Question

The temperature at which the velocity of oxygen will be half of that of hydrogen at NTP is

• A. $1092{}^{\circ }C$
• B. $1492{}^{\circ }C$
• C. $273K$
• D. $819{}^{\circ }C$

Answer 1

Answer: (D)

$v_{O_2}=\frac{1}{2}{v}_{H_2}$
$\therefore$ $\sqrt{\frac{3RT}{32}}=\frac{1}{2}\sqrt{\frac{3R\times 273}{2}}$
Or $\frac{T}{32}=\frac{273}{8}$
$\therefore$ $T=4\times 273$
$T=1092K=1092-273$
$T=819{}^{\circ }C$