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

1642


[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.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s