For example, Charles' law does not work with Celsius. Why is this? I can only think that the gas constant is not a constant when Celsius is used rather than Kelvin.

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    $\begingroup$ I'd say, when Celsius is used instead of Kelvin, there is no such law and no such constant. Physical laws are awfully strict in that they apply to certain things and not to others. You just can't expect them to work after you've changed some physical quantity to another at random, even if the names of the said quantities happen to be kinda similar ("temperature"). $\endgroup$ – Ivan Neretin Nov 17 '16 at 18:32
  • $\begingroup$ Charles law works perfectly well with celsius: the algebra is slightly more complicated but the constant is exactly the same. $\endgroup$ – matt_black Nov 17 '16 at 19:04
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    $\begingroup$ I'm voting to close this question as off-topic because this is a physics question with hardly any relevance to chemistry. $\endgroup$ – Jan Nov 17 '16 at 20:10
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    $\begingroup$ I disagree that it is a physics question per se- it is confusion about temperature scales and relative vs absolute. $\endgroup$ – Jon Custer Nov 17 '16 at 23:25

Charles' Law certainly DOES work with temperature in degrees Centigrade!

If we express it (relative to $T=0$ absolute) as $V/T = k$ and if we convert $T$ (K) to $t$ (°C) as $t = T-273$ then obviously Charles' Law would have the form $V/(t+273) = k$.

More generally if, instead of $T$ we use some linear function of $T$, say $Y= mT + b$, then $$V/T = Vm/(Y-b) = k$$ or $$V/(Y-b) = k/m,$$ where $k$, $b$, and $m$ are constants. We can substitute $k'$ for $k/m$ and get $V/(Y-b) = k'$, but we can't get rid of the $-b$ that way.

The claim that a law (even if it is an approximate one like Charles') "doesn't work" with a different unit of measure confuses a formula with a law. Formulas are great for solving high school level problems, but a deeper understanding is expected in college and beyond. Any strictly increasing function can be used as a temperature scale, for instance $X = π^T$ could be a temperature scale.

The formula we'd use to express Charles' Law would look very different, but the relationship between volume and $\log(X)/\log(π)$ would be just as constant (and since $\log(π)$ is a constant, we'd have $V/\log(X) = k''$.)


If you consider temperature to be the measure of mean kinetic energy, then to get the correct linear scale, you'd need an absolute temperature. The Kelvin scale (or the Rankine scale) provides that absolute temperature where in the limit of zero temperature, the system is in its lowest energy state.

  • $\begingroup$ Why do you need absolute temperature? $\endgroup$ – JobHunter69 Nov 18 '16 at 0:30
  • $\begingroup$ Because the third law of thermodynamics should work. You want properties to scale directly with temperature in a linear relationship (which they do with an absolute scale) instead of an affine relationship (linear if you shift the scale). You might as well just shift the scale and make the system linear. Then you just can multiply without that pesky addition. $\endgroup$ – Zhe Nov 18 '16 at 3:01

Charles’ law, for those in need of a refreshment (like me) states that an increase in temperature results in an increase in volume and vice-versa; thus, that temperature and volume are directly proportional.

Typically, this would be an experiment in physics class, measuring the temperature of a gas cylinder and the volume it requires under constant pressure. A number of different temperatures and volumes would be recorded and then filled into a graph. It turns out that that graph gives a linear relationship between the two. Assuming temperature to be on the $x$ axis and volume on the $y$ axis, the equation of the linear relationship would be:

$$V = k\vartheta + V_0\tag{1}$$

Wherein $k$ is a constant (which can be calculated depending on $p, n$ and the gas constant $R$) and $V_0$ is the $y$-axis-bisection of the graph. Since $V_0$ is also a constant, you still observe a linear relationship between $V$ and $\vartheta$. The difficulty is that you need to determine two constants and that you need both for any algebra that you want to perform using this law.

If you extrapolate, you realise that the $x$-axis will be bisected at a temperature of $\vartheta \approx -273~\mathrm{^\circ C}$. You could go ahead and decide to modify your equation so that $V_0 = 0$; that requires modifying what you insert as $\vartheta$ in such a way that you effectively move the graph sideways in the coordinate system. There is a number of general rules for doing so; what you want is to move the graph to the right by $273~\mathrm{^\circ C}$ so the resulting equation becomes:

$$V = k \cdot (\vartheta + 273~\mathrm{^\circ C})\tag{2}$$

And then all some clever people did was to replace $\vartheta + 273~\mathrm{^\circ C}$ with the ‘new’ temperature $T$ using the new unit $\mathrm{K}$; the relationship is such that $0~\mathrm{K}\ \hat{\approx}\ {-273}~\mathrm{^\circ C}$. This leads us to our final equation:

$$V = k \cdot T\tag{3}$$

Equations $(1), (2)$ and $(3)$ are all mathematically equivalent to each other. Thus, they describe the same physical consequence. Thus, unless you learnt equation $(3)$ and equation $(3)$ only as Charles’ law, said law is valid no matter which choice of temperature scale you make. Even Farenheit temperatures would obey the law.


The other answers sort of address your confusion, but not very directly. So let me give it a shot.

For gas laws, the gas constant is always a constant, regardless of which temperature scale you need. It's also a constant regardless of whether you are using it in an ideal gas law or some other law. It's even a constant when it shows up in places that aren't gas laws at all! That's because it's a fundamental constant which relates the statistical properties of molecules to macroscopic phenomena like pressure and temperature.

The catch is, in order to use it, the variables have to be on an absolute scale. The reason for this is a little hard to explain, but I guess an easy way to think about it is that it defines how things change relative to each other. For example, if you double the temperature while holding pressure constant, the volume will also double (Charles's Law). Volume is already on an absolute scale - zero volume means it takes up no space. Pressure is the same way - zero pressure means no force exerted on the walls of the container (on average). Temperature is really the only one that has an arbitrary zero point on a lot of scales, where "zero" doesn't necessarily mean "no kinetic energy."

If you double the temperature as measured in an absolute scale, you are doubling the average kinetic energy of the molecules. If you double the temperature as measured on the (for example) Celsius scale, the kinetic energy doesn't necessarily increase by the same amount. Take a negative temperature. What happens when you double -100? Or look at room temperature. If you go from 20 to 40 degrees Celsius, the absolute temperature only increased by about 7% (20 / 293).

Each combination of pressure, volume and temperature units has its own version of the gas constant, but these are all really just unit conversions of the "fundamental" constant, which is the Boltzmann constant multiplied by Avogadro's number. The key thing to remember is that besides making sure the units match the measurements you have, the temperature scale must be an absolute scale.


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