Question
Define degrees of freedom. How will you account for the five degrees of freedom in a diatomic molecule? Using the idea of equi-partition of energy, find the value of $\gamma$ for the same.
✓
Answer
The number ofdegrees offreedom ofa dynamical system is defined as total number of coordinates required to describe comiletely the position and configuration of the syslem. The diatomic molecules have two atoms in it. The molecule is capable of translatory motion of its centre of mass and also it can rotate about its cenfie of mass. Therefore it has three translational degrees of freedom and two rotational degrees of freedom. So diatomic molecule has in all five degrees of freedom.$\text{U}=5\times\Big(\frac{1}{2}\text{K}_{\text{B}}\text{T}\Big)\times\text{N}_{\text{A}}=\frac{5}{2}\text{RT}$
$\text{C}_{\text{v}}=\Big(\frac{\text{dU}}{\text{dT}}\Big)$
$\text{C}_{\text{v}}=\frac{\text{d}}{\text{dT}}\Big(\frac{5}{2}\text{RT}\Big)=\frac{5}{2}\text{R}$
$\text{C}_{\text{v}}=4.96\text{ cal mol}^{-1}\text{K}^{-1}$
also, $\text{C}_{\text{P}}=\text{C}_{\text{v}}+\text{R}=\frac{5}{2}\text{R + R}=\frac{7}{2}\text{R}$$\text{C}_{\text{P}}=\frac{7}{2}\times1.985$
$\text{C}_{\text{P}}=6.95\text{ cal mol}^{-1}\text{ K}^{-1}$
$\gamma=\frac{\text{C}_\text{P}}{\text{C}_{\text{v}}}=\frac{\frac{7}{2}\text{R}}{\frac{5}{2}\text{R}}=1.4$
$\text{C}_{\text{v}}=\Big(\frac{\text{dU}}{\text{dT}}\Big)$
$\text{C}_{\text{v}}=\frac{\text{d}}{\text{dT}}\Big(\frac{5}{2}\text{RT}\Big)=\frac{5}{2}\text{R}$
$\text{C}_{\text{v}}=4.96\text{ cal mol}^{-1}\text{K}^{-1}$
also, $\text{C}_{\text{P}}=\text{C}_{\text{v}}+\text{R}=\frac{5}{2}\text{R + R}=\frac{7}{2}\text{R}$$\text{C}_{\text{P}}=\frac{7}{2}\times1.985$
$\text{C}_{\text{P}}=6.95\text{ cal mol}^{-1}\text{ K}^{-1}$
$\gamma=\frac{\text{C}_\text{P}}{\text{C}_{\text{v}}}=\frac{\frac{7}{2}\text{R}}{\frac{5}{2}\text{R}}=1.4$
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