So it dawned on me the other day that the ideal gas constant $$R = \pu{8.31 J mol-1 K-1}$$ has the same units as molar entropy. Is there some deeper meaning behind this or is it just a coincidence that this occurs?
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$\begingroup$ Well, R is in the equation for the entropy change of an ideal gas at constant temperature. Is that what you were asking? $\endgroup$– Chet MillerCommented Apr 8, 2019 at 19:40
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$\begingroup$ What is that, equation as it may answer my question. But, i was asking why does R have the units for entropy - Is it representing entropy? $\endgroup$– H.LinkhornCommented Apr 8, 2019 at 19:42
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2$\begingroup$ Both entropy and the ideal gas constant are in some sense related to the Boltzmann constant. $\endgroup$– ZheCommented Apr 8, 2019 at 20:12
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$\begingroup$ Could you elaborate on that please. How are they related? $\endgroup$– H.LinkhornCommented Apr 8, 2019 at 20:13
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$\begingroup$ $\Delta S=R\ln{(V_f/V_i)}=-R\ln{(P_f/P_i)}$ $\endgroup$– Chet MillerCommented Apr 8, 2019 at 20:14
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3 Answers
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The gas constant is equal to Avogadro's constant times Boltzmann's constant, the latter serving as a proportionality constant between the average thermal (kinetic) energy of the particles in an ideal gas and the temperature:
$$\left(\frac{\partial \bar U}{\partial T}\right)_p=\frac{3}{2}k_\mathrm{B}$$
The entropy can be regarded as a proportionality constant between the change in free energy $G$ with change in temperature of a system at constant pressure, since
$$\left(\frac{\partial G}{\partial T}\right)_p=-S$$
Note also that the average entropy of a particle can be written as
$$S=k_\mathrm{B} \log\Omega$$
So both $R$ and $S$ can be regarded as proportionality constants between energy and temperature.
The Wikipedia page on the equipartition theorem may provide some enlightenment on the origin of these proportionalities.
answered Apr 8, 2019 at 21:02
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The fundamental equation is Boltzmann's for the entropy $S =k_B\ln(\Omega)$ where $k_B$, Boltzmann's constant, has units J/molecule/K. When we use $R$ instead of $k_B$ it is trivially because molar units are used to define $S$.
In the equation $\Omega$ is the number of arrangements or configurations (or 'complexions' to use an old word) of distinguishable 'particles' among all the available energy levels. In that sense entropy is a measure of the uniformity of population in these levels.
(If there are $N$ distinguishable particles then $\displaystyle \Omega =\frac{N!}{n_1!n_2!n_3!\cdots}$ where there are $n_i$ particles in level $i$. The log is evaluated by assuming all the numbers $N,\, n_i$ etc. are large numbers so that Stirling's approximation for $\ln(N!) $ etc. is valid.)
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In the ideal gas, $3R/2$ is the thermodynamic entropy of one Mol.
The factor $\frac{3}{2}$ comes from $ST=N\bar{E}=N\frac{3k}{2}T=\frac{3R}{2}T$.
The energy of 1 Mol of ideal gas is either given by $3pV/2$ or $3RT/2$ or $ST$.
(In the ideal gas) the entropy does not depend on temperature. That is why one writes the $T$ separate in $TS$. For non-ideal gases $S$ changes with temperature: $E=\iint dSdT$.
For the ideal gass $C_V = S$ (always per Mol here). $ST = \int dQ_{\text{rev}} = C_V \int dT = S \int dT$. Because of constant volume, all the energy goes into $T$. $C_V$ and $S$ stay constant. On the $pV$ side this amounts to an increase in pressure.
