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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.
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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.
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
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
Charles’ Law: V = kT or V/T = k. Direct relationship: linear & passes through origin Boyle’s Law: PV = k Inverse relationship. hyperbola Compare Charles’ & Boyle’s Laws
Charles’ Law: Problems V1 = V2 T1 T2 Given any 3 variables, you can find the 4th.
Problem 1 • 150 mL of a gas at constant pressure. • Temperature increases from 20C to 40C. • 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