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
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 = nRT# , where:
#P# is pressure; either#"atm"# or#"bar"# is the most common/useful.#V# is volume of the gas in#"L"# .#n# is the mols of the gas.#T# is the absolute temperature in#"K"# !- Two common gas law
#R# 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.
