Question

Calculate the total thermal energy in a gram of lead at room temperature, assuming that none of the degrees of freedom are "frozen out" (this happens to be a good assumption in this case).

Short answer

Total thermal energy is 35.3 J.

Step-by-step solution

Given information

Temperature T = 298 K

Number of degree of freedom f= 6 (3 kinetic + 3 potential)

Explanation

Total thermal energy is calculated for system is

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

Where

N = number of molecules

k = Boltzmann constants

T = temperature

f = number of degree of freedom

lead has degree of freedom of f= 6

molecular mass of lead is 207.2 u

we can calculate number of molecules as

$N=\frac{\text{mass}}{\text{molecular mass}}$

Here mass is 1 gram =1 x 10³kg

molecular mass is 207.2 u = 207.2 x 1.6 x 10^-27 kg

Substitute the values, we get

$$\begin{gathered} N=\frac{1\times {10}^{-3}\mathrm{k}}{207.2\times \left(1.6\times {10}^{-27}\right)\mathrm{kg}} \\ N=2.91\times {10}^{21} \end{gathered}$$

Now substitute the values in equation (1) to get thermal energy

$$\begin{gathered} U_{\text{thermal}}=\frac{6}{2}\times (2.91\times {10}^{21})\times (1.38\times {10}^{-23}J{K}^{-1})\times \left(298K\right) \\ {U}_{\text{thermal}}=35.3\mathrm{J} \end{gathered}$$

Total thermal energy is 35.3 J.