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Question
State the law of equipartition of energy and hence calculate the molar specific heat of mono-atomic and di-atomic gases at constant volume and constant pressure.
Solution
Law of equipartition of energy states that for a dynamic system in thermal equilibrium the total energy of the system is shared equally by all the degrees of freedom. The energy associated with each degree of freedom per molecule is `1/2`kBT, where kB is Boltzmann’s constant.
For example, for a mono-atomic molecule, each molecule has 3 degrees of freedom. According to the kinetic theory of gases, the mean kinetic energy of a molecule is `3/2`kBT.
Specific heat capacity of Mono-atomic gas:
The molecules of a mono-atomic gas have 3 degrees of freedom.
The average energy of a molecule at temperature T is `3/2"K"_"B""T"`
The total internal energy of a mole is: `3/2"N"_"A""K"_"B"T"`, where NA is the Avogadro number.
The molar specific heat at constant volume Cv is
For an ideal gas,
Cv (mono-atomic gas) = `"dE"/"dT" = 3/2"R"`
For an ideal gas, Cp - Cv = R
where Cp is molar specific heat at constant pressure.
Thus, Cp = `5/2"R"`
Specific heat capacity of Di-atomic gas:
The molecules of a di-atomic gas have 5 degrees of freedom, 3 translational, and 2 rotational.
The average energy of a molecule at temperature T is `5/2"K"_"B""T"`
The total internal energy of a mole is: `5/2"N"_"A""K"_"B""T"`
The molar specific heat at constant volume Cv is
For an ideal gas,
Cv (di-atomic gas) = `"dE"/"dT" = 5/2"R"`
For an ideal gas, Cp - Cv = R
where Cp is the molar specific heat at constant pressure.
Thus, Cp = `7/2"R"`
A soft or non-rigid di-atomic molecule has, in addition, one frequency of vibration which contributes two quadratic terms to the energy. Hence, the energy per molecule of a soft diatomic molecule is
`E=(1/2"k"_"B""T")+2(1/2"k"_"B""T")+2(1/2"k"_"B""T")=7/2"k"_"B""T"`
Therefore, the energy per mole of a soft diatomic molecule is
E = `7/2"k"_"B""T"xx"N"_"A"=7/2"R"`
In this case, Cv = `"dE"/"dT"=7/2`R and
Cp = Cv + R = `7/2`R + R = `9/2`R
Notes
For a monatomic gas adiabatic constant, `gamma = (C_p)/(C_V) = 5/3.`
For a diatomic gas adiabatic constant, `gamma = 7/5 or 9/7`
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