# Conversion of $R = 8317\ \mathrm{\frac{J}{kg\cdot mol\cdot K}}$ to $8317\ \mathrm{\frac{J}{kg\cdot mol\cdot {^\circ C}}}$

I'm trying to understand the unit conversions when one unit is obtained by shifting the value of another one by some constant. In particular, the unit conversions from Kelvin to Celsius scale:

$$T(\mathrm{^\circ C}) = T(\mathrm K) - 273.16.$$

I want to find the value of $$R = 8317\ \mathrm{\frac{J}{kg\cdot mol\cdot K}}$$ in the units of $$\mathrm{{\frac{J}{kg\cdot mol\cdot ^\circ C}}}$$.

What I did was to consider $$\frac{1}{\mathrm K}$$ as $$\frac{1}{1\ \mathrm K} = \frac{ 1}{274.16\ \mathrm{^\circ C}},$$ so that $$R = 30.3363\ \mathrm{\frac{J}{kg\cdot mol\cdot {^\circ C}}},$$ but according to the book that I using,

The dimension of temperature in the units of the gas constant is the size of its increment, not its value, that is, the degree size. Thus, $$R = 8317\ \mathrm{\frac{J}{kg\cdot mol\cdot K}} = 8317\ \mathrm{\frac{J}{kg\cdot mol\cdot {^\circ C}}}.$$

However, (maybe because I'm not a native English speaker, or the explanation is vague) I cannot understand the given explanation why it is the case, i.e the value of the constant is not affected by the shift $$-273.16$$ value.

Edit:

However, I still cannot understand how what I did above is not compatible with what we normally do. For example, if we have a quantity $$X= 1000g$$, then to convert $$X$$ to kg, what we would to is $$X = 1000g = 1000g \frac{1kg }{1000g } = 1 kg,$$ since $$1kg = 1000g$$, the fraction in the RHS is just a scalar constant 1, and multiplying a quantity with $$1$$ does not change the value of that quantity.

Similarly, I follow the same logic $$R = 8317\ \mathrm{\frac{J}{kg\cdot mol\cdot K}} = 8317\ \mathrm{\frac{J}{kg\cdot mol\cdot K}} * \frac{1K }{ 274.16 ^\circ C} = 30.3363\ \mathrm{\frac{J}{kg\cdot mol\cdot {^\circ C}}}.$$

• Your numeric value for 1 Kelvin is a little off. – Jasper May 30 at 10:44
• @Jasper should it have been 274.15 ? – onurcanbektas May 30 at 10:46
• duckduckgo.com/?q=1+kelvin+to+celsius – Jasper May 30 at 10:48
• @Jasper Thanks for pointing out, but that value is irrelevant to the main question. – onurcanbektas May 30 at 10:51
• And your question highlights exactly why engineers use Kelvin when dealing with temperature changes for steam etc... – user207455 May 30 at 11:51

Short Answer: $$1\ \mathrm{^\circ C}$$ is $$274\ \mathrm K$$. Does that mean $$2\ \mathrm{^\circ C}$$ is $$2 \times 274\ \text{K} = 548\ \mathrm K$$? It is not, because those two scales don't have the same origin (For $$\text{kg}$$ and $$\text{g}$$ the origins are same, as $$0\ \text{kg} = 0\ \text{g}$$).

This is why the substitution of $$\frac{1}{\mathrm K}$$ by $$\frac{274} {\mathrm{^\circ C}}$$ does not make sense.

In the unit of $$R$$, the physical meaning of $$\text{K}$$ is change in temperature (see below), and change in $$1\ \mathrm K =$$ change in $$1\ \mathrm{^\circ C} \neq$$ change in $$274 \ \mathrm{^\circ \text{C}}$$. In this sense $$1\ \mathrm K$$ is replaced by $$1\ \mathrm{^\circ \text{C}}$$.

However this is a kind of misuse of notation (although quite common), and here $${^\circ \text{C}}$$ has to be understood as the change in that scale. There is nothing more in that.

Details: $$R$$ is the universal gas constant, and has same dimension as molar specific heat.

For example, "the molar specific heat of monoatomic ideal gas is $$1.5 R$$" means, you have to provide $$1.5 \times 8.314\ \mathrm J$$ heat to $$1\ \mathrm{mol}$$ ideal gas to increase its temperature by $$1\ \mathrm K$$ (or equivalently by $$1\ \mathrm{^\circ C}$$).

So $$R$$ has units joule per mole per kelvin or joule per mole per change in degree centigrade.

In this sense, the $$\mathrm K$$ can be replaced by $$\mathrm{^\circ C}$$ in units of $$R$$, where we have to interpret degree Celsius as change in temperature.

• Please see my edit. – onurcanbektas May 30 at 10:29
• 2C != 248 K is probably the most important part of this answer. – Jasper May 30 at 10:50
• @Jasper Indeed; sometimes even if you don't understand something why something does not work, seeing that if it were to work, we would get absurd things really the only key to grasp that fact. Mathematicians basically call this the method of contradiction. – onurcanbektas May 30 at 11:13

The universal gas constant tells you how much the energy of a gas changes when you change the mass, the number of particles, or the temperature.

Let's say that we wanted to convert $$R$$ to $$\frac{J}{g*mol*K}$$. The value of $$R$$ in these units would be different than $$8317$$, because changing the gas's mass by 1 gram is different than changing the gas's mass by 1 kilogram. Specifically, adding 1 gram to the gas's mass is 1000 times smaller of a change than adding 1 kilogram of mass, so the corresponding value of $$R$$ should be 1000 times smaller, namely, $$8.317$$ $$\frac{J}{g*mol*K}$$.

Now, when we want to convert from degrees Celsius to Kelvin, the question to ask is: "How much more (or less) does the gas's temperature change when we add 1 degree Celsius, as opposed to 1 Kelvin?" And the answer is that adding 1 degree Celsius to the temperature is exactly the same as adding 1 Kelvin, because one is merely a shifted version of the other.

An analogy might help: suppose you were measuring the speed of a car going down your street. You do this by dividing the distance that the car travels by the time it takes to do so. Suppose you did this in two ways: in one trial, you start the stopwatch a few seconds before the car turns onto your street, and note the times at which it passes two markers. In another trial, you start the stopwatch an hour before the car turns onto your street, and note the times at which it passes two markers. Since you're interested only in the difference of the two times you recorded, it really doesn't matter when you start the stopwatch, so you'll get the same answer either way.

• please see my edit. – onurcanbektas May 30 at 10:30
• @onurcanbektas A change of one Kelvin is not equal to a change of 274.16 degrees Celsuis. If I heat a cup of water from 23 degrees Celsius to 24 degrees Celsuis, that is the same as heating it from 296.15 K to 297.15 K. When I change the temperature of the cup by 1 degree Celsius, its temperature changes by 1 Kelvin. Proportionality constants measure how things change, so the relative width of changes in the different units is what's relevant. – probably_someone May 30 at 10:33
• Why do you consider the change instead of the actual value ? – onurcanbektas May 30 at 10:37
• @onurcanbektas Because proportionality constants measure how one value changes in response to a change in another value. For example, velocity measures how an object's position changes over a given change in time. – probably_someone May 30 at 10:38
• Which proportionaliy constant ? They are just units. Even it you see them as some coefficient, "proportionality constants measure how one value changes in response to a change in another value" is not clear to me why it is the case. Can you elaborate on that point a bit ? – onurcanbektas May 30 at 10:48

The units are multiplicative. When you have 10 m it means you have 10 times the distance of one meter from some point. In distance, the unit is fully relative. In celsius/kelvin, the point was chosen apriori and all the values are refering to this absolute point.

I don't know what exactly your R is, but it is most probably again relative. It tells you how the energy changes when temperature changes by 1 °C/1 K and this change is same in kelvin as in degrees Celsius.

The magnitude of a degree Kelvin is the same as the magnitude of a degree Centigrade. A change in temperature in degrees Kelvin equals a change in temperature in degrees Centigrade.

Bottom line: Anything per degree Kelvin is the same as per degree Centigrade.

Hope this helps.

However the units for the gas constant are either $$\frac{J}{kmol \cdot K}$$ (value is then about 8310) or $$\frac{J}{mol \cdot K}$$ (with a value of about 8.31). Having the product of kg and mole in the denominator does not make sense. Its is puzzling to me that not only the OP but some answers have the same wrong units. It is even more puzzling if these quotes in the OP are reproduced exactly from an actual book.

I think your textbook choose a bad example because the universal gas constant is not usually written in units of degrees Celsius. Even the Wikipedia page does not list units with degree Celsius. The conversion factor is based on the fact that:

$$\frac{\Delta T[^oC]}{\Delta T[K]}=1$$ And not

$$\frac{T[C]}{T[K]}=1$$

For example for the equation $$Q=mC_p\Delta T$$ , the units of $$C_p$$ can be per Kelvin or per degree Celsius by using the first conversion formula for $$\Delta T$$. But for the universal gas constant as used in the ideal gas law:

$$PV=nRT$$

Where $$T$$ is absolute temperature you cannot have $$R$$ with both temperature scales and have the same magnitude because in one case you will substitute $$1 ^oC$$ for $$T$$ and in the other case you will substitute $$274.15 K$$ so the other values will be different. So $$R$$ in units of $$\frac{J}{mol ^oC}$$ is wrong.