# Is internal energy equal to (3/2)nRT for an ideal monoatomic gas?

I came across a Khan Academy video providing the following relation for the internal energy of the system $$U$$:

$$U=\frac32nRT,\label{eqn:1}\tag{1}$$

where $$n$$ is the amount of gas, $$R$$ is the universal gas constant and $$T$$ is the temperature of the system.

I know internal energy is a state variable and only its change can be measured, not the value at a particular state. Also, isn't \eqref{eqn:1} given for the average kinetic energy

$$\bar{E_\mathrm{k}} = \frac 32nRT,\tag{2}$$

not for the internal energy $$U$$? I'm confused.

• Internal Energy is (3/2)nRT plus some constant that cannot be measured. Commented Feb 13, 2018 at 17:06
• @IvanNeretin That constant can be approximated to be zero is it? Because in high school physics we just use 3/2 nRT. Commented May 7, 2019 at 13:45
• "Approximated" is a wrong word. Let's say it can be assumed to be 0. Commented May 7, 2019 at 13:46

Molecules have internal energy due to intermolecular interactions, as well as translational kinetic energy, rotational energy, vibrational energy, electronic energy (and if you care to include them, nuclear energy and the mass-energy of the protons, neutrons and electrons themselves).

By saying "ideal", energy due to intermolecular interactions is eliminated.

By saying "monoatomic", energy due to rotation and vibration is eliminated.

So only translational kinetic energy and electronic energy remains.

(3/2)nRT is the translational kinetic energy, and since almost all atoms are in the ground electronic state at low temperature, it is a good expression for internal energy as long as the temperature is low enough that essentially all atoms are in the electronic ground state.

From a statistical standpoint, the mean energy of a system is given by

$$\langle E \rangle = E \cdot P(x) = \frac{\int\limits_{-\infty}^{+\infty} E \mathrm e^{- \beta E}}{\int\limits_{-\infty}^{+\infty}\mathrm e^{-\beta E}},\tag{1}$$

where $$\beta = 1/(k_\mathrm{B}T)$$ and $$P(x)$$ is the probability of the system being at a particular energy $$E$$. Now, if your energy dependence is quadric in some variable, this is if $$E=ax^2$$, where $$a$$ is just some constant, the mean energy becomes (thanks to some pretty cool math and Gaussian integrals)

$$\langle E \rangle = \frac{\int\limits_{-\infty}^{+\infty} {ax^2\mathrm e^{- \beta ax^2}}}{\int\limits_{-\infty}^{+\infty}\mathrm e^{- \beta ax^2}} = \frac{1}{2 \beta}= \frac{1}{2}k_\mathrm{B}T.\tag{2}$$

You should note that this result is independent of $$x$$ and $$a$$. Actually, it is easy to show that if instead of one variable, the energy depends on $$n$$ quadratic variables, often called the modes of the system, each mode contributes the same amount of energy $$k_\mathrm{B}T/2$$ to the system (all you have to do is repeat the calculations for $$\left.E = \sum_{i=0}^n {a_i}x_i^2\right)$$ — this is known as the equipartition theorem. Hence, the mean energy of a system with $$n$$ quadratic modes becomes

$$\langle E \rangle = \frac{1}{2}nk_\mathrm{B}T.\tag{3}$$

In an ideal monoatomic gas the internal energy of the system $$U$$ is just its kinetic energy $$E_\mathrm{k}$$ (there is no energy due to vibration, rotation and intermolecular interactions), and therefore

$$E=E_\mathrm{k}=\frac{1}{2}mv^2 = \frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2 + \frac{1}{2}mv_z^2.\tag{4}$$

The energy is the sum of three quadratic modes, so the internal energy of the system is simply (substituting $$n$$ by $$3$$ in our expression for $$U$$)

$$U = \frac{3}{2}k_\mathrm{B}T = \frac{3}{2}nRT.\tag{5}$$

P.S. This might have been a bit more calculus then you'd asked for, but it is the reason this expression exists. I always find it helpful to know where things come from, so I hope this helps you as well.

• There seems to be an erroneous "n" in your very last expression. This answer does not seem to address the key point of the question, which is whether $U=3/2RT$ is literally valid, or whether it is actually $U=3/2RT+constant$ and only $\Delta U=3/2R \Delta T$ is valid (without making arbitrary assumptions about the energy reference level). Even for an ideal gas, I'm under the impression that the constant might be required. Commented Jan 30, 2022 at 21:09
• @electronpusher Think it's more like missing an N for the second expression since he's talking about internal energy of the whole system with N particles. Earlier equations were for individual particles. Can't comment on the supposed constant though; what does it even describe? Commented Oct 6, 2023 at 17:57