19
$\begingroup$

Today in my thermodynamics lecture, my teacher told me that it is not possible to find the absolute value of internal energy and so we have to calculate the change in internal energy.

So, my question is why isn't it possible to find the absolute value of internal energy? Is this statement true?

$\endgroup$
2
  • 2
    $\begingroup$ How would you find the internal energy of hydrogen gas? Even when it is cooled way down (to absolute zero), it still contains energy (the atoms never stop vibrating). There is no known way to determine the internal energy. $\endgroup$
    – LDC3
    Nov 15 '14 at 14:39
  • $\begingroup$ The internal energy of an ideal gas is the ensemble average determined by statistical mechanics and is $U=(3/2)NkT$, or in general $U=(N/Z)\Sigma_j\exp(-\epsilon_j/(kT))$ where Z is the partition function, k Boltzmann const, $\epsilon_j$ energy of level j, and T temperature. $\endgroup$
    – porphyrin
    Mar 23 '17 at 15:54
13
$\begingroup$

In order to answer this question, one needs to define what the absolute energy of a system is. Energy can be trapped in a system in ways not yet discovered or fully understood. Think of the energy associated with mass ($E=mc^2$), which is a result that was not known to the early founders of thermodynamics.

We need to define reference points, which we can use to compare our products with. Usually this "zero" point is defined as the energy of the individual atoms in their most energetically favorable state (phosphorus is an exception, since red phosphorus is much more available), such as $\ce{H2}$ and $\ce{He}$.

$\endgroup$
8
$\begingroup$

I also heard that and it's not exactly true, you could find total value using mentioned earlier $E=mc^2$, but it shouldn't be called absolute, as mass-energy depends on frame of reference - masses measured between inertial and non-inertial frames of reference are different. Also, as pointed in comments, you can't account for zero point energy of vacuum this way, although I wouldn't call it a problem as this energy is the same when the object of measurement is absent.

Still if you weighed something and multiplied this mass by $c^2$ you would get its total energy. However it would be a very rough estimate, one gram is equivalent to about 90 terajoules of energy, which is really large amount. Even if you would measure mass with great precision, and mass would be very small, it would still be a rough estimate.

$\endgroup$
12
  • 1
    $\begingroup$ @CurtF. Technically, "frame of reference" refers to your spacetime coordinate system, not the geometry of a particular object. Moreover, to the extent that general relativity can be ignored, the absolute internal energy measured at a given reference frame is indeed well-defined. Here, the absolute reference state of zero energy is the vacuum, and, for example, an electron with zero momentum has an absolute energy of $m_e c^2 = 511\mathrm{keV}$. $\endgroup$
    – higgsss
    Jan 5 '16 at 13:29
  • 7
    $\begingroup$ The absolute energy of a $H_2$ molecule of zero momentum is given by $E_{\mathrm{H}_2}=2m_e c^2 + 2 m_p c^2−2×(\mbox{binding energy of an H atom})−(\mbox{H-H bonding energy})$. This value divided by $c^2$ is the mass of this molecule, but notice that among the four terms, the last two is negligibly small compared to the first two. On the other hand, only the last two terms are of interest in chemistry. $\endgroup$
    – higgsss
    Jan 5 '16 at 13:31
  • 2
    $\begingroup$ @CurtF. Relative to the vacuum state. Quark confinement means that it takes infinite energy to free the quarks from, e.g., a proton, while a proton has a finite energy ($m_p c^2 = 938 \mathrm{MeV}$) relative to the vacuum. $\endgroup$
    – higgsss
    Jan 5 '16 at 13:38
  • 2
    $\begingroup$ I think we disagree about the meaning of "absolute". "Relative to the vacuum state" still sounds like you are defining energies relative to a reference state to me, but I can see how others might disagree. $\endgroup$
    – Curt F.
    Jan 5 '16 at 13:40
  • 3
    $\begingroup$ @CurtF. As you point out, there should always be a reference state. But nothing can have a lower energy than vacuum, and "the energy relative to the vacuum state" has an absolute sense. $\endgroup$
    – higgsss
    Jan 5 '16 at 13:46
7
$\begingroup$

In classical thermodynamics the change in internal energy is defined by the first law as $$\Delta U = q + w$$ so that only the difference in $U$ is known; $q$ is the heat absorbed by the 'system' and $w$ the work done on the system.

For example in a closed system (no exchange of matter with environment) we can write for a reversible change \begin{align} \mathrm{d}q &= T\mathrm{d}S \\ \mathrm{d}w &= -p\mathrm{d}V \end{align} and then if the only form of work on a gas is volume change $$ \mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V$$ and this is the fundamental equation for a closed system. Thus only difference in internal energy are measurable from thermodynamics, and this follows from the first law. (Even if you integrate this equation from say state $a$ to $b$ the result will be $U(b)-U(a)=\ddot {}$ in other words $\Delta U$.)

Thermodynamics was developed before the nature of matter was known, i.e. it does not depends on matter being formed of atoms and molecules. However, if we use additional knowledge about the nature of molecules then the internal energy (and entropy) can be determined from statistical mechanics.

The internal energy ($U$ not $\Delta U$) of a perfect monoatomic gas is the ensemble average and is $$U=(3/2)NkT$$ or in general $U=(N/Z)\Sigma_ j\exp(−\epsilon_j/(kT))$ where $Z$ is the partition function, $k$ Boltzmann constant, $\epsilon_j$ energy of level $j$, and $T$ temperature and $N$ Avogadro's number. The absolute value of the entropy $S$ (for a perfect monoatomic gas) can also be determined and is given by the Sakur-Tetrode equation.

$\endgroup$
-1
$\begingroup$

The statement that it is impossible to calculate the absolute energy at an ambient T seems to be intuitively incorrect, difficult possibly, but not impossible. Since U is a state function the value of U for a given set of conditions is invariable and can be set as an arbitrary zero and changes in U calculated. So all one must do is work in reverse to absolute zero. This is done for entropies using the 3rd Law. The general approach in thermodynamics is that it is entirely accurate to work with the changes in state functions and there is NO NEED to calculate absolute internal energies from absolute zero.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.