Question

Thermodynamics deals with the macroscopic properties of materials. Scientists can make quantitative predictions about these macroscopic properties by thinking on a microscopic scale. Kinetic theory and statistical mechanics provide a way to relate molecular models to thermodynamics. Predicting the heat capacities of gases at a constant volume from the number of degrees of freedom of a gas molecule is one example of the predictive power of molecular models. The molar specific heat Cv of a gas at a constant volume is the quantity of energy required to raise the temperature T of one mole of gas by one degree while the volume remains the same. Mathematically, Cv=1nΔEthΔT, where n is the number of moles of gas, ΔEth is the change in internal (or thermal) energy, and ΔT is the change in temperature. Kinetic theory tells us that the temperature of a gas is directly proportional to the total kinetic energy of the molecules in the gas. The equipartition theorem says that each degree of freedom of a molecule has an average energy equal to 12kBT, where kB is Boltzmann's constant 1.38×10^−23J/K. When summed over the entire gas, this gives 12nRT, where R=8.314Jmol⋅K is the ideal gas constant, for each molecular degree of freedom.
Required:
a. Using the equipartition theorem, determine the molar specific heat, Cv , of a gas in which each molecule has s degrees of freedom. Express your answer in terms of R and s.
b. Given the molar specific heat Cv of a gas at constant volume, you can determine the number of degrees of freedom s that are energetically accessible. For example, at room temperature cis-2-butene, C4H8 , has molar specific heat Cv=70.6Jmol⋅K . How many degrees of freedom of cis-2-butene are energetically accessible?

Answer 1

Answer:

Explanation:

From the information given:

a.

Using the equipartition theorem, the average energy of a molecule dor each degree of freedom is:

$U = \dfrac{1}{2}k_BT$

$U = \dfrac{1}{2}nRT$

For s degree of freedom

$U = \dfrac{1}{2}snRT$

However, the molar specific heat $C_v = \dfrac{1}{n} \dfrac{dU}{dT}$

Therefore, in terms of R and s;

$C_v = \dfrac{1}{n} \dfrac{d}{dT} \begin{pmatrix} \dfrac{1}{2} snRT \end {pmatrix}$

$C_v = \dfrac{Rs}{2}$

b.

Given that:

Cv=70.6Jmol⋅K and R=8.314Jmol⋅K

Then; using the formula  $C_v = \dfrac{Rs}{2}$

$70.6 \ J/mol.K = \dfrac{(8.314 \ J/mol.K)\times s}{2}$

$70.6 \ J/mol.K \times 2= (8.314 \ J/mol.K)\times s$

$s= \dfrac{70.6 \ J/mol.K \times 2}{ (8.314 \ J/mol.K) }$

s = 16.983

s $\simeq$ 17