Question

It is found that the most probable speed of molecules in a gas at equilibrium temperature

Answer 1

Answer:

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

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:

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