Charles Law

(or is it Charles's Law?)

 

Charles’s law states that if a given quantity of gas is held at a constant pressure, its volume is directly proportional to the absolute temperature. Think of it this way. As the temperature of the gas increases, the gas molecules will begin to move around more quickly and hit the walls of their container with more force—thus the volume will increase. Keep in mind that you must use only the Kelvin temperature scale when working with temperature in all gas law formulas! Here’s the expression of Charles’s law that you should memorize:
 
  V1    = k =   V2   The subscripts 1 and 2 refer to two different sets of conditions, just as with Boyle's law.
T1     T2

Try using Charles’s law to solve the following problem.

Example
A sample of gas at 15ºC and 1 atm has a volume of 2.50 L. What volume will this gas occupy at 30ºC and 1 atm?
Explanation
The pressure remains the same, while the volume and temperature change this is the hallmark of a Charles’s law question.
So, ,  

Highlight to reveal Answer-->         2.50 L

=    V2     V2 = 2.63 L

288K

303K
This makes sense—the temperature is increasing slightly, so the volume should increase slightly. Be careful of questions like this—it’s tempting to just use the Celsius temperature, but you must first convert to Kelvin temperature (by adding 273) to get the correct relationships!

Charles Law works best at high temperatures.

 

Why must the temperature be absolute? 

If temperature is measured on a Celsius (non absolute) scale, T can be negative. If we plug negative values of T into the equation, we get back negative volumes, which cannot exist. In order to ensure that only values of V= 0 occur, we have to use an absolute temperature scale where T= 0. The standard absolute scale is the Kelvin (K) scale. The temperature in Kelvin can be calculated via Tk = TC + 273.15. A plot of the temperature in Kelvin. Charles' law predicts that volume will be zero at 0 K. 0 K is the absolutely lowest temperature possible, and is called absolute zero.

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Highlight to reveal Answers
1. Calculate the decrease in temperature when 2.00L at 20.0oC is compressed to 1.00L.
 
V1=2.00L   V1   =   V2  
T1=20.0oC + 273=298K T1     T2
V2=1.00L 2.00L  

=

 1.00L 
T2=? 298K     T2  

( 2.00L)(T2) =(1.00L)(298K)

T2= 149K +273=  422oC

2. 600.0mL of air is at 20.0oC. What is the volume at 60.0oC?

V1=600.0mL   V1   =   V2  
T1=20.0oC + 273=298K T1     T2
V2=? 600.0mL  

=

   V2  
T2=60.0oC +273= 333K 298K     333K 

( 600.0mL)(333K ) =(V2)(298K)

V2= 670.mL

AP Try These-Highlight to reveal Answers
1. A ballon filled to a volume 7.00 x 102mL at a temperature of of 20.0oC. The ballon then is cooledat constant pressure to a temperature of 1.00 x 102K. What is the final volume of this balloon?
V1=7.00 x 102mL   V1 

=

   V2  
T1=20.0oC+ 273=293K T1 T2
V2=?

7.00 x 102mL

=

   V2  
T2=1.00 x 102K

293K       

 1.00 x 102K

(1.00 x 102K)(7.00 x 102mL)=(V2)(293K)    

239mL = V2      
2. A sample of gas at 15oC and 1.0 atm has a volume of 2.58L. What  volume will this gas occupy at 38oC and 1 atm?
V1=2.58L   V1 

=

   V2  
T1=15oC+ 273=288K T1 T2
V2=?

2.58L

=

   V2  
T2=38oC + 273=311K

288K

 311K
P1=1 atm

(311K)(2.58L)=(V2)(288K)    
P2=1atm

2.79L = V2      
***Note pressure is constant