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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?

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  • $\begingroup$ Chemistry SE (in contrary to some other SE sites) strongly recommends plain text question titles for index/search reasons and due possible displaying issues in question title lists. $\endgroup$
    – Poutnik
    May 11, 2023 at 11:24
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    $\begingroup$ Try $R=\pu{8.20574E-2 L atm K-1 mol-1}$ displaying $R=\pu{8.20574E-2 L atm K-1 mol-1}$. // \pu{} and \ce{} are involved in the optional MathJax module mhchem, loaded by default on CH SE. It can be loaded explicitly via \require{mhchem}, but IMHO only \ce (chemical equations/expresions) works on other sites. \pu may have some conflict and never worked for me e.g. on Physics SE. // Formatting of chem expressions $\endgroup$
    – Poutnik
    May 11, 2023 at 11:47
  • $\begingroup$ @Poutnik I used unicode superscripts in my edit of the title. This sometimes works (not all superscript characters are available). $\endgroup$
    – Karsten
    May 11, 2023 at 13:51
  • $\begingroup$ @Karsten Unicode should work, but it may cause hard time to some browsers during editing or copypasting. Like replacing some more exotic characters like various flavours of white spaces or hyphens/dashes by some universal ones. AFAIK Chrome is famous by that. IMHO, the safe bet is rather using plain ASCII if possible. $\endgroup$
    – Poutnik
    May 11, 2023 at 13:58
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    $\begingroup$ It seems like the content of your book is more than 40 years old if it still uses "atm". You might want to consider finding a better book. $\endgroup$
    – Loong
    May 11, 2023 at 17:26

4 Answers 4

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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

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.
  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.
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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 $n=\pu{1 mol}$, $T = \pu{1 K}$, $p = \pu{1 Pa}$.

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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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Its that R is written in those units and have to be applied in the units as written. Use whichever value of R has the most common units to help avoid stoichiometry. But you have to use the units as written. Also you misread the table. Should be first two divided by second two. Not all 4 multiplied. So R has to be written as a constant in the units of 0.0821(L * atm / K * mol) “0.0821 Latm/Kmol” You cant use 0.0821. But also whatever equation that you tried won’t work because its nonsensical. Imagine it as it it’s supposed to be read. “0.0821 L*atm / mol * K = (x)L * (y)atm / (n)mol * (t)K

Where all variables are written in (). You’re missing every single variable.

Ok i read your latest response.

If everything is 1. Then it reads as

.0821Latm / Kn = 1L1atm / 1K1mol

Cross cancel whatever units can be cross canceled. You end up with 0.0821 L•atm / K•mol = 1L•1atm / K•mol

Or just R=PV/nT

You’re not applying the units to the constant which is why you’re confused.

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