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Charles’ Law

• At constant pressure, the volume of a gas varies directly with its Kelvin temperature

• Equations:

V1 V2 V2

V1

=T2

T1

=T1 T2

or or V1T2 = V2T1

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At Constant Pressure,

• when volume , temp

• When volume , temp

Temp. scale K not °CWhat error exists

on this graph?

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Solving Charles’ Law Problems

– Write the given information– Convert to Kelvin temp. scale– Isolate the missing variable – Plug-in the given values and solve

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Example:A volume of 20.0m3 of argon gas is kept under constant

pressure. The gas is heated from 22ºC to 283 ºC. What is the new volume of the gas?

Step 1: Write down the information you know

V2 = ?

V1 = 20.0 m3

T1 = 22ºC

T2 = 283 º C

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Step2) Convert Temperatures into KelvinoC + 273

T1= 22oC 22 + 273 = 296

T2= 283oC 283 + 273 = 556K

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Step 3) Isolate the unknown variable

V1T2 = V2T1 We don’t know V2

Divide both sides by T1

V1T2 = V2T1

T1 T1

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Step 4) Plug in the numbers you have and solve

(20.0 m3)(556 K) = V2 (Note: Kelvin

296 K units cancel)

37.6 m3 = V2