jon1653
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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.​

Respuesta :

hamzaahmeds

Answer:

[tex]\frac{T_2}{T_1} = 1[/tex]

Explanation:

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

[tex]v = \sqrt{\frac{3RT}{M} }[/tex]

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 :

[tex]v = \sqrt{\frac{3RT_1}{M} }[/tex]

For T = T₂ :

[tex]v = \sqrt{\frac{3RT_2}{M} }[/tex]

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

[tex]\sqrt{\frac{3RT_1}{M} }=\sqrt{\frac{3RT_2}{M} }\\\\\frac{T_2}{T_1} = 1[/tex]

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