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JEE Main

Exam: JEE Main
Subject: Physics
Chapter: Kinetic Theory of Gases

JEE Main · Physics · Kinetic Theory of Gases

The R.M.S. speed of oxygen molecules at temperature T (in kelvin)...

The R.M.S. speed of oxygen molecules at temperature

T

(in kelvin) is

vms^{- 1}

. As the temperature becomes

4T

and the oxygen gas dissociates into atomic oxygen, then the speed of atomic oxygen

Options

Aremains the same
Bbecomes $2v$
Cbecomes $\sqrt{2}v$
Dbecomes $2\sqrt{2}v$

Correct answer

D. becomes $2\sqrt{2}v$

Step-by-step solution

v ∝ \sqrt{{γ}_{1}T}

{v}_{2} ∝ \sqrt{2 \times {γ}_{2}{T}_{2}} = \sqrt{2 \times {γ}_{2} \times 4T} \left(∵ {T}_{2} = 4T\right)

\dfrac{v2}{v1} ===2\sqrt{\dfrac{2{γ}_{2}}{\gamma1}}

\dfrac{v2}{v1} ===2\sqrt{\dfrac{5 \times 5 \times 2}{3 \times 7}}

{v}_{2} =

2\sqrt{\dfrac{50}{21}}

v

{v}_{2} ≅

2\sqrt{2}

v

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