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I know that for reversible processes: entropy change = dq/t Here its unit should J/Kelvin but it is mentioned in my books that its unit is J/Mol-Kelvin. I do not understand why there is mole in denominator of unit. Also, if J/mole-Kelvin is the unit, then why it is an extensive property? Extensive properties vary based on quantity, but this is strictly per mole.

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    $\begingroup$ dS=dq/T, not dS=dq/dT. When you have moles in the denominator, you are talking about molar entropy which is an intensive property (as you noted). Since, in practice, when you use the tables, the number of moles is going to be exclusive to your system, it is necessary to tabulate the molar entropy ( or specific entropy) rather than the extensive entropy. $\endgroup$ Oct 10, 2023 at 20:52
  • $\begingroup$ thanks @ChetMiller for correction $\endgroup$
    – Ayush
    Oct 11, 2023 at 13:47

2 Answers 2

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In physical chemistry/thermodynamics they remind you that:

  • Intensive properties are independent of the quantity of material in the system, such as pressure $p$, temperature $T$, electric potential $\phi$, composition $x_j$, etc.
  • Extensive properties are dependent of the quantity of material in the system, such as volume $V$, entropy $S$, kinetic energy $K$, etc.

However, for many thermodynamic functions, we can turn them into intensive properties dividing by the total amount of the system. I'll show you the notation that I like. Let $X$ denote the thermodynamic property $Q$, $S$, $G$ or $V$, then $x$ is defined as \begin{equation} \boxed{x \equiv \frac{X}{n}} \tag{1} \end{equation} where $n$ is the total amount of the system. Here $x$ has the name of molar thermodynamic property, and may be $q$, $s$, $g$ or $v$

For example, $g$ is the molar Gibbs energy (intensive) in $\pu{J mol^-1}$ and $G$ is the Gibbs energy (extensive) in $\pu{J}$. $v$ is the molar volume (intensive) in $\pu{m^3 mol^-1}$ and $V$ is the volume (extensive) in $\pu{m^3}$.

Thus, for the following equations \begin{equation} \mathrm{d}S = \frac{\unicode{x0111}Q_\mathrm{rev}}{T} \quad \mathrm{d}s = \frac{\unicode{x0111}q_\mathrm{rev}}{T} \tag{2,3} \end{equation}

  • Eq. (2) denotes entropy $S$ in units of $\pu{J K^-1}$, and heat $Q_\mathrm{rev}$ in units of $\pu{J}$.
  • Eq. (3) denotes molar entropy $s$ in $\pu{J mol^-1 K^-1}$, and $q_\mathrm{rev}$ denotes molar heat (although this sounds weird) in $\pu{J mol^-1}$.
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  • $\begingroup$ Note that not all quantities depending on material quantity are extensive. There must be proportionality. X(an)=a.X(n). By other words, there are quantities that are neither intensive nor extensive. $\endgroup$
    – Poutnik
    Oct 11, 2023 at 6:33
  • $\begingroup$ For differentials of W and Q, it is better to use small "dyet"("d with stroke") Unicode symbols as for inexact differentials $đQ$ (U+0111 or by (Win)compose Compose / d ). It is sometimes substituted in CH SE by MJ \delta ($\delta Q$), but it is not exactly correct. OTOH, \delta looks better in context of MJ typesetting. $\endgroup$
    – Poutnik
    Oct 11, 2023 at 9:32
  • $\begingroup$ @Poutnik Hello Poutnik, sorry for the late reply. Thanks for the comments, you are right, the differentials are now correctly written. $\endgroup$ Oct 27, 2023 at 1:43
  • $\begingroup$ Also thanks @Maurice for the modification! $\endgroup$ Oct 27, 2023 at 1:43
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So the 'mol' or mole is just a number, and so you would be perfectly valid to give entropy in units of either J/mol/K or just J/K. You can interchange between the two by multiplying or dividing by the Avigadro constant.

Also you are right about the second part, if entropy of a substance was reported in units of J/mol/K then that is a standard entropy change which is an intensive property. Entropy just generally however does depend on the number of moles of substance and so is an extensive property unless the molar quantity is specified.

Hope that helps.

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    $\begingroup$ 1 mol is 1 of 7 fundamental SI units, so not dimensionless. Involving N_A would be a disaster. $\endgroup$
    – Poutnik
    Oct 10, 2023 at 21:27
  • $\begingroup$ Molar entropy $S_\text{m}[\pu{J K-1 mol-1}] = \frac{S [\pu{J K-1}]}{n [\pu{mol}]}$. Or differentially, $S_\text{m} = \left(\frac{\partial S}{\partial n_i}\right)_{p,T,n_j,i \ne j}$ $\endgroup$
    – Poutnik
    Oct 11, 2023 at 8:16

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