I'm familiar with the ideal gas law $$PV=nRT$$ but I don't think it applies to liquids like water. If I'm wrong, please correct me! If I'm right, then what equation of state applies to liquids such as water?
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5 Answers
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Supercritical fluids are well described by real and ideal gas laws.
A common equation of state for both liquids and solids is
$$V_m = C_1 + C_2 T + C_3 T^2 - C_4 p - C_5 p T$$
where $V_m$ is molar volume, $T$ is temperature, $p$ is pressure, and $C_1$, $C_2$, $C_3$, $C_4$, and $C_5$ are empirical constants, all positive and specific to each substance.
The Peng & Robinson equation of state has been found to be useful for both liquids and real gasses:
$$p = \frac{RT}{V_m - b} - \frac{a(T)}{V_m (V_m + b) + b (V_m - b)}$$
where $a$ and $b$ are empirical constants and $R$ the ideal gas constant.
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$\begingroup$ Is there a name for your first equation for $V_m$? $\endgroup$– PaulCommented Oct 13, 2012 at 13:10
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$\begingroup$ $V_m$ stands for molar volume. That is, the actual volume divided by the number of moles. $\endgroup$ Commented Oct 14, 2012 at 1:34
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For all intents and purposes, you can use an incompressible equation of state:
$$ V = constant $$
That's it. No matter what pressure and temperature, you have the same volume. It's not completely true, but in relation to gasses it is true enough to make it that pressure work is negligible in liquids compared to gasses, and for liquids, you can just deal with the heat content without considering any work done in the expansions and contractions required to change temperature.
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2$\begingroup$ what about change in volume with temprature? $\endgroup$ Commented Oct 13, 2012 at 12:26
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4$\begingroup$ @Prathyush: no change in volume with either temperature or pressure--- this is the approximation which works for 99% of heat engines. You only have non-negligible volume changes in gasses. The reason is that liquids have touching atoms, and don't compress or decompress well. They do compress a little bit, but this is not a significant amount of work during heating and cooling cycles, or pressure increase/decrease cycles. It's not like in gasses, where the work and the heat are always comparable. $\endgroup$ Commented Oct 13, 2012 at 15:28
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2$\begingroup$ @Paul: The equation of state is what I said V=constant, no dependence of V on P or T. So for this description, you need a Gibbs ensemble, P and T are independently varied and V is constant. If you use another ensemble and want to take the constant V limit, you have to give V a tiny dependence on pressure/temperature. $\endgroup$ Commented Oct 13, 2012 at 15:30
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3$\begingroup$ @RonMaimon: Ok... I think I understand now... If the fluid is incompressible, then pressure and temperature do not affect the volume occupied by a constant quanitity of mass. Compressibility necessarily implies that pressure and temperature affect the volume occupied by a constant quanitity of mass. Am I stating this correctly? $\endgroup$– PaulCommented Oct 13, 2012 at 16:04
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2$\begingroup$ @Paul: Yeah, you got it. The incompressible approximation is good enough for most day-to-day fluids, away from the critical point. $\endgroup$ Commented Oct 13, 2012 at 21:14
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Accurate equations of states (EOS) for real gases, liquids, or solids are (in contrast to nice theoretical models such as an ideal gas or debye solid) quite complex, and must be fitted to experimental data.
For example, a very accurate equation of state for water and steam can be found in
Wagner and Pruss,
The IAPWS Formulation 1995 for the Thermodynamic Properties of Ordinary Water Substance for General and Scientific Use, 1995
http://www.teos-10.org/pubs/Wagner_and_Pruss_2002.pdf
But the water EOS is not very intelligible, except for a computer.
answered Oct 15, 2012 at 12:11
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To add just a bit to Ron Maimon's answer, the fact is that we do not have anything like a general equation of state for liquids. They are simply too complex for present day techniques.
We can do a bit better for solids since solids have an idealized crystalline form. But even there solids are very complex.
But don't fret. The situation isn't too much better for real gases, especially around the critical point.
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There is one you can derive for pressure and for density.
Start with:
$$\chi_T=\frac{1}{\rho}\left(\frac{\partial\rho}{\partial P}\right)_T$$
Thus by integration between $\rho_0$ and $\rho$ in one side and by $P_0$ and $P$ in the other side you'll get $$\rho=\rho_0 \exp\left(\chi_T\left(P-P_0\right)\right)$$
On a liquid column you can use it in: $$\frac{dP}{dz}=-\rho g$$
To get: $$P=P_0 - \frac{1}{\chi_T}\ln\left(1-\chi_T\rho_0g\Delta z\right)$$
Under many circumstances though, this boils down to: $$P=P_0 + \rho_0 g \Delta z$$
Because $\chi_T$ is small enough to make $\rho_0 g \Delta z$ neglectable. For instance, for water between $0 °C$ and $30 °C$, $\chi_T \approx 10^{-10}$ and even in Mariana Trench $\rho_0 g \Delta z \approx 10^3\cdot 10 \cdot 11 \cdot 10^3 \approx 10^8$ and $\ln(1-x) \approx x$ for small values of $x$ and you get the previous result. So for liquid water even in those conditions, up to $1100$ bar, you can just use the basic formula we all learn in high-school. How cool! :)
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$\begingroup$ This is not an answer to the OP question. The question is about the equation of state of a liquid, not about the gas/liquid column in a gravitational field. $\endgroup$– QuilloCommented Jul 21, 2023 at 20:27
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$\begingroup$ @Quillo it does. Predicting density is part of using EOS. From that molar volume can be derived and many other properties. The liquid column is just an example of what you can do. It's an equation of state for liquid. A simple one, for sure, so what? Knowing the pressure at the bottom of a column bioreactor is of importance. Of course this would be nitpicky to consider, yet deriving pressure is of importance as of it is for gases. $\endgroup$– ParaH2Commented Jul 22, 2023 at 22:28
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$\begingroup$ The EOS of an isotropic fluid is typically something like $P=P(\rho ,T)$ or some equivalent thermodynamic expression. Where is the EOS in your answer? The only expression you have for the pressure depends in the coordinate $ z$ (i.e. it is not an EOS). $\endgroup$– QuilloCommented Jul 22, 2023 at 23:27
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$\begingroup$ @Quillo the expression of $\rho$ is one EOS. $P_0$ would be the reference point. Also "typically something like" does not mean that's what we want all the time. I used this EOS in an example with the water column. If $\chi_T$ was much bigger, it would have an higher influence on pressure down the column. Something you couldn't guess without the simple EOS I derived. This makes it slightly better than $V=constant$. Could derive another one from $$\chi_T=-\frac{1}{V}\cdot \left(\frac{\partial V}{\partial P}\right)_T$$. $\endgroup$– ParaH2Commented Jul 23, 2023 at 15:18
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$\begingroup$ I am sorry but I disagree, you seem confused about what an equation of state is. en.wikipedia.org/wiki/Equation_of_state $\endgroup$– QuilloCommented Jul 23, 2023 at 15:59