Question
The specific heat of argon at constant volume is $0.075 ~kcal ~kg^{-1}K^{-1}$, then what will be its atomic weight?
$[Given, R = 2 ~cal ~mol^{-1}K^{-1}]$

Answer

Argon is a monoatomic gas, so$\text{C}_{\text{V}}=\frac{3}{2}\text{R}=\frac{3}{2}\times2=3\text{cal mol}^{-1}\text{K}^{-1}$
$\text{C}_{\text{V}}=\text{Mc}_{\text{V}}$
$\Rightarrow\text{M}=\frac{\text{C}_{\text{V}}}{\text{c}_{\text{V}}}=\frac{3}{0.075}=40$

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