Question

The root mean square velocity of hydrogen is $\sqrt{5}$ times than that of nitrogen. If $T$ is the temperature of the gas, then:

• A. $T_{H_2}=T_{N_2}$
• B. $T_{H_2}>T_{N_2}$
• C. $T_{H_2}<T_{N_2}$
• D. $T_{H_2}=\sqrt{7}{T}_{N_2}$

Answer 1

The correct option is C $T_{H_2}<T_{N_2}$

$U_{rms}=\sqrt{\frac{3RT}{Mw}}=\frac{{\left(U_{rms}\right)}_{H_2}}{{\left(U_{rms}\right)}_{N_2}}=\sqrt{\frac{T_{H_2}\times M_{N_2}}{M_{H_2}\times T_{N_2}}}$
${\left(U_{rms}\right)}_{H_2}=\sqrt{5}{\left(U_{rms}\right)}_{N_2}$
$\frac{{\left(U_{rms}\right)}_{H_2}}{{\left(U_{rms}\right)}_{N_2}}\times \sqrt{5}=\sqrt{\frac{T_{H_2}\times 28}{T_{N_2}\times 2}}$
$\frac{\sqrt{5}}{1}=\sqrt{\frac{T_{H_2}\times 14}{T_{N_2}}}$
$\frac{5}{1}=\frac{T_{H_2}\times 14}{T_{N_2}}$
$\frac{5\times T_{N_2}}{1}=T_{H_2}\times 14$
$T_{N_2}>T_{H_2}$