Charles Law by Abhishek Jaguessar

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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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