Most probale velocity of hydrogen molecules at $T^{0} C$ is $v_{0}$ . At $(2 T + 273)^{0} C$ , all the molecules are dissociated into atoms. Then new most probable velocity will be:

\sqrt{2} v_{0}

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2 v_{0}

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3 v_{0}

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4 v_{0}

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Solution

The correct option is A $\sqrt{2} v_{0}$
Given initial temperature $T^{0} = T^{0} C = T + 273 K$
probable velocity $= v_{0}$
Final temperature $T_{f} = (2 T + 273)^{0} C = 2 T + 273 + 273 K = 2 \times (273 + T) = 2 \times T^{0}$
$v \propto T$
$v^{0} = k {\sqrt{T}}_{0}$
$v_{f} = k \sqrt{2} T^{0}$
$\frac{v_{f}}{v^{0}} = \frac{k \sqrt{2 T_{0}}}{k \sqrt{T_{0}}}$
$v_{f} = \sqrt{2} v_{0}$

Suggest Corrections

Similar questions

t^{0} C

V_{0}

{(2 t + 273)}^{o} C

V_{0}

t^{\circ} C

V_{o}

{(2 t + 273)}^{\circ} C

v = v_{0} + c t + g t^{2}

x = 0

t = 0

(t = 2 sec)

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