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Q.
$14\, g$ of nitrogen is contained in a vessel at $300\, K$. How much heat should be taken out of the gas to half the rms speed of its molecules ? $R = 2\, cal/mol\, K$ :
Thermodynamics
Solution:
Number of moles of nitrogen
$n =\frac{14}{28}=\frac{1}{2}$
$v_{r m s}^{2} \propto T$
Thus to reduce $r m s$ speed to half, temperature has to be reduced by $4$ times
$T_{f} =\frac{300}{4}=75\, K$
$\Delta T =300-75=225\, K$
$\Delta Q =n C_{p} \Delta T$
$=\frac{1}{2} \times \frac{5}{2} R \times 225$
$=562.5\,cal$