The figure below depicts the gas thermometer used in the experiment.

• When the air in the bottle is heated, the pressure in the bottle increases.
• The pressure in the tube on the bottle side of the alcohol increases along with the pressure in the bottle.
• The alcohol column therefore pushes on the air column in the sealed end of the tube.
• This push compresses the air in the column, so that the air in the column is very nearly at the same pressure as that in the bottle.
• There is a difference in the pressure in the air column and in the bottle, due to the different levels of the alcohol on the two sides of the U.
• This pressure difference is approximately 1 atmosphere per 15 meters of altitude difference.

In class one group observed air column length of 39.5 cm, 37.8 cm and 30.1 cm at temperatures of 0 C, 24 C and 60 C.

• Accepting the fact that the ratio of pressures is equal to the inverse ratio of column lengths, we conclude that the air column pressure in state B (60 C) is 1.32 that of the pressure in state A (0 C).
• The difference in the levels of the alcohol column changes between state A and state B by double the change in column length (one side goes down while the other goes up), or by approximately 19 cm. This results in a pressure difference of approximately 19 / 1500 atm, or .01 atm between the two sides of the alcohol column.
• This pressure difference might or might not be significant with the apparatus we are using.
• A plot of bottle pressure vs. temperature should be linear; since pressure is inversely proportional to column length a plot of 1 / L vs. temperature should be linear. You should check this out.

Ignoring the pressure differences due to the different heights of the alcohol column, we plot two points on the pressure vs. temperature graph, corresponding to the extreme temperatures observed. We then extrapolate from our two points, assuming a linear pressure vs. temperature relationship, to determine the temperature at which we expect pressure to be zero.

• The temperature at which pressure reaches 0, for an ideal gas, is called absolute zero.
• The present experiment uses air rather than an ideal gas, though the inaccuracies inherent in the apparatus used here are much greater than those incurred by using air rather than an ideal gas.

• It doesn't matter what pressure units we use. The zero of pressure will occur at the same temperature whether we measure pressure in atm, Newtons/m^2, psi, or any other consistent unit.
• We therefore use as our unit of pressure whatever the pressure was in the air column at zero Celsius.
• Since the pressure ratio is 1.32, the pressure at 60 C will be 1.32 * 1 = 1.32.
• Thus our graph points are (0 C , 1) and (60 C, 1.32).

• We extrapolate easily enough, using similar triangles.
• Our result is an estimate of - 187 C for absolute zero. The accepted value is - 273 Celsius.

The linear pressure vs. temperature relationships for ideal gas give us an accurate way of defining a linear temperature scale.

Gas thermometers are slow and relatively expensive, so are not practical for a lot of important applications.

The uncalibrated thermometer you used in the initial experiment used an alcohol column to measure temperature.

• Alcohol does not expand in a perfectly linear fashion.
• The figure below depicts a hypothetical column length vs. temperature curve for alcohol (this is not an actual depiction of the correct curve for alcohol).
• We see that while the column length always increases with temperature, the increase is not linear.

When we measured the temperature changes when a warm object of known mass and temperature was placed into a known mass of water at a known temperature, we assumed that the thermal energy gained by the water was equal to that lost by the rock.

Here we symbolize the process.

• We assume that the mass and temperature of the rock are m2 and T2, while those of the water are m1 and T1.
• We assume that the final temperature of the system consisting of the water and the rock is Tf.
• The change in temperature of the water is `dT1 = Tf - T1, while that of the rock is `dT2 = Tf - T2.

We will symbolize our assumption of energy conservation in terms of these variables.

Assuming a closed system, with no energy gain or loss from the system, we see that the changes `dQ1 and `dQ2 in the thermal energies of the water and the rock must add up to zero.

If we let c1 stand for the 4.19 Joules / (gram Celsius), indicating the thermal energy required to raise one gram of water by one Celsius degree, and let c2 stand for the analogous quantity for the rock (i.e., c1 and c2 are the specific heats of the water and the rock), we see that

• `dQ` = c1 m1 `dT1 and
• `dQ2 = c2 m2 `dT2.

Using these quantities we translate `dQ1 + `dQ2 into the equation in the second line from the bottom of the figure below.

In our experiment, we were given the value of c1 and we measured m1, m2, TF and T1.

Looking at the equation for energy conservation, we see that the only quantity we do not know is c2.

• We can easily solve the equation for c2.
• This equation therefore summarizes the means we used to reason out the specific heat c2 of the rock
• .