CATAGORY:  Thermodynamics
QUESTION:  Devise an experiment to prove Charles' Law and hence determine
a value for absolute zero.  I have considered using a fairly primitive
method of trapping air in a thin tube , putting it in a water bath and
then heating it, measuring the volume of the gas at various temperatures.
This is not very good - any suggestions

Back to catalog.

---

I remember doing this experiment when I was an undergraduate student.
The use of a tube is good BUT you want to multiply the effect.  What
you are trying to do is measure a change in volume for with a change
in length.  Your tube has a cross sectional area A so its volume is A
times l where lis the length the change in volume dV is then equal to
A times the change in length.

(1) dV = A dl


                   V = Al        dV = A dl
           _____________________________
          ()_____________________)______) A
           |---------------------|------|
                      l             dl
Since T ~ V then

     New volume    new temp.
     ---------- = ----------
     Old volume    old temp
     
Which translates to                   

(2)         V(T+dT)    T + dT
            -------- = ------
              V(T)       T
              
Substitution of V =Al and dV = Al gives

(3)     Al + Adl    T + dT
        -------- = -------
           Al         T
           
Cancelling the Area a gives
           
              

(4)      l + dl    T + dT
        -------- = -------
           l         T
or
            dl          dT
(5)     1 + ---  = 1 + ---
             l          T
             
Subtracting 1 from both sides gives

            dl      dT
(6)         ---  = ---
             l      T 
             
So 
                   dT
(6)         dl = l ---
                    T
                    
So if you have a small length bubble then dl will be very small since in
"classroom" conditions it is difficult to get a big change in dT /T
when t is in degrees Kelvin. 

Now how about this instead.  construct your tube apparatus as before
but attach a air filled container on one end.  The volume of the
container is fixed so any gas that comes out as a result of expansion
will fill the tube.

      V(box)     
     ______
    |\     \       V(tube) = Al        dV = A dl
    | \     \ ___________________________
    |  \     \____________________)______) A
     \  \     \-------------------|------|
      \  \_____\       l              dl
       \ |     |
        \|     |
         |_____|
                                 
Thus if Charles law works then again

     New volume    new temp.
     ---------- = ----------
     Old volume    old temp.
     
but this time we have

        V(box) + V(tube) + Adl    T + dT
(7)    ---------------------- =  -------
        V(box) + V(tube)            T 
        
or
                 Adl               dT
(8)    1 + ---------------- = 1 + ---
           V(box) + V(tube)        T
           
Subtracting 1 from both sides.

          Adl             dT 
(9)  ---------------  =  ---                          
     V(box) + V(tube)     T
     
Solving for dl

             V(box) + V(tube)  dT
(10)   dl =  ---------------- ---      
                    A          T
                    
                    
Now lets compare the difference between dl from (6) with no box to dl
from (10) with a box.  Take the tube diameter to be 2mm so it has a
1mm (inner) radius then A = pi r^2 = 3.14 mm^2.  Also suppose that the
length of your bubble l is 10mm. If the temperature change is say 50K
and room temperature is say 293K then dl from (6) is

                 50K
(11)  dl = 10mm ----- = 1.7mm 
                 293K

Which may be difficult to measure.

On the other hand suppose that we now use a box that is 2cm x 2cm x
2cm on a side to give V(box) = 8 x 10^3 mm^3.  Then (10 ) gives us

           8 x 10^3 mm^3 + 3.14mm^2 x 10mm   50K
(12) dl = --------------------------------  ----
                   3.14 mm^2                293K


 =   436mm ~ .44 m  (this is 256 times larger than with (6)!)

Thus a relatively small temp difference is greatly magnified so that
the data you take will be more accurate and you can do a better
extrapolation of the V vs T graph as T -> 0.