Why is it incorrect to say that a sample of helium at 50°C is twice as hot as a sample of helium at 25°C?

1 Answer
Jun 24, 2017

Because degrees celsius does not handle ratios in the same way that kelvin does, and gas laws must use kelvin units to line up with the kelvin units in the universal gas constant...

e.g. the ideal gas law is:

PV = nRTPV=nRT, where:

  • PP is pressure; either "atm"atm or "bar"bar is the most common/useful.
  • VV is volume of the gas in "L"L.
  • nn is the mols of the gas.
  • TT is the absolute temperature in "K"K!
  • Two common gas law RR values are:

R \ \ \ = "0.082057 L"cdot"atm/mol"cdot"K"

= "0.083145 L"cdot"bar/mol"cdot"K"


To prove it, let's say celsius temperatures double.

(50^@ "C")/(25^@ "C") stackrel(?" ")(=) (50 + "273.15 K")/(25 + "273.15 K")

Clearly, these are not equal:

2 ne 1.0839

If kelvin temperatures double...

("50 K")/("25 K") stackrel(?" ")(=) (50 - 273.15^@ "C")/(25 - 273.15^@ "C")

2 ne 0.8993

Or, let's even try different kelvin temperatures...

("300 K")/("150 K") stackrel(?" ")(=) (300 - 273.15^@ "C")/(150 - 273.15^@ "C")

2 ne -0.2180

It's nonphysical and inconsistent! Makes no sense! The ratio of temperatures should always be done in kelvins.