Is really change in temperature inversely proportional to primary kinetic energy of ideal gas molecules?

Question

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.

Derivation:

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, $\Delta V = \frac {V_0\Delta E_k}{E_0}$ — (i)

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

So, $\Delta V \propto V_0\Delta t$ — (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.

Question:

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.

Answer 1

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.

Answer 2

$\dots$ 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.