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Charles’ Law

Charles’ Law. At constant pressure, the volume of a gas varies directly with its Kelvin temperature Equations: V 1 V 2 V 2. T 2. =. =. V 1 T 2 = V 2 T 1. or. or. T 1. T 2. V 1. T 1. At Constant Pressure,. when volume , temp When volume , temp.

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Charles’ Law

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  1. Charles’ Law • At constant pressure, the volume of a gas varies directly with its Kelvin temperature • Equations: V1V2V2 T2 = = V1T2 = V2T1 or or T1 T2 V1 T1

  2. At Constant Pressure, • when volume , temp • When volume , temp What error exists on this graph? Temp. scale K not °C

  3. Solving Charles’ Law Problems • Write the given information • Convert to Kelvin temp. scale • Isolate the missing variable • Plug-in the given values and solve

  4. 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 V1 = 20.0 m3 T1 = 22ºC T2 = 283 º C V2 = ?

  5. Step2) Convert Temperatures into Kelvin oC + 273 T1= 22oC  22 + 273 = 296 T2= 283oC  283 + 273 = 556K

  6. Step 3) Isolate the unknown variable V1T2 = V2T1 We don’t know V2 Divide both sides by T1 V1T2 = V2T1 T1 T1

  7. 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

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