The partition function for a gas is given by

Q(N, V, T) = \(\frac{1}{N!}\left(\frac{2\pi m}{h^2\beta}\right)^{3N/2}\) (v - Nb)N \(e^{\frac{\beta aN^2}{V}}\)

The internal energy of the gas is

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  1. \(\frac{3}{2}Nk_BT+\frac{2aN}{V}\)
  2. \(\frac{1}{2}Nk_BT-\frac{aN^2}{V}\)
  3. \(\frac{3}{2}Nk_BT-\frac{aN^2}{V}\)
  4. \(\frac{3}{2}NRT-\frac{2aN}{V}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{3}{2}Nk_BT-\frac{aN^2}{V}\)
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Concept:

  • The Maxwell–Boltzmann distribution concerns the distribution of an amount of energy between identical but distinguishable particles.
  • It represents the probability for the distribution of the states in a system having different energies. A special case is the so-called Maxwell distribution law of molecular velocities.
  • The final expression for the internal energy U of a system following the Boltzmann distribution of energy is:

 

U = \(\rm k_BT^2\left(\frac{\partial ln Q}{\partial T}\right)_{V} \).......(1)

Explanation:-

  • The partition function for a gas is given by

Q(N, V, T) = \(\frac{1}{N!}\left(\frac{2\pi m}{h^2\beta}\right)^{3N/2}\) (v - Nb)N\(\frac{\beta aN^2}{V}\).........(2)

  • By taking ln both the sides of equation (2) we got,

\(lnQ(N,V,T)=ln\left ( \frac{1}{N!} \right )+\frac{3N}{2}ln\left ( \frac{2\pi m}{h^2} \right )+\frac{3N}{2}ln(\frac{1}{\beta})\)

\(+ Nln\left ( V-Nb \right )+\frac{\beta \alpha N^2}{V}\)

or, 

\(lnQ(N,V,T)=ln\left ( \frac{1}{N!} \right )+\frac{3N}{2}ln\left ( \frac{2\pi m}{h^2} \right )+\frac{3N}{2}ln({k_BT}{})\)

\(+ Nln\left ( V-Nb \right )+\frac{ \alpha N^2}{k_BTV}\)

or, \(\left ( \frac{\partial lnQ}{\partial T} \right )=\frac{\partial }{\partial T}\left [ ln\left ( \frac{1}{N!} \right )+\frac{3N}{2}ln\left ( \frac{2\pi m}{h^2} \right )+\frac{3N}{2}ln({KT}{})+ Nln\left ( V-Nb \right )+\frac{ \alpha N^2}{k_BTV} \right ]\)

or, 

\(\left ( \frac{\partial lnQ}{\partial T} \right )=\left [0+0+\frac{3N}{2}\times \frac{1}{k_BT}\times k_B+ 0+\frac{ \alpha N^2}{k_BV}\frac{\partial }{\partial T} \left ( \frac{1}{T} \right )\right ]\)

or, 

\(\left ( \frac{\partial lnQ}{\partial T} \right )=\left [\frac{3N}{2}\times \frac{1}{T}+\frac{ \alpha N^2}{k_BV}\left ( \frac{-1}{T^2} \right )\right ]\)

or,

\(\left ( \frac{\partial lnQ}{\partial T} \right )=\left [ \frac{3N}{2T}-\frac{ \alpha N^2}{k_BVT^2}\right ]\)

  • Now, putting the value of \(\left ( \frac{\partial lnQ}{\partial T} \right )\) in equation (1) we get,

\(\rm k_BT^2\left [ \frac{3N}{2T}-\frac{ \alpha N^2}{k_BVT^2}\right ]\)

or, U = kBT2 × \(\frac{3N}{2T}\) -  kBT2 × \(\frac{ \alpha N^2}{k_BVT^2}\)

or, U = \(\frac{3}{2}Nk_BT-\frac{aN^2}{V}\)

Conclusion:-

Hence, the internal energy of the gas is \(\frac{3}{2}Nk_BT-\frac{aN^2}{V}\)
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