Engineering
Physics
Degree of Freedom and Internal Energy of Gas
Question

State the law of equipartition of energy and hence calculate the molar specific heat of mono- and diatomic gases at constant volume and constant pressure.

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Solution
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The Law of equipartition of energy states that for a dynamical 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 12kT, where k is the Boltzmann’s constant.

For example, for a monoatomic molecule, each molecule has 3 degrees of freedom. According to the kinetic theory of gases, the mean kinetic energy of a molecule is 32kT.

Specific heat capacity of Monatomic gas:

The molecules of a monatomic gas have 3 degrees of freedom.

The average energy of a molecule at temperature T is 3KBT2

The total internal energy of a mole is: 3KBT2×NA

The molar specific heat at constant volume CV is

For an ideal gas,

CV (monatomic gas) =dUdT=3RT2

For an ideal gas, CP – CV = R

where CP is molar specific heat at constant pressure.

Thus, CP=5R2

Specific heat capacity of Diatomic gas:

The molecules of a monatomic gas have 5 degrees of freedom, 3 translational, and 2 rotational.

The average energy of a molecule at temperature T is 5KBT2

The total internal energy of a mole is: 5KBT2×NA

The molar specific heat at constant volume CV is

For an ideal gas,

CV (monatomic gas) =dUdT=5RT2

For an ideal gas, CP – CV = R

where CP is molar specific heat at constant pressure.

Thus, CP=7R2

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