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(A) The root mean square (rms) velocity $V$ of a gas molecule is given by the formula:$V = \sqrt{\frac{3 k_B T}{m}}$where $k_B$ is the Boltzmann const

Vedclass · Accepted Answer

$\frac{V^2}{T} = 	ext{constant}$

Vedclass · Answer

(A) The root mean square (rms) velocity $V$ of a gas molecule is given by the formula:$V = \sqrt{\frac{3 k_B T}{m}}$where $k_B$ is the Boltzmann constant,$T$ is the absolute temperature,and $m$ is the mass of one molecule.Squaring both sides,we get:$V^2 = \frac{3 k_B T}{m}$Since $k_B$ and $m$ are constants for a given gas,we can write:$V^2 \propto T$Or,$\frac{V^2}{T} = \frac{3 k_B}{m} = 	ext{constant}$Therefore,the correct relation is $\frac{V^2}{T} = 	ext{constant}$.

When the rms velocity of a gas is denoted by $V$,which of the following relations is true? ($T=$ absolute temperature of the gas)

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