0
$\begingroup$

I was wondering how to re-write that ideal gas formula $PV = nRT$ applicable to Celsius or Fahrenheit scale. I know $R$ has unit $\frac{\mathrm J}{\mathrm{mol\cdot K}}$ and $T$ has units in K, that means this formula is derived only for absolute temperature $T$, is that true?

I want to re-write it for Celsius scale. If I change $T = C+273$ and apply $C$ to the formula, does that beget me the right conversion? I think there's more than just converting the temperature scale.

$\endgroup$
2
  • $\begingroup$ But you can keep in Celsius for temperature differences, relative is OK, absolute is not, it must be K. $\endgroup$
    – user108787
    Sep 22, 2016 at 1:03
  • $\begingroup$ Note that there are 10-20 different versions of the ideal gas constant, depending on what pressure, volume, and temperature units you are using. $\endgroup$ Jul 15, 2019 at 1:02

3 Answers 3

4
$\begingroup$

It only works for absolute temperature. You can convert your temperature in Celsius or Fahrenheit to Kelvin first, then plug in the result in the formula, or you can modify the formula to do that for you: $$PV=nR(T_{°C}+273.15)K$$ Where $T_{°C}$ is the number of the temperature in Celsius and $K$ just stands for Kelvin. Example: if the temperature is $25°C$, then $T_{°C}=25$ and $PV=nR(25+273.15)K=nR(298.15K)$.

$\endgroup$
14
  • 2
    $\begingroup$ :-( "K" looks like another variable here and superfluous. Just $PV = nR (T_C+273)$ where $T_C$ is the temperature in Celsius. $\endgroup$
    – MaxW
    Sep 22, 2016 at 2:02
  • $\begingroup$ @MaxW Agree! Please edit the formula to remove the $K$. $\endgroup$
    – garyp
    Sep 22, 2016 at 2:40
  • $\begingroup$ It's important to be consistent with the units. If you divide $PV$ by $nR(T_{ºC}+273.15)$, you get $1$ Kelvin. $\endgroup$
    – Wood
    Sep 22, 2016 at 2:40
  • $\begingroup$ The point is that no other variable in the equation has units. It is just weird to add the K. The K looks like a variable not a unit of measure. $\endgroup$
    – MaxW
    Sep 22, 2016 at 2:43
  • $\begingroup$ ALL other variables have units. For example, $P=100 Pa$ would be correct, not $P=100$. $\endgroup$
    – Wood
    Sep 22, 2016 at 2:45
3
$\begingroup$

Yes, the formula relies on absolute temperature (generally in Kelvins). But by adding 273 to the temperature in degrees Celsius, you are converting to Kelvin, which is the only change you need to make the formula work.

$\endgroup$
2
  • $\begingroup$ Note that Celcius and Fahrenheit cannot be used. The temperature must be in Kelvin or Rankin. (Absolute versions of Celsius and Fahrenheit) $\endgroup$ Sep 22, 2016 at 1:04
  • $\begingroup$ You could use some simple conversions to change any temperature scale to Kelvin. Some conversions are messier than other, but it is possible. $\endgroup$
    – MaxW
    Sep 22, 2016 at 2:05
0
$\begingroup$

You could use some simple conversions to change any temperature scale to Kelvin to work in the ideal gas law.

$\text{K} = {}^\circ\text{Celsius} + 273.15$

$\text{K} = (^\circ\text{Fahrenheit}-32)\dfrac{5}{9} + 273.15$

$\text{K} = \dfrac{^\circ\text{Rankine}}{1.8}$

When I first learned the equation pressure was in units of mm of Hg, volume was in liters, and temperature was Kelvin.

Note that since Rankine is an absolute temperature scale too you could use a different value for R which is a constant which depends on the units of pressure, volume and temperature. It would be odd to use English units for temperature and metric units for pressure and volume, so you could convert those values into a different value for R too. So with an appropriate value of R you could have pressure in $\mathrm{lbs/inch}^2$, volume in cubic feet, and temperature in Rankine.

$\endgroup$
1
  • 5
    $\begingroup$ Note that according to the BIPM the modern name for the unit is "kelvin," not "degrees kelvin." $\endgroup$
    – rob
    Sep 22, 2016 at 3:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.