I'm learning AP Thermodynamics and the textbook I'm learning from states that the properties of an ideal gas are independent of the type of gas but the kinetic theory of gases states that the temperature of the gas and the average translational kinetic energy of a molecule are related by the formula:

$$ \frac{1}{2}m\langle v^2\rangle = \frac{3}{2}kT $$

If temperature is defined to be the average kinetic energy of the molecules inside a gas, surely molecules with greater mass would have more kinetic energy for a given velocity and therefore increase the temperature of the gas via this equation and hence the temperature would depend on the type of gas?

Wouldn't the pressure of the gas also be dependent on the type of gas since heavier molecules would have more momentum for a given velocity when they strike the walls of the container and therefore exert a greater force per unit area?

  • $\begingroup$ Most answers here are right. Heavy molecules will move slower carrying the same momentum as lighter molecules moving faster. I will just like to add that even if that is the case, both with the same ideal gas law, that does not mean that there are no thermodynamical consequences. A gas with $N$ identical molecules does not have the same entropy as a gas with $N/2$ molcules of type $A$ and $N/2$ of type $B$, thus some work can be extracted going from 1 to the other. $\endgroup$
    – Mauricio
    Apr 7 at 18:53
  • $\begingroup$ Temperature is also stored in other modes of energy like vibrational energy, not just translational kinetic energy. Complex molecules like water have higher modes of vibrational energy. $\endgroup$
    – mcodesmart
    Apr 12 at 16:25

5 Answers 5


Here is a way to think about it: put two types of ideal gases in the same box and say that initially all molecules $A$ have higher kinetic energy that molecules $B$, either because they have large mass, larger velocity, or both. As a result of molecular collisions energy will transfer from $A$ to $B$. It turns out that in equilibrium both types of molecules will have the same kinetic energy on average: $$ \frac{1}{2} m_A \langle v_A^2\rangle = \frac{1}{2} m_B \langle v_B^2\rangle $$ The proof is given in statistical mechanics but the basic idea is that kinetic energy is transferred from high-energy molecules to low-energy molecules. As a result in equilibrium we have the same kinetic energy on average in every molecule regardless of its mass.

  • 2
    $\begingroup$ "It turns out that in equilibrium both types of molecules will have the same kinetic energy on average" — that is actually the non-trivial result of the equipartition theorem. In thermal equilibrium, temperatures are equal. The equipartition theorem states that quadratic terms in the Hamiltonian contribute the same to to the full average energy (namely $\frac12k_{\rm B}T$). Luckily, the kinetic energy is quadratic (e.g. $p^2/2m$), so equipartition for simple molecules implies "equilibrium" = "equal temperature" = "equal average kinetic energy". $\endgroup$
    – printf
    Apr 7 at 23:43

You are correct at saying that for a given velocity gases with different molecule masses have different pressure and temperature. That just means, that if you have two gases with different molecular masses at equilibrium with each other, they have different average speed of the molecules. At the same temperature the molecules of a heavier gas move more slowly. The sentence "the properties of an ideal gas are independent of the type of gas" does not mean that under the same conditions the microscopic properties (like average speed of the molecules) are the same for all gases. It means that the macroscopic properties are governed by the same laws independent of the gas type.

In the kinetic theory of gases it is shown that the pressure of the ideal gas is proportional to the average kinetic energy, because during a unit time the number of molecules colliding with a unit area of a wall is proportional to $v$, and each molecule carries a momentum proportional to $mv$, so the product is proportional to $m\langle v^2\rangle$. Thus, both pressure $P$ and the temperature $T$ are proportional to the average kinetic energy, i.e. they are proportional to each other, $P\propto T$. It can be shown that the proportionality coefficient is independent of the type of the gas, so you basically get from this the ideal gas law, $P = A\cdot T$, where the proportionality coefficient is $A = R\cdot n/V$. Here $n$ is a number of moles and $V$ is a volume of the gas.


The Maxwell(-Boltzmann) distribution is given by: $$ W(v_x,v_y,v_z)=Z^{-1}e^{-\frac{m(v_x^2+v_y^2+v_z^2)}{2k_BT}}, Z=\iiint_{-\infty}^{+\infty}\text{d}v_x\text{d}v_y\text{d}v_ze^{-\frac{m(v_x^2+v_y^2+v_z^2)}{2k_BT}} $$ From this, one can easily obtain that (see Gaussian integral): $$\langle v_x^2\rangle=\langle v_y^2\rangle=\langle v_z^2\rangle=\frac{k_BT}{m}.$$ That is, molecules of higher mass have lower average speeds. The above relation can be recast as: $$\frac{m\langle v_x^2\rangle}{2}=\frac{m\langle v_y^2\rangle}{2}=\frac{m\langle v_z^2\rangle}{2}=\frac{k_BT}{2}.$$


I'm learning AP Thermodynamics and the textbook I'm learning from states that the properties of an ideal gas are independent of the type of gas.

That's a poor characterization. It is not the individual thermodynamic properties of an ideal gas that are independent of the type of gas, but rather the relationships among those properties.*

So yes, at a given average squared velocity, the temperature of a gas is proportional to its molecular mass. And at a given volume, the pressure of an ideal gas is also proportional to its molecular mass. This is in no way inconsistent with the ideal gas law, $$P V = n R T$$ In fact, the ideal gas law would predict that any factor that increases the temperature of an ideal gas at constant volume would also increase the pressure proportionally.

* There's a bit more to being an ideal gas than just that, but that's for another answer.


You are correct that the kinetic theory of gases relates temperature to the average kinetic energy of gas molecules. However, the key to understanding the ideal gas law and the properties of an ideal gas is recognizing that ideal gases are a simplified, theoretical model that does not perfectly represent real gases.

The ideal gas law (PV = nRT) treats all gases as if they are composed of point particles that only interact through perfectly elastic collisions. It assumes that the particles have no volume and do not exert any attractive or repulsive forces on one another. Because of these simplifications, the ideal gas law does not take into account differences in molecular mass or other specific properties of the gas particles.

In reality, as you pointed out, the mass of gas molecules does affect their kinetic energy and the pressure they exert. Real gases deviate from ideal behavior, particularly at high pressures and low temperatures where intermolecular forces and the size of the particles become significant.

In many cases, however, the ideal gas law provides a useful approximation for real gases because, under a wide range of temperatures and pressures, the behavior of many gases is relatively independent of the specific properties of the molecules. This is particularly true for monatomic gases (e.g., helium, neon, argon) and some diatomic gases (e.g., nitrogen, oxygen) when they are not too dense or too cold.

When dealing with real gases that deviate significantly from ideal behavior, scientists use more accurate models, such as the van der Waals equation or other equations of state that take into account molecular size and intermolecular forces. These models provide a more accurate representation of the behavior of real gases, including the dependencies on molecular mass and other specific properties of the gas particles.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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