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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?

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    $\begingroup$ 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. $\endgroup$
    – andselisk
    Feb 2, 2022 at 4:00
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    $\begingroup$ 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? $\endgroup$ Feb 2, 2022 at 11:16
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    $\begingroup$ 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. $\endgroup$ Feb 2, 2022 at 11:26
  • $\begingroup$ 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. $\endgroup$ Feb 9, 2022 at 19:13

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