varies With Temperature
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 wetry to find the exact
relationship of V vs. T.
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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…
Temperature scalesIs 20°C twice as hot as 10°C?
Is 20 kg twice as heavy as 10 kg?
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Temperature vs. Volume Graph (fig.7,8 pg.430)
5
10
15
20
25
30
Volu
m
e(mL
)
Temperature (°C)
0 100 – 273
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• 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 startedat absolute zero. Thus, K = °C + 273.
62°C = ? K: K=°C+273 = 62 + 273 = 335 K
• Notice that kelvin is represented as K not °K.
The Kelvin Temperature Scale
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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
Kelvin Practice
273 373
– 173 243
27 130298 – 273
Absolute zero is – 273°C or 0 K
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• Looking back at the temperature vs. volumegraph, notice that there is a direct relationship.
• It can be shown that V/T = constant
Read pages 432-3. Answer these questions:
1. Give Charles’s law in words & as an equation.Charles’s Law: as the temperature of a gasincreases, the volume increasesproportionally, provided that the pressure andamount of gas remain constant,
V1/T
1= V
2/T
2
Charles’s Law
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2. A sample of gas occupies 3.5 L at 300 K. What
volume will it occupy at 200 K?
3. If a 1 L balloon is heated from 22°C to 100°C,what will its new volume be?
4. Do questions 16, 17, 19 on page 434
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
V1 = 1 L, T1 = 22°C = 295 K
V2 = ?, T2 = 100 °C = 373 KV1/T1 = V2/T2, 1 L / 295 K = V2 / 373 K
V2 = (1 L/295 K) x (373 K) = 1.26 L
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