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Equipartition Theorem

The temperatu...

Question

The temperature of an ideal gas is increased from $120 K$ to $480 K$ . If at $120 K$ the root mean square velocity of the gas molecules is $v$ , at $480 K$ it becomes

4 v

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2 v

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\frac{v}{2}

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\frac{v}{4}

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Solution

The correct option is B $2 v$
$v_{r m s} = \sqrt{\frac{3 R T}{M}}$
$v_{r m s} \propto \sqrt{T}$
When temperature is increased from $120 K$ to $480 K$ (i.e., four times), the root mean square speed will become $\sqrt{4}$ or $2$ times, i.e., $2 v$ .

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The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K the root mean square velocity of the gas molecules is v, at 480 K it becomes

The correct option is B 2vvrms=√3RTMvrms∝√TWhen temperature is increased from 120 K to 480 K (i.e., four times), the root mean square speed will become √4 or 2 times, i.e., 2v.

The temperature of an ideal gas is increased from $120 K$ to $480 K$ . If at $120 K$ the root mean square velocity of the gas molecules is $v$ , at $480 K$ it becomes

The correct option is B $2 v$
$v_{r m s} = \sqrt{\frac{3 R T}{M}}$
$v_{r m s} \propto \sqrt{T}$
When temperature is increased from $120 K$ to $480 K$ (i.e., four times), the root mean square speed will become $\sqrt{4}$ or $2$ times, i.e., $2 v$ .