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Mr. ShieldsMr. Shields Regents Chemistry Regents Chemistry U05 L06 U05 L06

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Charles LawCharles Law

Jacques Charles (1746 – 1823)

- Charle’s Law (1787)

- Volume vs. Temp relationship of a gas

V/T = kV/T = k(constant P and n)(constant P and n)

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Charles LawCharles Law

In 1787 Charle’s law was published

This law looked at the variables of Volume (V) andTemperature (T)

V/T = k V, T = variables P, n = Constants

Charles’s law states that the volume of a gas atCharles’s law states that the volume of a gas atconstant p,n varies constant p,n varies DIRECTLYDIRECTLY with it’s with it’sabsolute Temperature absolute Temperature (in (in KelvinKelvin!)!)

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Developing Charles LawDeveloping Charles LawSince velocity will increases with increasing temp thenThe number of collisions with the containers wall perunit time must be increasing

And increasing the rate of collisions with the containersWalls increases pressure

BUT

Charles law assumes constant P with increasing temp.

So how does P stay constant if T increases ?

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Developing Charles LawDeveloping Charles LawRemember Boyles law?

Boyles law says we can reduce the pressure if weIncrease the volume

PV=k As V P

In other words, if there is a driving force that increasespressure we can decrease the pressure by increasing thecontainer volume.

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Developing Charles LawDeveloping Charles LawSo if T increases we can hold pressure constant if weIncrease volume at the same time T increases

This is just what Charles Law says

V/T = k A direct relationship (p,n const)

Or in a more useful form for calculations:

V1/T1 = V2/T2 (why is this true?)

So what does the relationship V/T look like?

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Trial Trial T (°C)T (°C) V (mL) V (mL)

1 1 --

124124.00.00

50.0050.00

2 2 --

100100.00.00

58.0558.05

3 3 -75.00-75.00 66.4466.44

4 4 0.000.00 91.6191.61

5 5 25.0025.00 100.0100.000

6 6 30.0030.00 101.6101.688

7 7 45.0045.00 106.7106.711

8 8 100.0100.000

125.1125.177

9 9 225.0225.000

167.1167.111

10 10 323.0323.000

200.0200.000

Below is a data table relating temperature to volume

Note that all V/T equals the same k(try trial 2,4,8)A graph of this data looks like this:

Charles Law

Note: Plot relates deg C to V butNote: Plot relates deg C to V butCalculations were fist done in deg. KCalculations were fist done in deg. K

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Why Kelvin for Calculations?Why Kelvin for Calculations?T and V are related such that for each doubling T and V are related such that for each doubling (or Halving) of the temperature (in Kelvin) Volume(or Halving) of the temperature (in Kelvin) VolumeDoubles (or halves) i.e a direct relationshipDoubles (or halves) i.e a direct relationship

NOTE WELL: This relationship is only true for Deg.NOTE WELL: This relationship is only true for Deg.KelvinKelvin NOTNOT degrees degrees CelsiusCelsius. . But WhyBut Why??

What would happen if the temp were 0 deg. CelsiusWhat would happen if the temp were 0 deg. Celsius

Division by zero is undefinedDivision by zero is undefined

1x1x 2x2x

100k 200k100k 200k

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Why Kelvin?Why Kelvin?

Let’s try it:Let’s try it:

If V1=1 L and T1 = 25 deg C and T2 equals –50 deg C If V1=1 L and T1 = 25 deg C and T2 equals –50 deg C what does V2 equal?what does V2 equal?

VV11/T/T11=V=V22/T/T22; 1/25 = V; 1/25 = V22/-50 Then V/-50 Then V22 = -2 L (????) = -2 L (????)

And What would happen if Temperature was NEAGATIVE?And What would happen if Temperature was NEAGATIVE?

Predicted Volume would be negativePredicted Volume would be negative

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Decreasing TemperaturesDecreasing TemperaturesWhat happens when a gas drops to VERY LOW TEMPWhat happens when a gas drops to VERY LOW TEMP(assuming the gas does not condense to a liquid)?(assuming the gas does not condense to a liquid)?

Charles law would predict the following:Charles law would predict the following: (1)(1)The direct relationship of V/T remains a straight lineThe direct relationship of V/T remains a straight line

(2) Extending the line to zero volume shows there(2) Extending the line to zero volume shows there is a is a lower limit to temperaturelower limit to temperature (i.e. Avg KE) (i.e. Avg KE)

The lower limit for temperature is The lower limit for temperature is 0 deg. Kelvin0 deg. Kelvin. .

This is called This is called ABSOLUTE ZEROABSOLUTE ZERO

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There is a direct relationship between T(kelvin), V and There is a direct relationship between T(kelvin), V and EE

EE

There is a There is a DirectDirectRelationshipRelationshipBetweenBetweenTTkelvinkelvin

And EnergyAnd Energy

T = 343 KT = 343 K

T = 171.5 KT = 171.5 K

0 K0 K

Earths coldestEarths coldestRecordedRecordedTemp:Temp:

-132 deg C-132 deg C

High AntarticHigh AntarticPolar PlateauPolar Plateau

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V vs. T at Different PV vs. T at Different P

No matter what the initial P is all lines intersect at Absolute zeroNo matter what the initial P is all lines intersect at Absolute zero

2 atm2 atm

1 atm1 atm

0.5 atm0.5 atm

Absolute zero: -273 deg C or 0 K

Assumes gases do not Condense to a liquid

V/T = k (p,n held const.)V/T = k (p,n held const.)

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Let’s try a few problems.Let’s try a few problems.

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If the temperature of the gas in fig.1 is 35 deg. C andIf the temperature of the gas in fig.1 is 35 deg. C andthe volume is 1/2L what is the temperature in fig.2 and the volume is 1/2L what is the temperature in fig.2 and What is the temp in fig 3? (P,n are constant)What is the temp in fig 3? (P,n are constant)

Fig 1 Fig 2 Fig. 3Fig 1 Fig 2 Fig. 3VV11/T/T11 = V = V22/T/T22

.5L/308 = 1L/T.5L/308 = 1L/T22

TT22 = 1L/(.5L/308) = 1L/(.5L/308)TT22 = 616 deg K = 616 deg K

VV11/T/T11 = V = V33/T/T33

.5L/308=2L/T.5L/308=2L/T33

TT33 = 2L/(.5L/308) = 2L/(.5L/308)TT33 = 1232 deg. K = 1232 deg. K

Problem 1 & 2Problem 1 & 2

½ L 1 L 2L½ L 1 L 2L

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Problem 3Problem 3

A sample of gas at 30 deg C and standard pressure A sample of gas at 30 deg C and standard pressure (1 atm) occupies a volume of 200 ml. Calculate the(1 atm) occupies a volume of 200 ml. Calculate thetemperature in deg. Celsius at which the gas will temperature in deg. Celsius at which the gas will occupy a volume of 5 ml at std pressure if P is heldoccupy a volume of 5 ml at std pressure if P is heldconstant. constant.

VV11/T/T11 = V = V22/T/T22

TT22 = V = V22/(V/(V11/T/T11))TT22 = 5/(200/303) = 5/(200/303)TT22 = 7.58 deg K = 7.58 deg K What’s this temperature in deg. C?What’s this temperature in deg. C?

TT22 = -265.42 deg. C (K = C + 273) = -265.42 deg. C (K = C + 273)