Answer:
Explanation:
From the information given:
a.
Using the equipartition theorem, the average energy of a molecule dor each degree of freedom is:
[tex]U = \dfrac{1}{2}k_BT[/tex]
[tex]U = \dfrac{1}{2}nRT[/tex]
For s degree of freedom
[tex]U = \dfrac{1}{2}snRT[/tex]
However, the molar specific heat [tex]C_v = \dfrac{1}{n} \dfrac{dU}{dT}[/tex]
Therefore, in terms of R and s;
[tex]C_v = \dfrac{1}{n} \dfrac{d}{dT} \begin{pmatrix} \dfrac{1}{2} snRT \end {pmatrix}[/tex]
[tex]C_v = \dfrac{Rs}{2}[/tex]
b.
Given that:
Cv=70.6Jmol⋅K and R=8.314Jmol⋅K
Then; using the formula [tex]C_v = \dfrac{Rs}{2}[/tex]
[tex]70.6 \ J/mol.K = \dfrac{(8.314 \ J/mol.K)\times s}{2}[/tex]
[tex]70.6 \ J/mol.K \times 2= (8.314 \ J/mol.K)\times s[/tex]
[tex]s= \dfrac{70.6 \ J/mol.K \times 2}{ (8.314 \ J/mol.K) }[/tex]
s = 16.983
s [tex]\simeq[/tex] 17