# Chemistry - Is not possible to find absolute value of internal energy?

## Solution 1:

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 $\ce{He}$.

## Solution 2:

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.

## Solution 3:

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.