Physics: It is found that the most probable speed of molecules in a gas at equilibrium temperature T2 is the same as the root-mean-square speed of the molecules in this gas when its equilibrium temperature is T1. Find T2/T1.

They make their decision without talking, but why they choose not to help the man is left unclear.Explanation:...

It is found that the most probable speed of molecules

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14.06.2021

$\frac{T_2}{T_1} = 1$

Explanation:

The root mean square velocity of the gas at an equilibrium temperature is given by the following formula:

$v = \sqrt{\frac{3RT}{M} }$

where,

v = root mean square velocity of molecules:

R = Universal Gas Constant

T = Equilibrium Temperature

M = Molecular Mass of the Gas

Therefore,

For T = T₁ :

$v = \sqrt{\frac{3RT_1}{M} }$

For T = T₂ :

$v = \sqrt{\frac{3RT_2}{M} }$

Since both speeds are given to be equal. Therefore, comparing both equations, we get:

$\sqrt{\frac{3RT_1}{M} }=\sqrt{\frac{3RT_2}{M} }\\\\\frac{T_2}{T_1} = 1$

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