Using the free energy of a monatomic ideal gas, derive the ideal gas law

Question

toriannucz · Accepted Answer

The entropy of the system in terms of the partition function and free energy is,

S = \frac{E}{T} + k_{B} \log_{e} Z

Thus, the Helmholtz free energy is,

F = E - S T

Substituting the known values,

F = E - (\frac{E}{T} + k_{B} \log_{e} Z) T = - k_{B} T \times \log_{e} Z

The partition function for the monoatomic ideal gas is,

Z = e^{- N} (\frac{V}{N})^{N} (\frac{2 π m}{h^{2} β})^{(\frac{2}{3} N)}

Substituting the known values,

F = - k_{B} T \times \log_{e} (e^{- N} (\frac{V}{N})^{N} (\frac{2 π m}{h^{2} β})^{(\frac{3}{2} N)}) = - k_{B} T \times (\log_{e} (e^{- N}) + \log_{e} (\frac{V}{N})^{N} + \log_{e} (\frac{2 π m}{h^{2} β})^{\frac{3}{2} N}) = - k_{B} T \times (- N + N \log_{e} (\frac{V}{N}) + (\frac{3}{2} N) \log_{e} (\frac{2 π m}{h^{2} β}) - 1)

As the relation between pressure and the Helmholtz free energy is,

P = - (\frac{d F}{d V})_{T, N} = \frac{k_{B} T N}{V}

As the gas constant is defined as,

R = N_{A} k_{B} k_{B} = \frac{R}{N_{A}}

where

N_{A}

is an Avogadro's number and

k_{B}

is a Boltzmann constant
Thus, the pressure equation become,

P = \frac{R T N}{N_{A} V} P V = \frac{N}{N_{A}} R T P V = n_{a} R T

where

n_{A}

is the number of moles of gas.
This equation represents the ideal gas equation.

Using the free energy of a monatomic ideal gas, derive