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Relationship between the ideal gas constant and entropy

What are the units of P and V when R is expressed in terms of J/(mol K)?Applying Avagadro's law: the relationship between moles and volume of a gasEntropy increase when heating a gasWhy is the ideal gas law so ubiquitous?Work and entropyWhy do Hydrogen and Helium have molar volumes higher than an ideal gas?Relation between constant-pressure and constant-volume heat capacities: Cp - Cv = nRDoesn't the isothermal irreversible expansion/compression of an ideal gas law violate the ideal gas law?Adiabatic processes and the Ideal Gas LawEntropy change of isothermal irreversible expansion of ideal gas

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

edited 6 hours ago

orthocresol♦

40.1k7115246

asked 6 hours ago

H.LinkhornH.Linkhorn

41419

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 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
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

 | 
show 3 more comments

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

edited 6 hours ago

orthocresol♦

40.1k7115246

asked 6 hours ago

H.LinkhornH.Linkhorn

41419

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 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
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

 | 
show 3 more comments

$begingroup$

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

edited 6 hours ago

orthocresol♦

40.1k7115246

asked 6 hours ago

H.LinkhornH.Linkhorn

41419

$endgroup$

share|improve this question

edited 6 hours ago

orthocresol♦

40.1k7115246

asked 6 hours ago

H.LinkhornH.Linkhorn

41419

share|improve this question

edited 6 hours ago

orthocresol♦

40.1k7115246

asked 6 hours ago

H.LinkhornH.Linkhorn

41419

edited 6 hours ago

orthocresol♦

40.1k7115246

edited 6 hours ago

orthocresol♦

40.1k7115246

asked 6 hours ago

H.LinkhornH.Linkhorn

41419

asked 6 hours ago

H.LinkhornH.Linkhorn

41419

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

edited 3 hours ago

orthocresol♦

40.1k7115246

answered 4 hours ago

Night WriterNight Writer

2,806223

$endgroup$

add a comment | 

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

answered 2 hours ago

porphyrinporphyrin

18.3k3157

$endgroup$

add a comment | 

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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

edited 3 hours ago

orthocresol♦

40.1k7115246

answered 4 hours ago

Night WriterNight Writer

2,806223

$endgroup$

add a comment | 

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

edited 3 hours ago

orthocresol♦

40.1k7115246

answered 4 hours ago

Night WriterNight Writer

2,806223

$endgroup$

add a comment | 

$begingroup$

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

edited 3 hours ago

orthocresol♦

40.1k7115246

answered 4 hours ago

Night WriterNight Writer

2,806223

$endgroup$

share|improve this answer

edited 3 hours ago

orthocresol♦

40.1k7115246

answered 4 hours ago

Night WriterNight Writer

2,806223

edited 3 hours ago

orthocresol♦

40.1k7115246

edited 3 hours ago

orthocresol♦

40.1k7115246

answered 4 hours ago

Night WriterNight Writer

2,806223

answered 4 hours ago

Night WriterNight Writer

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

answered 2 hours ago

porphyrinporphyrin

18.3k3157

$endgroup$

add a comment | 

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

answered 2 hours ago

porphyrinporphyrin

18.3k3157

$endgroup$

add a comment | 

$begingroup$

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

answered 2 hours ago

porphyrinporphyrin

18.3k3157

$endgroup$

share|improve this answer

answered 2 hours ago

porphyrinporphyrin

18.3k3157

answered 2 hours ago

porphyrinporphyrin

18.3k3157

answered 2 hours ago

porphyrinporphyrin

18.3k3157