Question

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

Answer 1

We know,

$U_{RMS}=\sqrt{\frac{3RT}{M}}$

So,

$U_{{25}^{0}C}=\sqrt{\frac{3R\times \left(298\right)}{M}}$

$U_{0^{0}C}=\sqrt{\frac{3R\times \left(273\right)}{M}}$

Now,

$\frac{U_{{25}^{0}C}}{U_{0^{0}C}}=\sqrt{\frac{298}{273}}$

$⟹U_{0^{0}C}=\sqrt{\frac{273}{298}}\times U_{{25}^{0}C}$

$⟹U_{0^{0}C}=0.9571\times U_{{25}^{0}C}$

Hence decrease in RMS speed is 42.7 m/s