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$$PV=nRT$$

Where $P$ is pressure in pascals ($\text{Pa}$)
$V$ is volume in cubic metres ($\text{m}^3$)
$n$ is amount of substance in moles ($\text{mol}$)
$R$ is the gas constant having units, joules per mole kelvin ($\text {J K}^{-1}\text{ mol}^{-1}$)
$T$ is the temperature in kelvins ($\text K$)

The melting and boiling points of pure water are $0\ ^\circ\text C $ and $100\ ^\circ\text C $ respectively. $1\ ^\circ\text C$ is $\dfrac 1 {100}^\text{th}$ of the difference in temperature.

1 kelvin and 1 degree Celsius represent the same difference in temperature.

However, why is it that in any formula (such as the one above) involving temperature (not temperature difference), plugging in values will result in a nice and correct temperature?

For example, $P=100\ \mathrm{}$, $V=8.314\ \mathrm{}$, $n=1\ \mathrm{}$, $R=8.314\ \mathrm{\ \ }$
Plugging in the values will result in $T=100\ $
$100$ kelvin.

But what is the significance of 100 kelvin? Why could it not have been any other unit in an absolute temperature scale? e.g. 100 degrees Rankine

In simple formulas such as

$$\text{Displacement = Velocity} \times \text{Time}$$ I can understand that metres would be obtained when metres per second is multiplied with seconds. It does not make sense to obtain inches from the multiplication between metres per second and seconds.

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    $\begingroup$ It may be useful to note that $R T$ has the units of energy, so the choice for units of T is really about what value you take $R$ to be. $\endgroup$
    – jim
    Commented Jun 9, 2019 at 15:17
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    $\begingroup$ I suggest you keep the units with the numbers. $R$ is not $8.314$. $R = 8.314\ \mathrm{J}\cdot\mathrm{K}^{-1}\cdot\mathrm{mol}^{-1} = 1.104\ \mathrm{cal}_{th}\cdot\mathrm{⁰R}^{-1}\cdot\mathrm{mol}^{-1}$ (for example). That's true equals, because it is still the same constant with the same value, just expressed in different units. $\endgroup$
    – Jan Hudec
    Commented Jun 10, 2019 at 13:43
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    $\begingroup$ This post is relevant: Is the Boltzmann constant really that important?. $\endgroup$
    – DanielSank
    Commented Jun 10, 2019 at 19:47

4 Answers 4

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You do not need to use the kelvins for the ideal gas formula. The important thing to know is that $PV \propto nT$ where $T$ is the absolute temperature and can be expressed in any arbitrary linear scale (as long as the temperature is still absolute), including the kelvin and the rankine scale.

However, if you choose the constant of proportionality to be $R$ given in J/(K mol), then the temperature's units must be in kelvins for the equation to make sense. But you're not forced to use such a constant. You could pick up any other constant and you could adjust the temperature scale you're using. In particular you could have picked the constant that would make the temperature in rankine units.

Edit: Since the title of the question refers to "formulas", let me mention a case I've seen up to several papers. With regard to thermoelectric materials, there is a $ZT$ factor that's used to gauge the quality of the material to be used as a thermoelectric one, for example to make a thermoelectric generator. The important thing is that $T$ must be the absolute temperature. Unfortunately many papers mention that $T$ is given in kelvins, while it is not necessarily the case. Fortunately enough, I would say most (good) papers refer to $T$ as being the absolute temperature and do not mention the units.

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    $\begingroup$ Follow-up question: Would the most natural absolute temperature unit be one where the constant of proportionality R=1? $\endgroup$
    – helpme
    Commented Jun 9, 2019 at 4:20
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    $\begingroup$ @helpme The answer is subjective and would only be "well suited" for the ideal gas relation. Not with, for example, the ZT relation I mention in my answer. $\endgroup$
    – untreated_paramediensis_karnik
    Commented Jun 9, 2019 at 7:44
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    $\begingroup$ @helpme: "It's subjective" is probably not a satisfactory answer to your question. Thankfully there is a satisfactory answer and that answer is simply, no. :-) You'd instead want to use PV = NkT rather than PV = nRT, where k is the per-molecule normalized version of R, putting N in the natural units of "number of molecules". So the more natural choice would seem to be k = 1. And indeed this is what Planck units do... and if you continue this you get the Planck temperature as the natural unit. $\endgroup$
    – user541686
    Commented Jun 9, 2019 at 11:02
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    $\begingroup$ The equation $PV\propto T$ (in the ideal gas limit) does not work for any scale. It only works for linear scales, that is scales where the reference points are linearly interpolated. $\endgroup$
    – Diracology
    Commented Jun 10, 2019 at 17:59
  • $\begingroup$ Good point @Diracology let me modify and improve the answer. $\endgroup$
    – untreated_paramediensis_karnik
    Commented Jun 10, 2019 at 19:32
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For example, 𝑃=100, 𝑉=8.314, 𝑛=1, 𝑅=8.314. Plugging in the values will result in 𝑇=100 kelvin.

But what is the significance of 100 kelvin? Why could it not have been any other unit in an absolute temperature scale? e.g. 100 degrees Rankine

The answer is already contained in the first part of your question, when you write the units of the Gas Constant $R$. In order to get temperature in different units, one should use a different value and units for $R$. Similarly for a different thermometric scale.

Edit (improved the examples):

For example, in the case of Rankine degrees and using the same units for all other quantities, the perfect gas formula keeps the same form, but with a gas constant $R^{\prime}=4.61915$. In the case of temperature, $\theta$, in Celsius degrees, becomes: $$ PV = n\tilde R\theta+ Q n, $$ where $Q = 2.271 \cdot 10^3 = 8.314*273.15$ J mol$^{-1}$, and $\tilde R$ has the same numerical value of $R=8.314$, although now its units should be read as J mol$^{-1}$ $^{\circ}$C$^{-1}$.

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  • $\begingroup$ I don't think you should have $R\prime$ and $\tilde R$ there. It is still the same constant, with the same value, even if expressed in different units (and then, you really should have units in the last equation, otherwise the dimensions don't match). $\endgroup$
    – Jan Hudec
    Commented Jun 10, 2019 at 13:47
  • $\begingroup$ @JanHudec I used different symbols for clarity. Even if the numerical values are the same, different units are used. I'll modify the way I am writing the last equation to avoid misunderstandings, $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Jun 10, 2019 at 17:46
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There is a point that the other answers haven't made, which I believe is important: Celsius and Farenheit both allow temperatures below (what they call) zero. If you plug those measurements into the formula, you get a value of (pressure times value) which is negative; this is not physically possible. Note also that negativeness cannot be compensated-for by a simple constant of proportionality (R). Kelvin at least has the virtue of never being negative-valued.

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Since a few weeks, the kelvin is a unit that is defined by setting Boltzmann's constant to exactly $1.380 649 \cdot 10^{-23}$ J/K (see https://physics.nist.gov/cgi-bin/cuu/Value?k ).

For now, this does not change the practical temperature scale for calibrating thermometers around ambient temperatures: https://www.bipm.org/en/committees/cc/cct/guide-its90.html

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