
4.02 Absolute Zero 
Purpose: To explore the concept of absolute zero 
Introduction: In principle, there is no upper limit to temperature. As thermal motion increases, a solid object first melts and then vaporizes. As the temperature is further increased, molecules break up into atoms, and atoms lose some or all of their electrons, forming a plasma. Plasmas exist in stars where the temperature is many millions of degrees Celsius. However, there is a definite lower limit to temperature. Experiments in the nineteenth century showed that there is a limit to coldness. At this lowest temperature, no more energy can be extracted from a substance. What is the lowest possible temperature? The name absolute zero has been given to the lowest possible theoretical temperature. In this activity, you will use Graphical Analysis to experience the prediction method that has led to absolute zero. 
Materials: Graphical Analysis 
Procedure:

Tutorial: 






Analyzing
the graph: A fixed mass of gas in an enclosed rigid container with an initial pressure of 5.0 Pascals and absolute temperature of 200. K is heated such that the gas reaches an absolute temperature of 400. K. What will the pressure be when the gas reaches the absolute temperature of 400. K (Notice the absolute temperature doubled)? 
The equation for a straight line: y =
m x 
Solving equation for constant of proportionality k: k = P/T 
Setting the constant equal to each other gives you the equation for GayLussac's Law: k = k 
P_{1} = k T_{1}  k = P_{1}/T_{1}  Therefore: 
P_{2} = k T_{2} 
k = P_{2}/T_{2} 
P_{1}/T_{1}= P_{2}/T_{2} 
Using the graph or the
equation, answer the following question: A fixed mass of gas at initial pressure of 5.0 Pascals and absolute temperature of 200. K will be at what absolute temperature when the pressure is 8.0 Pascals? P_{1} = 5.0 Pascals T_{1} = 200. K P_{2} = 8.0 Pascals T_{2} = ? P_{1}/T_{1} = P_{2}/T_{2} T_{2} = (P_{2}T_{1})/P_{1} T_{2} = [(8.0 Pascals)( 200. K)]/5.0 Pascals T_{2} = 320 K If you look at the graph you will see that the absolute temperature reading for 8.0 Pascals of pressure is approximately 320 K. 
Activity: 

Sample data is shown in Table 1 shown below. 
Table 1 


Trials  Temperature (degrees Celsius) 
Pressure (psi) 
1  100.  18.5 
2  23.0  14.7 
3  0  13.5 


Select Select the Axes Options tab.



Questions: 





4.02 Discussion Question: Go to the discussion area and post an answer to this question. Respond to at least one other person and explain why you agree or disagree with his/her comment.

4.02
ScorcherAbsolute Zero

Image: © 2003flvs 