14 g of nitrogen is contained in a vessel at 300 K. How much heat (in cal) should be taken out of the gas to half the rms speed of its molecules? $\mathrm{R}=2 \mathrm{cal}/\mathrm{mol} \mathrm{K}$

Number of moles of nitrogen, $\mathrm{n}=\frac{14}{28}=\frac{1}{2}\Rightarrow {\mathrm{v}}_{\mathrm{rms}}^2\propto \mathrm{T}$

Thus to reduce rms speed to half, temperature has to be reduced by 4 times 

 $\begin{array}{l}{\mathrm{T}}_{\mathrm{f}}=\frac{300}{4}=75\mathrm{K}\\ \mathrm{\Delta T}=300-75=22.5\mathrm{K}\\ \mathrm{\Delta Q}={\mathrm{nC}}_{\mathrm{p}}\mathrm{\Delta T}=\frac{1}{2}\times \frac{5}{2}\mathrm{R}\times 225=562.5\mathrm{cal}\end{array}$