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

Bihar CECEBihar CECE 2010Kinetic Theory

Solution:

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