How does entropy change with pressure?

Feb 11, 2016

Starting from the first law of thermodynamics and the relationship of enthalpy $H$ to internal energy $U$:

$\setminus m a t h b f \left(\Delta U = {q}_{\text{rev" + w_"rev}}\right)$

$\setminus m a t h b f \left(\Delta H = \Delta U + \Delta \left(P V\right)\right) = {q}_{\text{rev" + w_"rev}} + \Delta \left(P V\right)$

where:

• ${q}_{\text{rev}}$ and ${w}_{\text{rev}}$ are the most efficient (reversible) heat flow and work, respectively. ${w}_{\text{rev}} = - P \mathrm{dV}$.
• $P$ is pressure and $V$ is volume.

THE ENTHALPY MAXWELL RELATION

Using the differential form, we get:

$\mathrm{dH} = \partial {q}_{\text{rev" + delw_"rev}} + d \left(P V\right)$

Another relationship that relates with heat flow is the one for entropy and reversible heat flow:

$\setminus m a t h b f \left(\mathrm{dS} = \frac{\partial {q}_{\text{rev}}}{T}\right)$

Thus, utilizing this relationship and invoking the Product Rule on $d \left(P V\right)$, we get:

$\mathrm{dH} = T \mathrm{dS} - \cancel{P \mathrm{dV} + P \mathrm{dV}} + V \mathrm{dP}$

$\textcolor{b l u e}{\mathrm{dH} = T \mathrm{dS} + V \mathrm{dP}}$

which is what you would get for the Maxwell relation.

ENTHALPY VS ENTROPY

When we relate pressure then, to entropy, with $S = S \left(T , P\right)$:

$\mathrm{dS} = \frac{\mathrm{dH}}{T} - \frac{V}{T} \mathrm{dP}$

For an ideal monatomic gas, $P V = n R T$, so:

$\mathrm{dS} = \frac{\mathrm{dH}}{T} - \frac{n R}{P} \mathrm{dP}$

Finally, when we integrate this, we get:

${\int}_{{S}_{1}}^{{S}_{2}} \mathrm{dS} = {\int}_{{H}_{1}}^{{H}_{2}} \frac{\mathrm{dH}}{T} - n R {\int}_{{P}_{1}}^{{P}_{2}} \frac{1}{P} \mathrm{dP}$

Enthalpy is $\Delta H = {\int}_{{H}_{1}}^{{H}_{2}} \mathrm{dH} = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{P} \mathrm{dT}$, where ${C}_{P}$ is the heat capacity at constant pressure in $\text{J/K}$. Thus:

$\implies {\int}_{{T}_{1}}^{{T}_{2}} {C}_{P} / T \mathrm{dT} - n R {\int}_{{P}_{1}}^{{P}_{2}} \frac{1}{P} \mathrm{dP}$

Since ${C}_{P}$ for a monatomic ideal gas is ${C}_{V} + n R = \frac{3}{2} n R + n R = \frac{5}{2} n R$, with ${C}_{V}$ as the constant-volume heat capacity and the $\frac{3}{2}$ coming from the three linear degrees of freedom ($x , y , z$), this becomes:

$\textcolor{b l u e}{\Delta {S}_{\text{sys}} = \frac{5}{2} n R \ln | {T}_{2} / {T}_{1} | - n R \ln | \frac{{P}_{2}}{{P}_{1}} |}$

PRESSURE VS ENTROPY

Therefore, if pressure increases, a negative contribution is made on the change in entropy of an ideal gas, but depending on the change in temperature, the actual change in entropy for the system might be positive or negative.

(Regardless, the entropy of the universe is $\ge 0$.)

On the other hand, the change in volume of a liquid is appreciably low upon small increases in pressure that should substantially compress a gas, so the change in pressure of a liquid makes a smaller negative contribution to the change in entropy.

With solids, I would not expect pressure to significantly alter any entropy patterns they already have for small values for pressure that would otherwise be significant for gases.

For di/polyatomicsolids, we can consider either the complexity or the bond strength, as it relates to the number of "ways" it can exist. One of the most popular thermodynamics equations is:

$\setminus m a t h b f \left(\Delta S = {k}_{b} \ln \Omega\right)$

where:

• ${k}_{b}$ is the Boltzmann constant, $1.38 \times {10}^{-} 23$ $\text{J/K}$.
• $\Omega$ is the number of microstates consistent with a chosen macrostate, which is proportional to the number of observable "snapshots" of molecular motion.

The stronger the bond, the lower the magnitude of the entropy because the lower the number of microstates available to the solid.

We can see that here with some data.

Increasing charge magnitudes:

DeltaS_("NaF"(s))^@ = "51.46 J/mol"cdot"K"

DeltaS_("MgO"(s))^@ = "26.9 J/mol"cdot"K"

DeltaS_("AlN"(s))^@ = "20.2 J/mol"cdot"K"

Naturally the increase in charge magnitudes correlates with an increase in bond order or bond strength.

DeltaS_("NaCl"(s))^@ = "72.13 J/mol"cdot"K"

DeltaS_("NaBr"(s))^@ = "86.82 J/mol"cdot"K"

DeltaS_("NaI"(s))^@ = "98.53 J/mol"cdot"K"

Generally a greater difference in radii between cation/anion corresponds to a greater internuclear distance and thus a weaker bond.

And since I recall carbonates decomposing at higher temperatures (which I found weird when I first learned it), it doesn't hurt to check alkaline earth metal carbonates going down the periodic table (wikipedia).

Increasing distance in alkaline-earth-metal/carbon $n s \text{/} 2 s$ orbital ground-state energies (increasing the number of nonbonding orbitals):

DeltaS_("BeCO"_3(s))^@ = "52 J/mol"cdot"K"

DeltaS_("MgCO"_3(s))^@ = "65.7 J/mol"cdot"K"

DeltaS_("CaCO"_3(s))^@ = "93 J/mol"cdot"K"

And for something more complex like buckminsterfullerene, the entropy is significantly higher because it can assume more microstates for a given macrostate (more molecular motions possible, more vibrational modes, etc).