# How do we know -273.15 centigrade is 0 kelvin?

I am in no way doubting that humans know that -273.15 centigrade is 0 Kelvin. I'm just curious.

Suppose you know only two things:

1. A new scale for temperature known as the centigrade scale, calibrated against the freezing and boiling points of water using nifty glass thermometers (or any other standard thermometer for that matter).
2. Thermodynamics as it was in the mid 1800's - i.e. knowledge that temperature should be a strictly positive quantity.

How do you go from knowing those two things to saying "Aha! I have now inferred that absolute zero-temperature in the modern thermodynamic sense is, on the centigrade scale, at -273.15!"

• Hint: show the temperature of an object at 0C is 273.15 K. Then use 1K=1C by definition. – pathintegral Sep 16 '16 at 18:01
• What would happen if you plotted the $V$ vs $T$ relationship with various gasses where $T$ is in Celsius? – BowlOfRed Sep 16 '16 at 18:02
• Of course, in @BowlOfRed 's experiment, the pressure should be held constant. – Chet Miller Sep 16 '16 at 18:23
• The kelvin was originally defined in 1954 by designating the triple point of water as its second defining point and assigned its temperature to exactly 273.16 K. Later it was redefined such that "The kelvin, unit of thermodynamic temperature, is equal to the fraction 1⁄273.16 of the thermodynamic temperature of the triple point of water." There are moves afoot to redefine the Kelvin as the temperature scale for which Boltzmann's constant is $1.3806505 \times10^{−23}$ J/K exactly. – jim Sep 16 '16 at 18:34

I'm not sure how a precise value would have been obtained circa 1800's. Here's how I had my Chemistry students derive an approximate value with simple equipment.

The apparatus is a hollow steel ball about 10 cm in diameter connected to a pressure gauge. Also involved is a thermometer capable of -70 C to 100 C.

The bulb of this apparatus is then immersed in boiling water, ice water and a mixture of dry ice and acetone. The temperature and pressure are recorded for each treatment. Then the presure is plotted versus the temperature. Since P is proportional to T, a linear plot results. When the plot is extrapolated to P = 0, the temperature is reasonably close to -273 C.

Since temperature is a measure of the average kinetic energy of the particles, the thermodynamic temperature should be zero when the particles stop moving. When the particles are no longer moving the pressure should then become zero as well.

• This theory would hold together great if there were no phase changes for the gas between -70 C and absolute zero. – Chet Miller Sep 16 '16 at 19:24
• There is no "theory" here, merely application of the Ideal Gas Law and the definitions of temperature and pressure. The only way phase transitions would matter is by attempting to gather data points too low for the gas used. Air behaves relatively ideally at the temperatures used in the experiment. But, conceptually, the gas we're cooling to 0K is an ideal gas, not air. No intermolecular attractions to invoke phase changes. – bpedit Sep 16 '16 at 20:13
• You did your tests on real gases, not ideal gases. Even though the ideal gas law is a good approximation for the temperature range you considered, extrapolating this to -273 is not a very convincing way of showing that the kinetic energies of real substances go to zero at absolute zero (in my judgment). – Chet Miller Sep 16 '16 at 20:30
• Thousands of my students have been convinced. As have, I suspect, tens of thousands of others. The graph really is, if not precise, pretty convincing. And very doable in the 1800's which was my intrepretation of the OP's quiestion. I also intrepreted the question as asking how to arrive at the value in Celsius, not as proof that the kinetic energy of a real substance would be zero at 0K. I (we, including other Chemistry teachers) use that as an axiom inherent in the Ideal Gas Law and a consequence of the thermodynamic definition of temperature, at least as it was back then. – bpedit Sep 16 '16 at 23:00
• If the same extrapolation works for a number of real gases, that is persuasive. – sammy gerbil Sep 16 '16 at 23:35