Question

Root mean square (rms) speed of $O_{2}$ is $500 \,m / s$ at a constant temperature. Calculate the rms speed and the average kinetic energy of $H _{2}$ at the same temperature. (consider $R =8.33\, J\,k ^{-1} \,mol ^{-1}$

• A. $500\, m / s$ and $4.0\, kJ / mol$
• B. $2000 \,m / s$ and $4.0 \,kJ / mol$
• C. $500 \,m / s$ and $4.7 \,kJ / mol$
• D. $2000 \,m / s$ and $4.7\, kJ / mol$

Answer 1

Answer: (B)

Given, root mean square $(rms)$ speed of $O_{2}$ is $500\, m / s$ at a constant temperature.
Root mean square speed in given by the following expression
$u_{ rms }=\sqrt{\frac{3 R T}{M w}}$
For $H_{2}$ gas, $u_{ rms }\left(H_{2}\right)=\sqrt{\frac{3 R T}{M w_{H_{2}}}}$
( $M w$ of $H_{2}=2\, g / mol$ )
For $O_{2}$ gas, $u_{ rms }\left(O_{2}\right)=\sqrt{\frac{3 R T}{M w_{O_{2}}}}$
$\left( Mw\right.$ of $\left.O _{2}=32 \, g / mol \right)$
$\frac{u_{ rms \left(H_{2}\right)}}{u_{ rms }\left(O_{2}\right)}=\sqrt{\frac{3 R T}{M w_{H_{2}}}} \times \sqrt{\frac{M w_{O_{2}}}{3 R T}}$
(Temperature constant, $R=8.33\, JK ^{-1} \, mol ^{-1}$ )
$u_{ rms \left(H_{2}\right)}=\sqrt{\frac{M w_{o_{2}}}{M w_{H_{2}}}} \times u_{ rms \left(O_{2}\right)}=\sqrt{\frac{32}{2}} \times 500$
$=2000 \, m / s$
Average kinetic energy of $H _{2}$ can be calculate as
$=\frac{1}{2} m u^{2} $
${\left[u=2000\, m / s , m = H _{2}=2\, g / mol =0.002 \, kg / mol \right]}$
$=\frac{1}{2} \times 0.002 \times(2000)^{2} $
$=\frac{1}{2} \times \frac{2}{1000} \times 2000 \times 2000$
$=4000\, J\, mol ^{-1} $ or $ 4 \, kJ mol ^{-1}$