Charles’s LawCharles’s LawVolume & Temperature RelationshipVolume & Temperature Relationship

In 1787, Jacques Charles discovered that volume and temperature of a gas are related in mathematical terms.

In 1787, Jacques Charles discovered that volume and temperature of a gas are related in mathematical terms.

Increasing the temperature of a gas by

273OC , at a constant pressure, increases the

volume by 2.

Here's what Charles did:Here's what Charles did:  He put a gas into a container in which he could change the temperature and measure the volume. 

  He put a gas into a container in which he could change the temperature and measure the volume. 

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NASA LINK

When he divided the volume of the gas times it's temperature, he found it was equal to some arbitrary number (let's call it k, because he did).

When he divided the volume of the gas times it's temperature, he found it was equal to some arbitrary number (let's call it k, because he did).

5050100100150150200200250250300300350350

Volume (mL)Volume (mL)

Temperature (K)Temperature (K)100100mL / mL / 1010K = K = kk

10 K10 K

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Volume (mL)Volume (mL)

Temperature (K)Temperature (K)

10 K10 K

20 K20 K

If he changed the temperature of the gas, he found that the volume also changed.If he changed the temperature of the gas, he found that the volume also changed.

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Volume (mL)Volume (mL)

Temperature (K)Temperature (K)

10 K10 K

20 K20 K

100100mL / mL / 1010K = K = kk200200mL / mL / 2020K = K = kk

What is surprising is that if you divide the new volume by the new temperature, the answer is the same arbitrary number that you had in the first place (k!). 

What is surprising is that if you divide the new volume by the new temperature, the answer is the same arbitrary number that you had in the first place (k!). 

Here’s what You’ll do:Here’s what You’ll do:

-15-15ococ 55ococ 6565ococ

Data TableData TableWater Bath

Temp oCAverage Volume

V T oC

Absolute Temp K

V T K

Saltwater- Ice

Ice water

Room Temp.

Hot Water

Lab ProcedureLab Procedure1. Read Lab Instructions

2. Set Syringe at specified volume

3. Test & Record & Calculate Data

4.Return to seat to answer discussion questions

1. Read Lab Instructions

2. Set Syringe at specified volume

3. Test & Record & Calculate Data

4.Return to seat to answer discussion questions

Charles’s LawCharles’s Law

Charles's Law states that as Temperature increases in a gas, Volume increases!

T↑ V↑

This is the basis for his law.

V1 = V2T1 T2

Charles's Law states that as Temperature increases in a gas, Volume increases!

T↑ V↑

This is the basis for his law.

V1 = V2T1 T2

TEMPERATURE MUST BE IN KELVINSTEMPERATURE MUST BE IN KELVINS

V1 = V2T1 T2 V1 = V2T1 T2

In this equation:

T1 is the initial temperature of the gas

V1 is the initial volume of the gas

T2 is the final temperature of the gas

V2 is the final volume of the gas. 

This way, if you know the initial temperature and volume of a gas and know what the final temperature will be, you can predict what the volume will be after you change the temp.  Let's see an example.

In this equation:

T1 is the initial temperature of the gas

V1 is the initial volume of the gas

T2 is the final temperature of the gas

V2 is the final volume of the gas. 

This way, if you know the initial temperature and volume of a gas and know what the final temperature will be, you can predict what the volume will be after you change the temp.  Let's see an example.

Example ProblemExample Problem

Q: Find the final volume, in Liters, using Charles's Law and the following info: T1= 273 K

V1= 2.0L T2= 300 K

V2=?1) Write down formula: V1 = V2 T1 T22) Plug in numbers: (2.0) = (V2) (273) (300) 3) Solve for V2 using Algebra (multiply by 300 on both sides): (300)(2.0) = (300)(V2) (273) (300)4) Multiply and divide (the Kelvins cancel, so you're left with L): (300)(2.0)/(273)= 2.2L5) You're done! V2=1L

Q: Find the final volume, in Liters, using Charles's Law and the following info: T1= 273 K

V1= 2.0L T2= 300 K

V2=?1) Write down formula: V1 = V2 T1 T22) Plug in numbers: (2.0) = (V2) (273) (300) 3) Solve for V2 using Algebra (multiply by 300 on both sides): (300)(2.0) = (300)(V2) (273) (300)4) Multiply and divide (the Kelvins cancel, so you're left with L): (300)(2.0)/(273)= 2.2L5) You're done! V2=1L