Question

Find the number of oxygen molecules in 0.1 kg, whose speeds lie between 195 and 205 m/s at

0C.

Answer 1

Solution;

Given;

T=O°c=273K

m=0.1kg which is contained in 3.125mols

$v_1=195$

$v_2=205$

$v=\frac{v_1+v_2}{2}=\frac{195+205}{2}=200m/s$

The number of molecules is;

$dN=G(v)dvN$

But;

$G(v)dv=4πv^2(\frac{M}{2πRT})^{\frac32}e^{\frac{-Mv^2}{2RT}}dv$

Compute each term;

$4πv^2=4π×200^2=160000π$

$\frac{M}{2πRT}=\frac{0.032}{2π×8.314×273}=2.244×10^{-6}$

$e^{\frac{-Mv^2}{2RT}}=e^{\frac{-0.032×200^2}{2×8.314×273}}=0.7543$

$DV=205-195=10$

$G(v)dv=160000π×2.244×10^{-6}×0.7543×10=8.5082$

Then the number of molecules are;

$dN=8.5082×3.125×6.02×10^{23}=1.6×10^{25} molecules$