Charles’ Law.

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Presentation transcript:

Charles’ Law

Temperature Changes & Matter Solids & Liquids expand and contract as temperature changes. Change is usually v. small. Gases show large volume changes with temperature changes. What usually happens to V as T ?

Jacques Charles Balloonist. 1787 did expts on how volume of gases depends on temperature.

How do hot air balloons work?

Relationship between V and T Pressure & amount are constant. At high temperature, the gas particles move faster and collide with the walls more often. Pressure is constant, so volume has to increase.

Charles’ Law Tiger Graphic

Data for Volume-Temperature Trial Temperature (C) Volume (mL) 1 10 100 2 50 114 3 132 4 200 167 5 300 202

What did Charles do next?

Linear Relationship Plot Volume vs. C and you get a straight line. The relationship between Volume and C is linear. The equation of a line is: Y = mX + b.

Charles extrapolated the graph to 0 volume. At 0 mL, the X-intercept is -273 C.

Hints of Kelvin scale Charles extrapolated his data to see the temperature at which the volume was 0. 1st indication that the temperature -273 C might have a fundamental meaning. Why did Charles have to extrapolate his lines in this temperature range instead of taking data?

Plot Volume vs. Kelvin Temp. Get a straight line that passes through the origin. The relationship between the variables is direct. Y = mX or Y/X = m.

Charles’ Law: Verbal The volume of a gas at constant pressure varies directly with its absolute temperature.

Charles’ Law: Graphically Plot Volume vs. Kelvin Temperature Straight line that passes through the origin. V = kT or V = k T

Compare Charles’ & Boyle’s Laws Charles’ Law: V = kT or V/T = k. Direct relationship: linear & passes through origin Boyle’s Law: PV = k Inverse relationship. hyperbola

Charles’ Law: Problems V1 = V2 T1 T2 Given any 3 variables, you can find the 4th.

The low temperature region is always extrapolated. Why?

Balloons can expand & contract with the gas.

Problem 1 150 mL of a gas at constant pressure. Temperature increases from 20C to 40C. What is the new volume? Step 1: Convert T1 and T2 to Kelvin scale. Step 2: Rearrange equation: V1 = V2 becomes V1T2 = V2 T1 T2 T1 Step 3: Substitute and solve: 150 mL X 313 K 293 K = 160 mL