Question

Air is cooled form ${25}^{o}C$ to $0^{o}C$. Calculate the decrease in the r.m.s speed of the molecules.

Answer 1

$V_r$${}_m$${}_s$ = $\sqrt{\frac{3RT}{M}}$

Let the velocities be $v_1$ and $v_2$

$v_1$ - $v_2$ = $\sqrt{\frac{3RT\times (298-273)}{M}}$

= $\Delta v$ = $\sqrt{\frac{75}{3\times 298}}$ $v_1$

= $\Delta v$ = 0.29 $v_1$

The decrease in the r.m.s speed is about 29%.