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For a closed system at constant temperature and volume, the criterion for spontaneity is $dA < 0$. However, for a system at constant composition, the total differential of $A$ is given by $$dA = -PdV -SdT.$$ This seems to imply that $dA = 0$ for a system at constant volume, temperature, and composition. The Wikipedia page for Helmholtz Free Energy claims that this relationship holds even for non-spontaneous processes. The derivation of $A$ as criterion for spontaneity at constant volume and temperature is $$dA = dU - TdS - SdT = \delta q - TdS + \delta w - SdT.$$ If we consider only $PdV$ work, we obtain $$dA = \delta q - TdS - PdV - SdT$$ and at constant $T$ and $V$, we get $$dA = \delta q -TdS.$$ The criterion $dA <0$ for spontaneous processes then follows from the Clausius inequality. These two equations seem to be contradictory, so I was hoping someone could explain how this can be reconciled. Thank you so much!

Clarification: The way I derived the $dA = -PdV - SdT$ for an arbitrary process is as follows:

For a function $f(x_1,\dots,x_n)$, the complete differential of $f$ is $$df = f_1 dx_1 + f_2 dx_2 + \dots + f_n dx_n$$ where $$f_i = \left(\frac{\partial f}{\partial x_i}\right)_{x_j \: j\neq i}.$$ Consider the function $$g = f - f_1 x_1.$$ The differential of $g$ is $$dg = df - f_1 dx_1 -x_1df_1$$ so that $$dg = -x_1 df_1 + f_2 dx_2 + \dots + f_n dx_n.$$ Now, applying this to the special case of internal energy $U(S,V)$, if we define $$A = U - \left(\frac{\partial U}{\partial S}\right)_{V} S = U - TS$$ then the above implies that $$dA = -SdT - PdV$$ since $$dU = TdS - PdV.$$

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  • $\begingroup$ Consider review of search results for site:stackexchange.com OR site:libretexts.org thermodynamics closed system spontaneity criteria $\endgroup$
    – Poutnik
    Oct 10 at 20:57
  • $\begingroup$ Provide you reasoning what you think they are contradictory. as dS >= dq/T, dq - TdS <=0. 0 for reversible processes, <0 for spontaneous ones. $\endgroup$
    – Poutnik
    Oct 10 at 21:03
  • $\begingroup$ For a spontaneous process at constant temperature and volume, the fact that $dQ < TdS$ implies that $dA < 0$ with a strict inequality. However, the fact that $dA = - PdV -SdT$ seems to imply that for any constant volume and temperature (and constant composition), we have $dA = 0$ since $dV = 0$ and $dT = 0$. This is the main source of my confusion. $\endgroup$
    – Johnny Smith
    Oct 10 at 22:39
  • $\begingroup$ dA=-p.dV -SdT is derived with assumption dq=T.dS, i.e. for reversible, non spontaneous changes, so TdS and -TdS mutually cancel. $\endgroup$
    – Poutnik
    Oct 11 at 6:11
  • $\begingroup$ The link I provided to the Wikipedia article states that this relationship holds for non-reversible processes as well. That is where I am confused $\endgroup$
    – Johnny Smith
    Oct 11 at 10:37

1 Answer 1

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$$A = U -TS$$ $$\mathrm{d}A = \delta Q - p\mathrm{d}V + \delta W_\text{nonexp} - T\mathrm{d}S - S\mathrm{d}T$$ $$dS \ge \delta Q/T \implies T\mathrm{d}S \ge \delta Q $$ Assuming $\delta W_\text{nonexp}=0$: $$\mathrm{d}A = \delta Q - p\mathrm{d}V - T\mathrm{d}S - S\mathrm{d}T$$ As $T\mathrm{d}S \ge \delta Q$: $$\mathrm{d}A \le T\mathrm{d}S - p\mathrm{d}V - T\mathrm{d}S - S\mathrm{d}T$$ $$\mathrm{d}A \le - p\mathrm{d}V - S\mathrm{d}T$$

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  • $\begingroup$ This makes sense. Does the derivation of $dA = -PdV - SdT$ using Legendre transformations assume reversibility? I thought it holds in general $\endgroup$
    – Johnny Smith
    Oct 11 at 11:46
  • $\begingroup$ Because the statement that "the natural variables of A are V and T" seems to imply that A(V,T) is constant for any process at constant V and T so that dA = 0 $\endgroup$
    – Johnny Smith
    Oct 11 at 11:49
  • $\begingroup$ It uses the A definition, the Clausius inequality TdS>=dQ and ordinary mathematical operations. // AFAIK, the Leg. tr. was used and the source for decision of the way to define A and G as state functions. $\endgroup$
    – Poutnik
    Oct 11 at 11:54
  • $\begingroup$ Aren't all of those true in general? I don't see why that derivation is only valid for reversible processes $\endgroup$
    – Johnny Smith
    Oct 11 at 11:58
  • $\begingroup$ But I as a chemist am not familiar with Legendre transformations by the name, rather just Legendre orthogonal polynoms. $\endgroup$
    – Poutnik
    Oct 11 at 12:00

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