ABC26PH0199 · Statistical Thermodynamics

Subject: Physical Chemistry · Chapter: Statistical Thermodynamics · Topic: Partition Functions · Exam: CSIR-NET DEC 2012 · Marks: 2 · Difficulty: Easy

The rotational partition function of $H_{2}$ is :
(a)$\sum_{J=0,1,2 }(2 J+1) e^{-\beta h c B J(J+1)}$
(b)$\sum_{J=1,3,5 }(2 J+1) e^{-\beta h c B J(J+1)}$
(c)$\sum_{J=0,2,4 }(2 J+1) e^{-\beta h c B J(J+1)}$
(d)$\frac{1}{4}\left[\sum_{J=0,2,4 \ldots}(2 J+1) e^{-\beta h c B J(J+1)}+3 \sum_{J=1,3,5 \ldots}(2 J+1) e^{-\beta h c B J(J+1)}\right]$
Answer
Answer (as printed): D
Explanation
Rotational partition function $Z_{rot}=2(Z'_{even}Z_{even}+Z'_{odd}Z_{odd})$ ......(i) For $H_{2}$ molecule, spin $i$ equal to $1/2$. So, $Z'_{even}=\dfrac{i}{2i+1}$ and $Z'_{odd}=\dfrac{i+1}{2i+1}$ $Z_{even}=\displaystyle\sum_{J=0,2,4,\ldots}(2J+1)e^{-\beta Bhc J(J+1)}$ and $Z_{odd}=\displaystyle\sum_{J=1,3,5,\ldots}(2J+1)e^{-\beta Bhc J(J+1)}$ $H_{2}\ (i=1/2)$: ortho $Z'_{odd}=\dfrac{\frac{1}{2}+1}{2}=\dfrac{3}{2}$, para $Z'_{even}=\dfrac{\frac{1}{2}}{2}=\dfrac{1}{4}$ Putting the values of $Z'_{even}, Z_{even}, Z'_{odd}, Z_{odd}$ in equation (i): $\dfrac{1}{4}\left[\displaystyle\sum_{J=0,2,4,\ldots}(2J+1)e^{-\beta Bhc J(J+1)}+3\displaystyle\sum_{J=1,3,5,\ldots}(2J+1)e^{-\beta Bhc J(J+1)}\right]$

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