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. $ 0\,K $
• 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
$v_{r m s}=\sqrt{\frac{3 R T}{M}}$
Where $T$ is temperature, $M$ is molecular weight, $R$ is gas constant.
Let $v_{H}$ and $v_{H e}$ be velocities of hydrogen and Helium respectively.
$T$ and $T$' be the temperature of hydrogen and helium respectively
$\therefore v_{H e}=\frac{5}{7} v_{H} $
$\therefore \sqrt{\frac{3 R T'}{M_{H e}}}=\frac{5}{7} \sqrt{\frac{3 R T}{M_{H}}} $
$\Rightarrow \sqrt{\frac{T'}{T}}=\frac{5}{7} \sqrt{\frac{M_{H e}}{M_{H}}} $
$\Rightarrow \frac{T'}{T}=\left(\frac{5}{7}\right)^{2} \frac{M_{H e}}{M_{H}} $
$\Rightarrow \frac{T'}{T}=\frac{25 \times 2}{49 \times 1}=1$
$\Rightarrow T'=273 \,K$
In Centigrade
$T'=273-273=0^{\circ} C$
Note: Faster the motion of molecules of a gas, Higher will be the temperature of the gas.