Question

The root mean square velocity of the molecules in a sample of helium $\frac{5}{7}$ th that of the molecules in a sample of hydrogen at $0^{\circ }C$. Then, the temperature of helium sample is about :

• A. $100^{\circ }C$
• B. $273^{\circ }C$
• C. $0K$
• D. $0^{\circ }C$

Answer 1

Answer: (D)

Square-root of the mean-square velocity of the gas molecules is called the root mean square velocity and is denoted by
$v_{rms}=\sqrt{\frac{3RT}{M}}$
Where $T$ is temperature, $M$ is molecular weight, $R$ is gas constant.
Let $v_H$ and $v_{He}$ be velocities of hydrogen and Helium respectively.
$T$ and $T$' be the temperature of hydrogen and helium respectively
$\therefore v_{He}=\frac{5}{7}{v}_H$
$\therefore \sqrt{\frac{3R{T}^{\prime }}{M_{He}}}=\frac{5}{7}\sqrt{\frac{3RT}{M_H}}$
$\Rightarrow \sqrt{\frac{T^{\prime }}{T}}=\frac{5}{7}\sqrt{\frac{M_{He}}{M_H}}$
$\Rightarrow \frac{T^{\prime }}{T}={\left(\frac{5}{7}\right)}^2\frac{M_{He}}{M_H}$
$\Rightarrow \frac{T^{\prime }}{T}=\frac{25\times 2}{49\times 1}=1$
$\Rightarrow T^{\prime }=273K$
In Centigrade
$T^{\prime }=273-273=0^{\circ }C$
Note: Faster the motion of molecules of a gas, Higher will be the temperature of the gas.