Relationship between the ideal gas constant and entropy












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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?






gas-laws entropy







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$endgroup$












  • $begingroup$
    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 Miller
    8 hours ago










  • $begingroup$
    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.Linkhorn
    8 hours ago






  • 2




    $begingroup$
    Both entropy and the ideal gas constant are in some sense related to the Boltzmann constant.
    $endgroup$
    – Zhe
    8 hours ago










  • $begingroup$
    Could you elaborate on that please. How are they related?
    $endgroup$
    – H.Linkhorn
    8 hours ago










  • $begingroup$
    $Delta S=Rln{(V_f/V_i)}=-Rln{(P_f/P_i)}$
    $endgroup$
    – Chet Miller
    8 hours ago


















1












$begingroup$


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?






gas-laws entropy







share|improve this question











$endgroup$












  • $begingroup$
    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 Miller
    8 hours ago










  • $begingroup$
    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.Linkhorn
    8 hours ago






  • 2




    $begingroup$
    Both entropy and the ideal gas constant are in some sense related to the Boltzmann constant.
    $endgroup$
    – Zhe
    8 hours ago










  • $begingroup$
    Could you elaborate on that please. How are they related?
    $endgroup$
    – H.Linkhorn
    8 hours ago










  • $begingroup$
    $Delta S=Rln{(V_f/V_i)}=-Rln{(P_f/P_i)}$
    $endgroup$
    – Chet Miller
    8 hours ago
















1












1








1


1



$begingroup$


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?






gas-laws entropy







share|improve this question











$endgroup$




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?






gas-laws entropy




gas-laws entropy






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 9 hours ago









orthocresol♦

40.1k7115246




40.1k7115246










asked 9 hours ago









H.LinkhornH.Linkhorn

41419




41419












  • $begingroup$
    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 Miller
    8 hours ago










  • $begingroup$
    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.Linkhorn
    8 hours ago






  • 2




    $begingroup$
    Both entropy and the ideal gas constant are in some sense related to the Boltzmann constant.
    $endgroup$
    – Zhe
    8 hours ago










  • $begingroup$
    Could you elaborate on that please. How are they related?
    $endgroup$
    – H.Linkhorn
    8 hours ago










  • $begingroup$
    $Delta S=Rln{(V_f/V_i)}=-Rln{(P_f/P_i)}$
    $endgroup$
    – Chet Miller
    8 hours ago




















  • $begingroup$
    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 Miller
    8 hours ago










  • $begingroup$
    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.Linkhorn
    8 hours ago






  • 2




    $begingroup$
    Both entropy and the ideal gas constant are in some sense related to the Boltzmann constant.
    $endgroup$
    – Zhe
    8 hours ago










  • $begingroup$
    Could you elaborate on that please. How are they related?
    $endgroup$
    – H.Linkhorn
    8 hours ago










  • $begingroup$
    $Delta S=Rln{(V_f/V_i)}=-Rln{(P_f/P_i)}$
    $endgroup$
    – Chet Miller
    8 hours ago


















$begingroup$
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 Miller
8 hours ago




$begingroup$
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 Miller
8 hours ago












$begingroup$
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.Linkhorn
8 hours ago




$begingroup$
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.Linkhorn
8 hours ago




2




2




$begingroup$
Both entropy and the ideal gas constant are in some sense related to the Boltzmann constant.
$endgroup$
– Zhe
8 hours ago




$begingroup$
Both entropy and the ideal gas constant are in some sense related to the Boltzmann constant.
$endgroup$
– Zhe
8 hours ago












$begingroup$
Could you elaborate on that please. How are they related?
$endgroup$
– H.Linkhorn
8 hours ago




$begingroup$
Could you elaborate on that please. How are they related?
$endgroup$
– H.Linkhorn
8 hours ago












$begingroup$
$Delta S=Rln{(V_f/V_i)}=-Rln{(P_f/P_i)}$
$endgroup$
– Chet Miller
8 hours ago






$begingroup$
$Delta S=Rln{(V_f/V_i)}=-Rln{(P_f/P_i)}$
$endgroup$
– Chet Miller
8 hours ago












2 Answers
2






active

oldest

votes


















1












$begingroup$

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



$$S=k_mathrm{B} logOmega$$



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.






share|improve this answer











$endgroup$





















    1












    $begingroup$

    The fundamental equation is Boltzmann's for the entropy $S =k_Bln(Omega)$ 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.)






    share|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      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



      $$S=k_mathrm{B} logOmega$$



      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.






      share|improve this answer











      $endgroup$


















        1












        $begingroup$

        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



        $$S=k_mathrm{B} logOmega$$



        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.






        share|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          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



          $$S=k_mathrm{B} logOmega$$



          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.






          share|improve this answer











          $endgroup$



          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



          $$S=k_mathrm{B} logOmega$$



          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.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 7 hours ago









          orthocresol♦

          40.1k7115246




          40.1k7115246










          answered 7 hours ago









          Night WriterNight Writer

          2,806223




          2,806223























              1












              $begingroup$

              The fundamental equation is Boltzmann's for the entropy $S =k_Bln(Omega)$ 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.)






              share|improve this answer









              $endgroup$


















                1












                $begingroup$

                The fundamental equation is Boltzmann's for the entropy $S =k_Bln(Omega)$ 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.)






                share|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The fundamental equation is Boltzmann's for the entropy $S =k_Bln(Omega)$ 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.)






                  share|improve this answer









                  $endgroup$



                  The fundamental equation is Boltzmann's for the entropy $S =k_Bln(Omega)$ 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.)







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 6 hours ago









                  porphyrinporphyrin

                  18.3k3157




                  18.3k3157






























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