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​The entropy in terms of internal energy {U-U(0)}, and canonical partition function (Q) is given by S=(U-U(0))/T+k ln⁡Q. Assuming that the atoms are d
Question

​The entropy in terms of internal energy {U-U(0)}, and canonical partition function (Q) is given by S=(U-U(0))/T+k ln⁡Q. Assuming that the atoms are distinguishable, the corresponding expression for a monatomic perfect gas is
[Here n, m, NA, k and h represent number of moles, mass of atom, Avogadro constant, Boltzmann constant and Planck constant, respectively. U(0) in internal energy at T=0].

A.

S=52nR+nRln(V(2πmkT)3/2nNAh3)S = \frac{5}{2} nR + nR \ln \left( \frac{V (2 \pi m k T)^{3/2}}{n N_A h^3} \right)​​

B.

S=32nR+nRln(V(2πmkT)3/2h3)S = \frac{3}{2} nR + nR \ln \left( \frac{V (2 \pi m k T)^{3/2}}{h^3} \right)​​

C.

S=52nR+nRln(V(2πmkT)3/2h3)S = \frac{5}{2} nR + nR \ln \left( \frac{V (2 \pi m k T)^{3/2}}{h^3} \right)​​

D.

S=32nR+nRln(V(2πmkT)3/2nNAh3)S = \frac{3}{2} nR + nR \ln \left( \frac{V(2\pi mkT)^{3/2}}{n N_A h^3} \right)​​

Correct option is B

​The Sackur–Tetrode equation is an expression for the entropy of a monatomic ideal gas.

The Sackur–Tetrode equation expresses the entropy S of a monatomic ideal gas in terms of its thermodynamic state—specifically, its volume V , internal energy U, and the number of particles N:

where k​is the Boltzmann constant, m is the mass of a gas particle and h is the Planck constant.

For n mole of monoatomic gas, the internal energy is given as:

Translational Partition Function

An atom has two kinds of energy, translational and electronic energies. The partition function can be written as q= qtqe

The translational energy of an atom in a volume V is given by

Since the three exponential functions and summation over n1,n​2 and n3 are identical, we can write the above expression as

Since the translational energy levels are very close to each other, the summation in the above expression may be replaced by integration.

From equation i and ii we get,

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