Question
Find $\gamma$ for polyatomic gas and hence determine its value for a triatomic gas in which the molecules are linearly arranged.
✓
Answer
The energy of a polyatomic gas having 'n' degrees of freedom is given by.
$\text{E}=\text{n}\times\frac{1}{2}\text{KT}\times\text{N}=\frac{\text{n}}{2}\text{RT}$
$\therefore\text{C}_\text{v}=\frac{\text{dE}}{\text{dT}}=\frac{\text{n}}{2}\text{R}$
$\therefore\text{C}_\text{p}=\text{C}_\text{v}+\text{R}=\frac{\text{n}}{2}\text{R}+\text{R}=\Big(\frac{\text{n}}{2}+1\Big)\text{R}$
$\therefore\gamma=\frac{\text{C}_\text{p}}{\text{C}_\text{v}}=\frac{\frac{\text{n}}{2}+1}{\frac{\text{n}}{2}}=1+\frac{2}{\text{n}}.$
In case of a triatomic gas, n = 7
$\therefore\gamma=\gamma=1+\frac{2}{7}=\frac{9}{7}.$
$\text{E}=\text{n}\times\frac{1}{2}\text{KT}\times\text{N}=\frac{\text{n}}{2}\text{RT}$
$\therefore\text{C}_\text{v}=\frac{\text{dE}}{\text{dT}}=\frac{\text{n}}{2}\text{R}$
$\therefore\text{C}_\text{p}=\text{C}_\text{v}+\text{R}=\frac{\text{n}}{2}\text{R}+\text{R}=\Big(\frac{\text{n}}{2}+1\Big)\text{R}$
$\therefore\gamma=\frac{\text{C}_\text{p}}{\text{C}_\text{v}}=\frac{\frac{\text{n}}{2}+1}{\frac{\text{n}}{2}}=1+\frac{2}{\text{n}}.$
In case of a triatomic gas, n = 7
$\therefore\gamma=\gamma=1+\frac{2}{7}=\frac{9}{7}.$
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