Question

Air is cooled from ${25}^{o}C$ to $0^o\left(C\right)$ The decrease in rms speed of the molecules is:

• A. $4.3$
• B. $5$
• C. $4.8$
• D. $3$

Answer 1

The correct option is A $4.3$

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

$U_{rms}\propto \sqrt{T}$

at $T_1={25}^{o}C=298K$

$\Rightarrow U_{rm{s}_1}=\sqrt{\frac{3R{T}_1}{M}}=\sqrt{\frac{3R}{M}\times 298}$

$\Rightarrow U_{rm{s}_2}=\sqrt{\frac{3R{T}_2}{M}}=\sqrt{\frac{3R}{M}\times 273}$

$\Rightarrow U_{rm{s}_1}-U_{rm{s}_2}=\sqrt{\frac{3R}{M}}(\sqrt{298}-\sqrt{273})\times \frac{\sqrt{298}}{\sqrt{298}}$

$\Rightarrow U_{rm{s}_1}-U_{rm{s}_2}=0.043\sqrt{\frac{3R}{M}\times 298}=0.043U_{rms}$

$\Delta U_{rm{s}_2}=0.043\times 100=4.3$%