I started thermodynamics mostly through independent study and basically built up my own definitions of terms that appeared to fit with what was going on. They seemed to work but my question is whether or not this is how they are actually supposed to be viewed.

Equipartition theorem. 'It is possible to show that at equilibrium molecules share an equal amount of energy between multiple independent coordinates or degrees of freedom which fully describe their states, so long as the energy term is quadratic for each coordinate.' The simplest example is an ideal gas, whose energy is $1/2mv_x^2+1/2mv_y^2 + 1/2mv_z^2$ by pythagoras theorem. The $x,y,z$ terms are statistically the same because rotating the coordinates doesn't change pythagorean distance.

'Temperature is the rate of energy transfer away from a point due to particular collisions of molecules (not net transfer)'. To assign a number to this value, we calculate the work done in a given collision on a molecule. Collisions occur in one degree of freedom and the velocity of approach is statistically the same as separation. so for instance, on average for a ideal gas

$kT = 2 \times \frac{1}{2}mv_x^2$.

Where $k$ is Boltzmann constant. $\frac{1}{2}mv_x^2$ is roughly $E/a$ where E is the energy of a particle and $a$ is the number of degrees of freedom. I have been able to apply calculus to this to deduce lots of useful facts, such as that the heat capacity of an ideal gas is $\frac{3}{2M_r}kN_a$, at equilibrium $PV = NkT$ and if little heat energy is transferred between an equilibrium gas and its surroundings, $T V^{\frac{2}{a}} = constant$

'Entropy is the average number of degrees of freedom that contain energy, for a given molecule in a system'. It is

s = $\frac{a_{mean}}{n}$

where $n$ is the number of moles. This can change with temperature. For example, increasing the temperature of oxygen gas will 'free up' a greater proportion of molecules, causing $a_{mean}$ to increase. $s$ of the universe is always increasing, much like water spreads through an ice cube tray, energy spreads through existing empty coordinates

'Enthalpy change is the amount of energy transferred (per mole) from the surroundings to the system, measured at constant temperature and pressure'. When negative, it goes into freeing up the coordinates of surrounding molecules if the 'universe' remains at roughly the same temperature. If enthalpy change is zero and the entropy change of the system is positive, all the energy released in the reaction frees up coordinates within the system. We can therefore say $\Delta S_{surroundings} = - \frac{\Delta H}{T}$

'Gibbs free energy (per mole) is proportional (-ve) to the amount of energy which has gone into freeing new coordinates in the 'universe' before and after an event at constant temperature and pressure'. Dimensionally it can be seen as $Ts - T(s_{system} + s_{surroundings}) = E_i - E_f$ where $E_i$ and $E_f$ are initial and final heat energies of the world. This leads to $-T(\Delta s_{system} + \Delta s_{surroundings}) = G < 0$ so

$\Delta H-T\Delta s_{system} = G < 0$

for any feasible reaction. Not sure about this because it suggests that energy has entered the closed system

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    $\begingroup$ None of this works for me, but I wouldn't know were to begin explaining why. $\endgroup$ Commented May 1, 2018 at 19:05
  • $\begingroup$ Your definition of temperature does not match any of the standard ones. It doesn't work for gases that are essentially collisionless (like most dilute gases or all photon gases). $\endgroup$ Commented May 1, 2018 at 19:32
  • $\begingroup$ chester miller, start with temperature? $\endgroup$
    – user86425
    Commented May 1, 2018 at 22:01
  • $\begingroup$ physics.stackexchange.com/questions/337549/… So looking at some other questions I found this. Scroll to OrangeSherbet's answer, it does bear similarities with mine. I understand that there are probably more useful, intuitive and computable definitions of entropy but is there something fundamentally wrong here? $\endgroup$
    – user86425
    Commented May 1, 2018 at 22:07
  • 1
    $\begingroup$ I suppose you could improve it to the energy per degree of freedom via the equipartition theorem as mentioned in zutchens1's answer $\endgroup$
    – user86425
    Commented May 2, 2018 at 18:47

1 Answer 1


I would like to define these terms, and then you can compare how your definitions compare.


I can think of a dozen definitions of temperature. One of the definitions I particularly like is that temperature is the willingness of an object to give up energy to its surroundings (I have paraphrased this from Dan Schroeder). This provides a general definition of temperature regardless of the specific situation.

Temperature can also be defined in terms of the internal energy of the system (think, for example, the equipartition theorem, $E = \frac{1}{2}fNkT$). As you allude to in the case of an ideal, non-relativistic, non-degenerate gas, the temperature is related to the average speed, and thus the average energy of a molecule. In other words, if you input energy to the gas, its average energy and temperature increases, and it is more likely to spontaneously give up energy to its surroundings, because it wants to reach thermal equilibrium with said surroundings.


If you have been studying some basic statistical mechanics then you will be familiar with the idea of multiplicity. The multiplicity, $\Omega$, represents the total number of ways to arrange a system in terms of its microstates and macrostates. The simple case typically taught is the Einstein model of the elemental solid, which connects atoms via springs. The multiplicity determines the number of ways to distribute the energy quanta along its oscillators for a given macrostate, and in total for a given solid.

Entropy is simply a reformulation of multiplicity: $S = k\ln\Omega$, and for this reason, it is highly related to a variety of thermodynamic quantities including temperature and heat.

The second law of thermodynamics says that entropy, and thus multiplicity, tends to increase for the universe. If any process increases the entropy of the universe (the other alternative is to keep it constant), then that process is said to be irreversible, because it takes more energy to recover the initial state than it did to arrive at the current situation (e.g, you cannot recover energy losses due to friction in a piston).


We define the enthalpy of a system as

$$ H = U + PV $$.

Let's not complicate this. We know that $U$ is the internal energy of the system, and that $PV$ represents mechanical work done on/by the surrounding atmosphere. As such, the enthalpy tells you the energy you would need to both create a system, and push the atmosphere out of the way (to make room for your new system).

Gibbs free energy

Basically, on that last step, when we were creating our system, we did two things: we (a) gave the system some internal energy and (b) pushed the surroundings out of the way to make room. Turns out, we goofed because we totally forgot that the surroundings already have a temperature, and based on the discussion we already about temperature, we know that the surroundings might therefore spontaneously give up some energy to us.

Thus, when we are making our system, we still need $U + PV$ overall, but part of that of energy is going to come from the surroundings, and thus we have to provide $TS$ less of what we thought. You can think of this quantity $TS$ as heat. So, the actual energy, $G$, we have to provide is

$$ \Delta G = \Delta U + P\Delta V - T\Delta S = \Delta H - T\Delta S. $$

This is what we call the Gibbs free energy.

A dumb analogy

Gibbs free energy is basically like a tax return, if you take $\Delta H$ to be the tax you have to pay based on your income. At the end of the year, turns out you paid too much money, so the government gives you back an amount $T \Delta S$. So the total amount you paid the government was $\Delta H - T\Delta S$.

  • $\begingroup$ The equipartition states that the energy will be divided evenly among its quadratic degrees of freedom. In the case of a three-dimensional solid, each atom has three kinetic and three potential degrees of freedom, thus $f=6$ and $E = 3NkT$. The temperature only cares about the total internal energy. $\endgroup$
    – zh1
    Commented May 2, 2018 at 17:23
  • $\begingroup$ Why not simply define temperature as $T=(\partial U/\partial S)_V$, as that thing that's equalized between two systems when only heat transfer is allowed? You use the exact definitions for the other parameters (which I think is great). When the ideal gas is used to define temperature, though, people tend to incorrectly generalize the associated relationships to solids, resulting in misconceptions. $\endgroup$ Commented May 2, 2018 at 19:06
  • $\begingroup$ You are then defining temperature in the case of constant volume; I was trying to be general. $\endgroup$
    – zh1
    Commented May 2, 2018 at 19:32

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