91Ó°ÊÓ

Temperature T = 298 K

Degree of freedom of Helium is f=3

Total thermal energy is for a system contains N molecules each with f degree of freedom is given as

$U_{\text{thermal}}=Nf\left(\frac{1}{2}kT\right)..............................\left(1\right)$

Where

N = number pf molecules

k = Boltzmann constants

f = degree of freedom

T = Temperature

A liter of helium at room temperature and pressure of atm = 101325 Pa

Using ideal gas law P V=n k T, Find the thermal energy of

$$\begin{gathered} U_{\text{thermal}}=3NkT\frac{1}{2} \\ {U}_{\text{thermal}}=\frac{3}{2}PV..............................\left(2\right) \end{gathered}$$

Substitute the values given we get

$$\begin{gathered} U_{\text{thermal}}=\frac{3}{2}PV \\ =\frac{3}{2}×(101325Pa)×(1×{10}^{-3}{m}^3) \\ {U}_{\text{thermal}}=151.98\mathrm{J} \end{gathered}$$

For air , Number of degree of freedom is f=5
Find the thermal energy by substituting values in equation 2.

$$\begin{gathered} U_{\text{thermal}}=\frac{5}{2}PV \\ =\frac{5}{2}×\left(101325Pa\right)×(1×{10}^{-3}{m}^3) \\ {U}_{\text{thermal}}=253.31\mathrm{J} \end{gathered}$$