I was trying to derive Charles law. While deriving I got two results and one of these was unexpected. Those are, $$\Delta t \propto \Delta E_k$$

Where, $\Delta t$ is change in temperature and $\Delta E_k$ is average change in kinetic energy of ideal gas molecules. Now it is obvious because temperature is directly proportional to average kinetic energy of the ideal gas molecules.

But the unexpected result is below.

$$\Delta t \propto \frac {1}{E_0}$$

Where, $E_0$ is average primary kinetic energy of ideal gas molecules.


Suppose, we have $n$ moles of ideal gas particles in a volume $V_0$. If there is a equal distribution of particles per volume, then the gas has $n/V_0$ moles of particles per unit volume. If the average kinetic energy of each particle is $E_0$, then the total kinetic energy of particles per unit volume is $\frac {n}{V_0}E_0$

Let's assume with pressure and amount of substance remaining constant, the gas' volume changes into $KV_0$. So, it has $n/KV_0$ moles of particles per unit volume. Now, the average kinetic energy of each particle changes because they have got change in volume by colliding with changed kinetic energy going far from each other or coming near. Let's suppose the average kinetic energy of each particle is $E_1$ and the total kinetic energy per unit volume is $\frac {n}{KV_0}E_1$.

Notice that, the total kinetic energy must be equal to apply a constant pressure on container walls because pressure is constant for both cases. So,

$\frac {n}{V}E_0 = \frac {n}{KV}E_1$

Or, $E_0 = \frac {E_1}{K}$

Or, $E_0 = \frac {V_0E_1}{V}$ [suppose, $K = \frac {V}{V_0}$]

Or, $V - V_0 = \frac {V_0(E_1 - E_0)}{E_0}$

Or, $\Delta V = \frac {V_0\Delta E_k}{E_0}$ β€” (i)

Now according to Charles law, $\Delta V = \frac {V_0\Delta t}{273}$

So, $\Delta V \propto V_0\Delta t$ β€” (ii)

Comparing eq. (i) and (ii),

$\Delta t \propto \frac {\Delta E_k}{E_0}$

$\therefore \Delta t \propto \frac {1}{E_0}$, when average change in kinetic energy of particles is constant.


Is the relation $\Delta t \propto \frac {1}{E_0}$ really true? What do actually make whether true or false at microscopic level?

I think it's true because the more kinetic energy a molecule has, the attraction between its atom's nucleus and electrons becomes lower. So, we need less energy to take electrons to a higher energy state and hence more kinetic energy of molecules. The reverse happens when primary kinetic energy is lower.

  • 2
    $\begingroup$ This is a somewhat convoluted derivation - I suppose you are not yet familiar with Boltzmann distribution or 1st law of thermodynamics? Some things are missing: the work done by the force changing the volume (which changes the kinetic energy) and that the pressure depends on the particle concentration (not only on their average kinetic energy.) Still, now it is a complete question, so I vote to reopen it. $\endgroup$
    – Roger V.
    Commented Jan 26, 2023 at 9:14
  • $\begingroup$ @RogerVadim I thought I included the fact of particle concentration in my question. Isn't it? $\endgroup$ Commented Jan 26, 2023 at 11:06
  • $\begingroup$ Yes, indeed. Still, I do not understand the problem: $E_0=\frac{V_0E_1}{V}$ is already Charle's law, if we keep in mind that average kinetic energy is the temperature. The rest is just manipulating this law. $\endgroup$
    – Roger V.
    Commented Jan 26, 2023 at 11:23
  • $\begingroup$ @RogerVadim I am taught that 𝘊𝘩𝘒𝘳𝘭𝘦𝘴' 𝘭𝘒𝘸 states 'At constant pressure for same amount of gas, its volume increases or decreases by $1/273$ of its volume at 0°C for 1°C temperature increase or decrease' $\endgroup$ Commented Jan 26, 2023 at 15:18
  • 1
    $\begingroup$ The usual definition is When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be in direct proportion. See Charles law. It doesn't contradict what you say, but you seem to be discussing consequences of this law (or its particular aspect), rather than the law itself. $\endgroup$
    – Roger V.
    Commented Jan 26, 2023 at 15:23

2 Answers 2


As the other answer already pointed out, what you actually derived is

$$ \frac{\Delta t}{t_0} = \frac{\Delta E}{E_0} $$

In other words, the relative increase in kinetic energy will lead to an identical relative increase in temperature (assuming the volume is held constant). This is completely the expected result given the fact that the temperature is by definition proportional to the kinetic energy.

You have confused yourself here by mixing up absolute and relative increases: of course, if you have the same increase $\Delta E$ for a higher initial energy $E_0$, then your relative increase $\Delta t / t_0$ will obviously be smaller.

NB.: you should quickly forget about the idea that the electron levels have anything to do with this. The kinetic energy of a gas molecule has only to do with its total mass and its velocity as a whole.

  • $\begingroup$ What a simple fact I just ignored! Thank you for ponting it out. $\endgroup$ Commented Feb 3, 2023 at 14:44

$\dots$ What do actually make it true at microscopic level?

For starters, let me first say that regardless of the correctness of your relation, it generally cannot hold at a microscopic level since temperature is an emergent physical quantity, which is hinted at already by your equation $\Delta t \propto \Delta E_k$. In other words, the very concept of temperature is a macroscopic one.

Is the relation $\Delta t \propto \frac{1}{E_0}$ really true? $\dots$

Well your relation isn't entirely correct. Why? Since $E_0$ is characteristic of the particles in question, it remains constant for a given Pressure, Volume $V_0$ and Temperature $t_0$. Thus it only changes if you change your particles or demand a different $t_0$ (equipartition theorem). What does it mean for a quantity to be proportional to a constant? It means that it is a constant. This is in contradiction to the first proportionality relation you have derived. In fact, what you actually derive is, $$ \frac{\Delta t}{t_0} = \frac{\Delta E_k}{E_0}. $$ However, you conveniently ignore $t_0 = 273 K$ when you state your relations since it is a constant. The same can be said for the primary energy $E_0$ as well, which would lead to $\Delta t \propto \Delta E_k$. Essentially what you end up with is a constant slope for a $t-E_k$ graph, where the constant is a characteristic of the temperature $t_0$ and primary kinetic energy $E_0$. In particular, we know from the equipartition theorem that, $t_0/E_0 = 2/(3k_B)$, where $k_B$ is the Boltzmann constant.

To make sense of this, we see that for a higher value of $E_0$ at a given non-zero temperature $t_0$, this slope becomes flatter. This is logically consistent with your first relation, since if your particles originally have extremely high energies, you wouldn't be able to observe an appreciable change in temperature. However, for lower energies, since $E_1 = KE_0$, the slope is steeper, and you get a well-defined change in temperature with respect to the change in energy.

  • $\begingroup$ I wanted something treating the temperature as heat energy at microscopic level and give an argument like my last paragraph. $\endgroup$ Commented Jan 28, 2023 at 13:59
  • $\begingroup$ @DebanjanBiswas I think you didn't grasp the crux of the explanation. The second relation you derive isn't entirely correct since what you claim is that $\Delta t$ is proportional to a constant $1/E_0$, which is equivalent to saying that $\Delta t$ is itself a constant. That's of course, inconsistent with your first proportionality relation which is in fact, correct. The only claim you can make is about the slope $\Delta t/ \Delta E_k$ which must be a constant, i.e, $t_0/E_0$ which is what you hint at when you say $\Delta t \propto \Delta E_k$. $\endgroup$ Commented Jan 28, 2023 at 15:16
  • $\begingroup$ @DebanjanBiswas morover, if I may add. $E_0$ is characteristic of the particles in question and remains constant for a given Pressure, Volume $V_0$ and Temperature $t_0$. Thus the only change occurs if you change your system or demand a different $t_0$ as a result of the equipartition theorem. $\endgroup$ Commented Jan 28, 2023 at 15:27
  • $\begingroup$ Yes. If you need different average primary kinetic energy of particles, you need to heat it from different temperatures or you need different systems. And for these the law becomes true according to you. But it doesn't answer my question as I asked for the microscopic phenomenon treating temperature as heat energy. $\endgroup$ Commented Jan 28, 2023 at 16:46
  • $\begingroup$ @DebanjanBiswas I believe the question was (as you posed it) whether or not your expression was correct and if yes then how to explain it in those terms. The point is, it is not correct. Moreover, following up on what I mentioned, the $t_0/E_0$ constant is a characteristic of the system given $P$ and $V$ remain unchanged. By the equipartition theorem, this is typically $2/(3k_B)$ where $k_B$ is the Boltzmann constant. $\endgroup$ Commented Jan 28, 2023 at 16:58

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