Relationship between the ideal gas constant and entropyWhat are the units of P and V when R is expressed in...
What's the difference between repeating elections every few years and repeating a referendum after a few years?
Landlord wants to switch my lease to a "Land contract" to "get back at the city"
Add an angle to a sphere
What is the command to reset a PC without deleting any files
Why do UK politicians seemingly ignore opinion polls on Brexit?
Was there ever an axiom rendered a theorem?
How to answer pointed "are you quitting" questioning when I don't want them to suspect
Pristine Bit Checking
Where else does the Shulchan Aruch quote an authority by name?
Need help identifying/translating a plaque in Tangier, Morocco
Lied on resume at previous job
Why doesn't a const reference extend the life of a temporary object passed via a function?
Why do we use polarized capacitors?
Prime joint compound before latex paint?
How would photo IDs work for shapeshifters?
How to make payment on the internet without leaving a money trail?
When blogging recipes, how can I support both readers who want the narrative/journey and ones who want the printer-friendly recipe?
How to manage monthly salary
Re-submission of rejected manuscript without informing co-authors
Manga about a female worker who got dragged into another world together with this high school girl and she was just told she's not needed anymore
What is the meaning of "of trouble" in the following sentence?
Why is the design of haulage companies so “special”?
What does 'script /dev/null' do?
Can I legally use front facing blue light in the UK?
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
$begingroup$
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
$endgroup$
|
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
$endgroup$
$begingroup$
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
$endgroup$
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
edited 6 hours ago
orthocresol♦
40.1k7115246
40.1k7115246
asked 6 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
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
2
$begingroup$
Both entropy and the ideal gas constant are in some sense related to the Boltzmann constant.
$endgroup$
– Zhe
5 hours ago
$begingroup$
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
$begingroup$
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
$begingroup$
$Delta S=Rln{(V_f/V_i)}=-Rln{(P_f/P_i)}$
$endgroup$
– Chet Miller
5 hours ago
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
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.)
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "431"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f112376%2frelationship-between-the-ideal-gas-constant-and-entropy%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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.
edited 3 hours ago
orthocresol♦
40.1k7115246
40.1k7115246
answered 4 hours ago
Night WriterNight Writer
2,806223
2,806223
add a comment |
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.)
$endgroup$
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.)
$endgroup$
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.)
$endgroup$
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.)
answered 2 hours ago
porphyrinporphyrin
18.3k3157
18.3k3157
add a comment |
add a comment |
Thanks for contributing an answer to Chemistry Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f112376%2frelationship-between-the-ideal-gas-constant-and-entropy%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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