Suppose there's a 10 liter perfectly-insulated sealed tank of saturated liquid nitrogen at 4 bar. Therefore the fluid has properties,

$\rho=740 \,\frac{\text{kg}}{\text{m}^3}$, $m=7.4 \,\text{kg}$, $\hat{u}=-93\,\frac{\text{kJ}}{\text{kg}}$, and $U=-690\,\text{kJ}$.

If you drain all the liquid from the tank, then eventually in the tank you'll have 10 liters of saturated nitrogen vapor at atmospheric pressure with properties,

$\rho=4.6 \,\frac{\text{kg}}{\text{m}^3}$, $m=0.046 \,\text{kg}$, $\hat{u}=55\,\frac{\text{kJ}}{\text{kg}}$, and $U=2.5\,\text{kJ}$.

So the total internal energy has increased, despite removing most of the mass! Have I made an error in my calculations, or is this correct? Is there an intuitive way to think about this phenomenon?


Internal energy is a measure of the distribution of translation, rotation, and vibration in the molecules. A liquid has far less in translational and rotational energy per molecule.

As proof, consider that

$$\tilde{U}_{vap} - \tilde{U}_{liq} = \Delta_{vap}\tilde{U} = \Delta_{vap}\tilde{H} - \Delta_{vap}(p\tilde{V})$$

For an ideal gas, $\Delta_{vap}(p\tilde{V}) = RT_{vap}/M$ (SI units J/kg). Since vaporization is endothermic, the first term is positive. The term $RT_{vap}$ is typically smaller. For nitrogen with $T_{vap} \approx 77$ K and $\Delta_{vap}\tilde{H} \approx 6$ kJ/mol (NIST Webook), one has this in units of J/mol.

$$\Delta_{vap}\bar{U} \approx 6000 - 8.3(77) \approx 5400$$

You have to provide this energy to vaporize the liquid. You loose this internal energy when you condense the vapor.

To compare with what you have, take only the gas and condense it to liquid. The internal energy will be less by 9.3 kJ just for the 46 g as below.

$$\Delta_{cond}{U} \approx (46/28)(-5.4) = -9.3$$

This is for a transformation entirely at 1 bar pressure. Changing pressure on an ideal gas does not change its internal energy. It does change however change enthalpy, so the $\Delta(pV)$ term has to be done differently. Basically, you have to do the equivalent of vaporization at 4 bar and then an expansion to 1 bar. The specific difference in your case between liquid at 4 bar and gas at 1 bar is about -4.1 kJ/mol. Without deeper consultation, I might suspect the difference is from the loss at the expansion of the gas since, for real gases or liquids, $U(V,T)$, meaning ultimately also that $\tilde{U}(p,T)$.

So at the end, we have no surprise to find that a liquid has a lower internal energy than a gas even though the liquid has a higher mass than the gas. The difference has no reference to how the tank was emptied, since internal energy is a state property/function.


Starting with Liquid Nitrogen at: $$ \rho = 740 \frac{kg}{m^3}, m = 7.4 kg, \hat{u}= -93 \frac{kJ}{kg}, P = 4 bar, U=-690 kJ $$ Two things: 1) You can't drain all the liquid from the tank because you've said that $Q = \frac{m_{liquid}}{m_{vapor}}=0 $ (saturated liquid)

2) If the specific internal energy of the nitrogen is $\hat{u} = -93 \frac{kJ}{kg}$, then the specific internal energy of the liquid that you drain from the tank will be $\hat{u} = -93 \frac{kJ}{kg}$. If you drain 7.54 kg of liquid from the tank, as you did before, you'll reach this thermodynamic state: $$ \rho = 4.6 \frac{kg}{m^3}, m = 0.046 kg, \hat{u}= -93 \frac{kJ}{kg}, P = 0.23 bar, U=-4.27 kJ $$ So the pressure is actually lower, but the specific internal energy is the same. I looked this up using density and specific internal energy. The problem is that this still looks like you've added internal energy to the tank, because it's a smaller negative number. Internal energy is given on a relative scale. As Jeffrey J Weimer pointed out, the absolute internal energy of the fluid is the sum of kinetic and potential energies of all the fluid particles, which can never be negative. Maybe the simplest answer is that you certainly removed energy from the tank by removing mass, but engineering thermodynamics was mostly invented before the atomic model, or the idea of absolute zero temperature, or so many important models in physics. Really, engineers don't use internal energy, they use the change in internal energy: $$ \delta U = \delta W + \delta Q$$ Which is saying that you can increase the internal energy of the fluid by adding heat to the fluid, or doing work on the fluid, but those are your only choices. https://en.wikipedia.org/wiki/Internal_energy

  • $\begingroup$ The first law equation for engineers is stated as $\Delta U = q - w$ (the Clausis model) or $dU = \delta q - \delta w$. Work is taken as positive when done by the system. The Kelvin model uses $dU = \delta q + \delta w$ with work done by the system as negative. And engineers do calculate absolutes in internal energy (or entropy or enthalpy or ...) even without knowing what they really mean fundamentally. $\endgroup$ – Jeffrey J Weimer Feb 13 '19 at 1:01

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