Question

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

• A. $100^{\circ} C$
• B. $273^{\circ} C$
• C. $173\,K$
• D. $0^{\circ} C$

Answer 1

Answer: (D)

$0^{\circ} C$

Answer 2

$\frac{v_{H e}}{v H_{2}}=\frac{\sqrt{\frac{3 R T_{H e}}{M_{H e}}}}{\sqrt{\frac{3 R T_{H_{2}}}{M H_{2}}}}$ $=\sqrt{\frac{M_{H_2}}{M_{H e}} \cdot \frac{T_{H e}}{T_{H_2}}} \frac{v_{H e}}{v_{H_2}} \sqrt{\frac{2}{4} \times \frac{T_{H e}}{273}}$ $\frac{5}{7}=\sqrt{\frac{T_{H e}}{2 \times 273}}$ $\Rightarrow T_{H e}=\frac{25 \times 273 \times 2}{49}$ $\approx 278\, K =5^{\circ} C \cong 0^{\circ} C$