# 11.3.5 Internal Energy Relationship with Temperature

In this section, we will discover the quantitative link between a gas’s internal energy and its temperature. We start with the equation of state ($\displaystyle pV=nRT$ ) and the pressure formula ($\displaystyle \displaystyle p=\frac{1}{3}\rho \left\langle {{{c}^{2}}} \right\rangle$ ), written in their alternative form.

$\displaystyle pV=NkT$

$\displaystyle \displaystyle pV=\frac{1}{3}Nm\left\langle {{{c}^{2}}} \right\rangle$

Combining them, we obtain

\displaystyle \begin{aligned}\frac{1}{3}Nm\left\langle {{{c}^{2}}} \right\rangle &=NkT\\m\left\langle {{{c}^{2}}} \right\rangle &=3kT\\\frac{1}{2}m\left\langle {{{c}^{2}}} \right\rangle &=\frac{3}{2}kT\\\left\langle {\frac{1}{2}m{{c}^{2}}} \right\rangle &=\frac{3}{2}kT\end{aligned}

Recognize that $\displaystyle \displaystyle \left\langle {\frac{1}{2}m{{c}^{2}}} \right\rangle$  represents the average translational KE of gas particles in the gas. So

$\displaystyle \displaystyle \left\langle {KE} \right\rangle =\frac{3}{2}kT$

Voila. We now have an understanding of temperature at a molecular level. The thermodynamic temperature is directly proportional to the average (translational) kinetic energy of the gas particles!

Remember that the internal energy $\displaystyle U=P{{E}_{{total}}}+K{{E}_{{total}}}$ ? An ideal gas does not have any potential energy. So for an ideal gas, U can be simplified to be[1]

$\displaystyle \displaystyle U=K{{E}_{{total}}}=\frac{3}{2}NkT=\frac{3}{2}nRT$

It turns out that the internal energy of an ideal gas depends only on its temperature. Or in mathematical jargon, we say that U is a function of T only.

Video Explanation

Derivation of <KE>=3/2kT

Concept Test

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[1] Actually, \$ $\displaystyle \displaystyle U=\frac{3}{2}nRT$  is valid only if the gas is ideal (if not PEtotal won’t be zero) and monatomic (if not KEtotal will include some rotational KE in addition to translational KE). If you’re interested, go read section 11.A.