Question

At certain temperature RMS velocity of a gas is
12,240 m/s, then it's most probable velocity
would be​

Answer 1

Answer:

what????????

Explanation:

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Answer 2

Given:

At certain temperature RMS velocity of a gas is

12,240 m/s.

To find:

Most probable velocity at the same temperature.

Calculation:

$\sf{v_{rms} = \sqrt{ \dfrac{3RT}{m} } = 12240 \: m {s}^{ - 1} }$

$\sf{ = > \sqrt{3 } \times \sqrt{ \dfrac{RT}{m} } = 12240 \: }$

$\sf{ = > \sqrt{ \dfrac{RT}{m} } = \dfrac{12240 }{ \sqrt{3} }\: }$

We know that at the same temperature, the most probable Velocity expression will be :

$\sf{v_{mp} = \sqrt{ \dfrac{2RT}{m} } }$

$\sf{ = > v_{mp} = \sqrt{2} \times \sqrt{ \dfrac{RT}{m} } }$

$\sf{ = > v_{mp} = \sqrt{2} \times \dfrac{12240}{ \sqrt{3} } }$

$\sf{ = > v_{mp} = 1.41 \times \dfrac{12240}{ 1.732} }$

$\sf{ = > v_{mp} = 9964.43 \: m {s}^{ - 1} }$

So, final answer is:

$\boxed{\rm{ \red{v_{mp} = 9964.43 \: m {s}^{ - 1} }}}$