Is Internal Energy = (3/2)nRT for a ideal monoatomic gas?

Internal Energy is a state variable and its value at a particular state cannot be measured; only the change in internal energy can be measured. So how come we write that
$$U=\frac32nRT$$
where
$U$ = internal energy of the system,
$n$ = no of moles of the gas,
$R$ = Universal gas constant,
$T$ = Temperature of the system.

Till now, I knew that it was 'Avg Kinetic energy' that followed the above relation that is $\mathrm{K.E.} = (3/2)nRT$, but today while looking at a video in Khan Academy, I came across the relation $U=(3/2)nRT$ and got entirely confused.

• Internal Energy is (3/2)nRT plus some constant that cannot be measured. Feb 13 '18 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. May 7 '19 at 13:45
• "Approximated" is a wrong word. Let's say it can be assumed to be 0. May 7 '19 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: $$\langle E \rangle =E \cdot P(x)=\frac{\int_{-\infty}^{+\infty} E {e^{- \beta E}}}{\int_{-\infty}^{+\infty} {e^{- \beta E}}}$$ Where $$\beta =\frac{1}{k_B T}$$ and $$P(x)$$ is the probabily of the system being at a particular energy, $$E$$. Now, if your energy dependance 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): $$\langle E \rangle = \frac{\int_{-\infty}^{+\infty} {ax^2 e^{- \beta ax^2}}}{\int_{-\infty}^{+\infty} {e^{- \beta ax^2}}} = \frac{1}{2 \beta}= \frac{1}{2}k_BT$$

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 ($$\frac{1}{2}k_BT$$) to the system (all you have to do is repeat the calculations for $$E=\sum_{i=0}^n {a_i}x_i^2$$) - this is known as the equipartition theorem. Hence, the mean energy of a system with $$n$$ quadratic modes becomes: $$\langle E \rangle = \frac{1}{2}nk_BT$$

In an ideal monoatomic gas, the internal energy of the system, $$U$$, is just its kinetic energy (there is no energy due to vibration, rotation and intermolecular interactions), and therefore: $$E=KE=\frac{1}{2}mv^2 = \frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2 + \frac{1}{2}mv_z^2$$ The energy is the sum of $$3$$ quadratic modes, so the internal energy of the system is simply (substituting $$n$$ by 3 in our expression for $$U$$): $$U=\frac{3}{2}k_BT = \frac{3}{2}nRT$$

PS: 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