Derive equation 6.92 and 6.93 for the entropy and chemical potential of an ideal gas.

Short Answer

Expert verified

S=NklnVNvQ+52-FintT

μ=-kTlnVZintNvQ

Step by step solution

01

Step 1. Given information

The Helmholtz free energy for an ideal gas is given by

F=-NkTlnV-lnN-lnvQ+1+Fint.........................(1)

Here, V,N,vQ,Fintare the volume, number of molecules, quantum volume and the internal free energy of the ideal gas.

The quantum volume is given by

vQ=h22πmkT32........................(2)

02

Step2. Calculation for the entropy of the gas

Take logarithm from both sides of equation (2).

lnvQ=lnh22πmkT32=-32lnT+32lnh22πmk32...........................(3)

The formula calculate the entropyS of the gas is given by

S=-FTV,N................................(4)

Substitute the value of the free energy from equation (1) into equation (4) and simplify to obtain the required entropy.

localid="1647062285813" S=-T-NkTlnV-lnN-lnvQ+1+Fint=NklnVNvQ+1+NkT32T-FintT=NklnVNvQ+52-FintT

03

Step 3. Calculation for chemical potential

The internal free energy of the gas is given by

Fint=-NkTlnZint............................(5)

The formula to calculate the chemical potential of the gas is given by

μ=-FNT,V.............................(6)

Substitute the value of the free energy from equation (1) into equation (6) and simplify to obtain the required chemical potential.

μ=N-NkTlnV-lnN-lnvQ+1+Fint=N-NkTlnV-lnN-lnvQ+1-NkTlnZint[fromequation(5)]=-kTlnVNvQ+1+NkT1N-kTlnZint=-kTlnVNvQ-kTlnZint=-kTlnVZintNvQ

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