# Why 3 and not 6 in U = 3/2RT from the equation for the internal energy of an ideal gas?

Internal energy of an ideal monatomic gas is

$$U=\frac{3}{2}RT.\tag{1}$$

While I understand the derivation, I do not understand why we multiply by $$3$$ rather than by $$6.$$ According to Khan Academy video the force of gas molecules hitting one of the six faces of the wall is

$$pV=Nmv_x^2.\tag{2}$$

Why don't we multiply by $$2$$ after we multiply by $$3$$ to account for the pressure arising in the $$y$$ and $$x$$ directions as well to account for the force applied to all six faces of the theoretical box?

• Probably the answer to Is Internal Energy = (3/2)nRT for a ideal monoatomic gas? could be more helpful since there is no boxes involved in derivation. Commented Feb 2, 2022 at 4:00
• Pressure is force divided by area. If you count double the force (hitting two faces of the wall), surely you need to double the area as well? Commented Feb 2, 2022 at 11:16
• Also, your expression is not right; it should read $pV = Nm\langle v\rangle ^2$. The angle brackets denote mean values (in general velocities are a distribution, e.g. Maxwell–Boltzmann), so there is no single velocity $v$ that you can use; but more importantly, $v$ is the magnitude of the 3D velocity vector ($v = |\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$), whereas $v_x$ is the $x$-component of the velocity. In general, you have that $\langle v^2 \rangle = 3\langle v_x^2 \rangle$, which is where the factor of 3 comes from. Commented Feb 2, 2022 at 11:26
• Err, that first bit of my last comment is totally wrong, in fact. :-/ It should not have gotten three upvotes. Sorry. The correct formula is $pV = Nm\langle v_x^2 \rangle$, which is the same as the one in the question, except that you do need the angle brackets. The rest is correct. Commented Feb 9, 2022 at 19:13