Charles’ LawCharles’ Law

Temperature Changes & MatterTemperature Changes & Matter

Solids & Liquids expand and contract as Solids & Liquids expand and contract as temperature changes. temperature changes. – Change is usually v. small.Change is usually v. small.

Gases show large volume changes with Gases show large volume changes with temperature changes.temperature changes.

What usually happens to V as T What usually happens to V as T ??

Jacques CharlesJacques Charles

Balloonist.

1787 did expts on how volume of gases depends on temperature.

How do hot air balloons work?How do hot air balloons work?

Relationship between V and TRelationship between V and TPressure & 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’ LawCharles’ Law

Tiger GraphicTiger Graphic

Data for Volume-TemperatureData for Volume-Temperature

TrialTrial Temperature (Temperature (C)C) Volume (mL)Volume (mL)

11 1010 100100

22 5050 114114

33 100100 132132

44 200200 167167

55 300300 202202

What did Charles do next?What did Charles do next?

0

50

100

150

200

250

0 50 100 150 200 250 300 350

10

50

100

200

300

Linear RelationshipLinear Relationship

Plot Volume vs. Plot Volume vs. CC and you get a straight and you get a straight line. line.

The relationship between Volume and The relationship between Volume and C C is is linearlinear..

The equation of a line is: Y = mX + b.The equation of a line is: Y = mX + b.

0

50

100

150

200

250

0 100 200 300 400 500 600 700

283

323

373

473

573

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

Hints of Kelvin scaleHints of Kelvin scale

Charles extrapolated his data to see the Charles extrapolated his data to see the temperature at which the volume was 0.temperature at which the volume was 0.

1st indication that the temperature -273 1st indication that the temperature -273 C C might have a fundamental meaning.might have a fundamental meaning.

Why did Charles have to extrapolate his lines Why did Charles have to extrapolate his lines in this temperature range instead of taking in this temperature range instead of taking data?data?

Plot Volume vs. Kelvin Temp.Plot Volume vs. Kelvin Temp.

Get a straight line Get a straight line that passes through that passes through the originthe origin..

The relationship between the variables is The relationship between the variables is directdirect..

Y = mX or Y/X = m.Y = mX or Y/X = m.

Charles’ Law: VerbalCharles’ Law: Verbal

The volume of a gas at constant pressure The volume of a gas at constant pressure varies directly with its absolute varies directly with its absolute temperature.temperature.

Charles’ Law: GraphicallyCharles’ Law: Graphically

Plot Volume vs. Kelvin TemperaturePlot Volume vs. Kelvin Temperature

Straight line that passes through the Straight line that passes through the origin. origin.

V = kT or V = kT or VV = k = kT

Compare Charles’ & Boyle’s LawsCompare Charles’ & Boyle’s Laws

Charles’ Law:Charles’ Law:

V = kT or V/T = k.V = kT or V/T = k.

Direct relationship: Direct relationship: linear & passes linear & passes through originthrough origin

Boyle’s Law: PV = kBoyle’s Law: PV = k

Inverse relationship. Inverse relationship. hyperbola hyperbola

Charles’ Law: ProblemsCharles’ Law: Problems

VV11 = = VV22

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.Balloons can expand & contract with the gas.

Problem 1Problem 1

150 mL of a gas at constant pressure.150 mL of a gas at constant pressure.Temperature increases from 20Temperature increases from 20C to 40C to 40C.C.What is the new volume?What is the new volume?

Step 1:Step 1: Convert T Convert T11 and T and T22 to Kelvin scale. to Kelvin scale.

Step 2:Step 2: Rearrange equation: Rearrange equation:

VV11 = = VV22 becomes becomes VV11TT22 = V = V22

TT11 T T22 T T11

Step 3:Step 3: Substitute and solve: Substitute and solve:150 mL X 313 K150 mL X 313 K 293 K293 K

= 160 mL