How to obtain the value of the gas constant R = 8.20574 × 10⁻² L atm K⁻¹ mol⁻¹

While reading a chemistry book, I saw a table of values of the gas constant for different units.


For example,


$$R=8.20574\times 10^{-2}\ \text{L}\cdot\text{atm}\cdot \text{K}^{-1}\cdot \text{mol}^{-1}\tag{1}$$


Now, $R$ comes from the ideal gas law


$$R=\frac{PV}{nT}$$


My question is how arrive at the number in (1).


Here are the calculations I came up with.


Suppose $n$ is in the unit mole and $T$ is in the unit $K$.


Let's check the unit of $PV$.


$$\mathrm{Pa\cdot m^3}$$


$$\mathrm{Pa\cdot 10^3 L}$$


$$\mathrm{\frac{101325\cdot Pa}{101325}\cdot 10^3 L}$$


$$\mathrm{atm \cdot \frac{10^3}{101325}L}$$


$$\mathrm{atm\cdot 0.009869232667\cdot L}$$


$$\mathrm{9.86\cdot 10^{-3} atm\cdot L}$$


According to these calculations, $R$ would have the unit


$$\mathrm{9.86\cdot 10^{-3}\ L\cdot atm\cdot K^{-1}\cdot mol^{-1}}$$


Now, clearly, given that (1) is present in both the book and the internet, I surmise there must be an error in the calculations above. But what is it?





> 
> My question is how arrive at the number in (1).
> 
> 
> 


You will never arrive to the value of $R$ by paper. The value of $R$ is a universal physical constant. It appears for **all** gases, independent of their chemical identity, when the pressure is low enough. It necessitates experimental data to be verified. It would be pretty cool if you could do that, like if we could arrive to the value of Planck's constant $h$ by a mere inspection of his black-body radiation formula.


Nevertheless, lets try to obtain this value for a substance. The approach is simple:


* We put a fixed amount of moles of water in a device.
* We mantain the pressure $p$ and temperature $T$.
* We measure the resulting volume. I will use here the molar volume, this is, this measured volume divided by the amount of moles of $v = V/n$.


Thus, the ideal gas law states that if we graph the pressure $p$ vs the molar volume $v$ we get
\begin{align}
 p = RT\left(\frac{1}{v}\right) \tag{1} \\ 
\end{align}
Eq. (1) predicts that a plot of $1/v$ vs $p$ should give an increasing linear function with a slope of $RT$.


Lets use isothermal data, along three different temperatures, and do linear regressions. We take care of three points:


* I will be cautious and pick on purpose high temperatures, so the ideal gas law will be approached better.
* I will pick the lowest pressures available for the same reason.
* I will put everything in SI units to get the value in $\pu{[J]/[mol]·[K]}$.


The data extracted is:
\begin{array}{c|c c c}
 p(\pu{Pa}) & V (\pu{m^3/mol}) \; (573.15 \; K) & V (\pu{m^3/mol}) \;
 (673.15 \; K) & V (\pu{m^3/mol}) \; (773.15 \; K) \\ \hline
 1000 & \pu{5.59554} & \pu{6.42793} & \pu{7.25446} \\
 10000 & \pu{0.55955} & \pu{0.64259} & \pu{0.72582} \\
 20000 & \pu{0.27959} & \pu{0.32121} & \pu{0.36282} \\
 30000 & \pu{0.18646} & \pu{0.21420} & \pu{0.24194} \\
 40000 & \pu{0.13984} & \pu{0.16065} & \pu{0.18141} \\
\end{array}


The results are as follows:


[![enter image description here](https://i.stack.imgur.com/HFDQY.jpg)](https://i.stack.imgur.com/HFDQY.jpg)


The results are encouraging, it seems that we have obtained straight lines. The fitted line has a slope $m$ and y-intercept $b$ of:
\begin{array}{c|c c c}
 \text{Temperature} & 573.15 \; \pu{K} & 773.15 \; \pu{K} & 873.15 \; \pu{K} \\ \hline
 m \; [\pu{J/mol}] & 5593.4 & 6457.9 & 7256.8 \\
 b \; [\pu{Pa}] & 0.33051 & -1.0556 & 0.81970 \\
\end{array}


Now, by Eq. (1), the constant should be


$$ \boxed{R = \frac{m}{T}} \tag{2} $$


We tabulate the results of Eq. (2) and the porcentual error with respect to the value informed by CODATA ($\pu{8.314462618 \; J/mol K}$)
\begin{array}{c| c c c}
 \text{Temperature} & 573.15 \; \pu{K} & 773.15 \; \pu{K} & 873.15 \; \pu{K} \\ \hline
 R \; [\pu{J/mol K}] & 8.3093 & 8.3114 & 8.3111 \\
 \% E\_\pu{r} & 0.062 \% & 0.036 \% & 0.040 \% \\
\end{array}


If what I did you think is **fun**, you can find $PVT$ data over the internet of other substances, and try this approach. You can then corroborate that $R$ is still the same. There are many other ways to get $R$, some are more simpler and elegant (like physicists would do). This is the one that came through my head when reading.


**References**


1. [R Value of NIST](https://physics.nist.gov/cgi-bin/cuu/Value?r).
2. The steam tables come from any thermodynamic book at the back of it. I used "Introduction to Chemical Engineering Thermodynamics", J.M. Smith, H.C. Van Ness, M.M. Abbott, McGraw-Hill, 7th edition, 2004.


 2

Your analysis does not consider that one of $n$, $p$, $V$, $T$ variables is dependent on the other three. By other words, with 4 variables and 1 equation, we get 3 degrees of freedom and 1 of 4 variables dependent. R is the constant involved in this dependency.


$R=\pu{8.20574E-2 L atm K-1 mol-1}$ is numerically equal to the volume (L) of $\pu{1 mol}$ of an ideal gas at $T=\pu{1 K}$ and pressure $p=\pu{1 atm}$.


It could be recalculated from $R=\pu{8.31446 J K-1 mol-1}$, numerically equal to the volume in $\pu{m3}$ for $n=\pu{1 mol}$, $T = \pu{1 K}$, $p = \pu{1 Pa}$.


 1

Here is my new understanding based on comments and answers to the original question.


In the original question, the value in (2) is derived by starting with 1 Pa multiplied by 1 cubic meter, then we divide by 1 mol multiplied by 1 K. But then this starting point has the gas constant at a value of 1, which is experimentally wrong.


If we start with the correct value of


$$8.3145 \mathrm{\frac{J}{mol}}$$


$$=8.3145\mathrm{\frac{kg\cdot m^2}{K\cdot mol\cdot s^2}}$$


$$=8.3145\mathrm{\ Pa\cdot m^3\cdot\frac{1}{K\cdot mol} }$$


and proceed from here, then we can use the original calculations for converting $\mathrm{Pa\cdot m^3}$ and we reach


$$8.3145\cdot 9.86\cdot 10^{-3} \mathrm{\frac{L\cdot atm}{K \cdot mol}}$$


$$=8.20577\cdot 10^{-2}\mathrm{\frac{L\cdot atm}{K \cdot mol}}$$


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