Question

air is cooled from 25° to 0° calculate the decrease in RMS velocity of molecules

Answer 1

hope it helps you out

Answer 2

Answer:

The decrease in RMS velocity of molecules will 0.957 times of speed at 25°.

Explanation:

The formula used for root mean square speed is:

$\nu_{rms}=\sqrt{\frac{3kN_AT}{M}}$

where,

$\nu_{rms}$ = root mean square speed

k = Boltzmann’s constant = $1.38\times 10^{-23}J/K$

T = temperature of air

M = atomic mass of air

1)  RMS velocity of molecules at 25°:

T = 25° = 298.15 K

$\nu_{rms}=\sqrt{\frac{3kN_A\times 298.15 K}{M}}$ ..[1]

2)  RMS velocity of molecules at 0°:

T = 0° = 273.15 K

$\nu_{rms}'=\sqrt{\frac{3kN_A\times 273.15K}{M}}$ ..[2]

Dividing [1] and [2]:

$\frac{\nu_{rms}}{\nu_{rms}'}=\frac{\sqrt{\frac{3kN_A\times 298.15 K}{M}}}{\sqrt{\frac{3kN_A\times 273.15K}{M}}}$

$\frac{\nu_{rms}}{\nu_{rms}'}=1.045$

${\nu_{rms}'}=\nu_{rms}\times 0.957$

The decrease in RMS velocity of molecules will 0.957 times of RMS of speed at 25°.