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The entropy of an ideal monatomic gas is(3/2)NkBlnE+,NkBlnV-NkBlnN to within an additive constant. Show that this implies the correct relationship between internal energy Eand temperature.

Short Answer

Expert verified

The expression for the internal energy of an ideal monatomic gas is in agreement with the kinetic theory of gases, and thus the correct relationship between E and T.

E=32NkBT

Step by step solution

01

Concept used

Equation of relationship between the internal energy and temperature for the entropy is

∂S∂E=1T

02

Put the given entropy in the equation

Assuming constant volume V for the ideal monatomic gas, the temperature T relates to the energy rate of change E of entropy S according to the equation:

∂S∂E=1T

Taking the partial derivative of our expression for S with respect to E we obtain:

∂S∂E=∂∂E32NkBlnE+NkBlnV-NkBlnN=32NkB1E=32NkBE

∂S∂E=32NkBE=1T

Rearranging for E we thus obtain:

32NkBE=1TE=32NkBT

This expression for the internal energy of an ideal monatomic gas is in agreement with the kinetic theory of gases, and thus the correct relationship between E and T.

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