Temperature function of a gas derivation help

Ljn

May 2019
1
0
I cant quite figure out how the so-called temperature function of a gas equation is derived from equations 1-4 in the attachment(from Planck's treatise on thermodynamics). Anything that helps me understand how this is arrived at would
d be much appreciated!
 

Attachments

Oct 2017
567
287
Glasgow
The derivation looks a little odd because they use the temperature scale in Celsius rather than Kelvin. That means there's various constants in the derivation which need to be considered and removed after the fact using substitutions.

So... from experiments on certain gases, which can be called ideal gases, there are certain laws which seem to be the case. There are three findings which are used in this derivation.

1:
If the temperature of the gas is held constant, this relationship is true:

\(\displaystyle pv = \theta\)

That is, pressure multiplied by volume gives a constant. If pressure goes up, volume goes down. If pressure goes down, volume goes up.

2:
Now, let's say that instead of temperature being held constant, let's instead keep pressure constant. The following relationship is true:

\(\displaystyle t = (v - v_0)P\)

where \(\displaystyle P\) is some other constant (like \(\displaystyle \theta\) in the previous formula, but not the same number).

In this old text, they're using the Celsius scale for temperature instead of Kelvin. As a consequence, it's important to consider the "normal volume", \(\displaystyle v_0\). If it were Kelvin, there would be no need for this quantity and it would just be set to zero. In this text, however, it is necessary and is equal to the volume of the gas when the temperature is 0 degrees Celsius.

They substitute this into the first equation to give:

\(\displaystyle p\left(\frac{t}{P} + v_0\right) = \theta\)
\(\displaystyle \frac{pt}{P} + p v_0 = \theta\)
\(\displaystyle p v_0 = \theta - \frac{pt}{P}\)

Now comes a trick. Remember the first equation? For a constant temperature, the pressure multiplied by the volume is a constant. What if the temperature is held constant at t=0? For this case only, you now have a particular constant, \(\displaystyle \theta = \theta_0\) and the equation reduces to

\(\displaystyle p v_0 = \theta_0\)

The other term is 0 because t=0. This gives a third formula to play around with.

3:
Now they say that when a gas is heated from 0 to 1 degree Celsius, the volume increases by the same fraction for all gases. That is:

\(\displaystyle v_1 - v_0 = \alpha v_0\)

where \(\displaystyle \alpha\) is the constant dictating the proportion of the increase in volume.

However, we also have the second equation (rearranged slightly):

\(\displaystyle v - v_0 = \frac{t}{P}\)

Therefore, when t=1:

\(\displaystyle v_1 - v_0 = \frac{1}{P}\)

and substituting the third experimental finding, we have

\(\displaystyle \alpha v_0 = \frac{1}{P}\)

or

\(\displaystyle P \alpha v_0 = 1\)

Now the idea is to use the four equations derived here and some simultaneous equations to eliminate the constants \(\displaystyle P\), \(\displaystyle v_0\) and \(\displaystyle v\).

Have a go :)