91Ó°ÊÓ

If the temperature changes by $∆T$,then the thermal energy for diatomic gas changes by equation $\left(20.30\right)$in the form

$\mathrm{Δ}{E}_{\mathrm{th}}=n{C}_V\mathrm{Δ}T$

Knowing the mass $M$and the molar mass $m$, we can get the number of moles by

$n=\frac{M}{m}=\frac{0.2\mathrm{g}}{2\mathrm{g}/\mathrm{mol}}=0.1\mathrm{mol}$

This thermal energy converts to heat..The temperature change from$50K$to $100K$. In this range,$C_V=\frac{3}{2}R$. So, the heat in equation will be

$\begin{array}{r}\mathrm{Δ}{E}_{\mathrm{th}}=n{C}_V\mathrm{Δ}T\\ =\frac{3}{2}nR\mathrm{Δ}T\\ =\frac{3}{2}\left(0.1\mathrm{mol}\right)(8.314\mathrm{J}/\mathrm{mol}â‹\dots \mathrm{K})(100\mathrm{K}−50\mathrm{K})\\ =62\mathrm{J}.\end{array}$

The temperature changes from $250K$to $300K$.In this range,$C_V=\frac{5}{2}R$.so,the heat equation will be

$\begin{array}{r}\mathrm{Δ}{E}_{\mathrm{th}}=n{C}_V\mathrm{Δ}T\\ =\frac{5}{2}nR\mathrm{Δ}T\\ =\frac{5}{2}\left(0.1\mathrm{mol}\right)(8.314\mathrm{J}/\mathrm{mol}â‹\dots \mathrm{K})(300\mathrm{K}−250\mathrm{K})\\ =104\mathrm{J}.\end{array}$

The temperature changes from $2250K$to $3000K$.In this range,$C_V=\frac{7}{2}R$.so,the heat equation will be

$\begin{array}{r}\mathrm{Δ}{E}_{\mathrm{th}}=n{C}_V\mathrm{Δ}T\\ =\frac{7}{2}nR\mathrm{Δ}T\\ =\frac{7}{2}\left(0.1\mathrm{mol}\right)(8.314\mathrm{J}/\mathrm{mol}â‹\dots \mathrm{K})(2300\mathrm{K}−2250\mathrm{K})\\ =150\mathrm{J}.\end{array}$