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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?

User Ad Rees
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Answer:

Step-by-step explanation:

From the information given:

a.

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


U = (1)/(2)k_BT


U = (1)/(2)nRT

For s degree of freedom


U = (1)/(2)snRT

However, the molar specific heat
C_v = (1)/(n) (dU)/(dT)

Therefore, in terms of R and s;


C_v = (1)/(n) (d)/(dT) \begin{pmatrix} (1)/(2) snRT \end {pmatrix}


C_v = (Rs)/(2)

b.

Given that:

Cv=70.6Jmol⋅K and R=8.314Jmol⋅K

Then; using the formula
C_v = (Rs)/(2)


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


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


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

s = 16.983

s
\simeq 17

User Gaganpreet Singh
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