Question

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

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

Answer 1

Answer: (D)

Given, $v_{o_2}=\frac{1}{2}{v}_{H_2}$
$\therefore \sqrt{\frac{3RT}{32}}=\frac{1}{2}\sqrt{\frac{3R\times 273}{2}}$
$\frac{T}{32}=\frac{273}{8}$
$\therefore T=4\times 273$
$T=1092K=1092-273$
$T=819^{\circ }C$