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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.
5050100100150150200200250250300300350350
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
5050100100150150200200250250300300350350
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.
5050100100150150200200250250300300350350
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
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