Question

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.​

Answer 1

`(T_2)/(T_1) = 1`

Step-by-step explanation:

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

`v = \sqrt{(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{(3RT_1)/(M) }`

For T = T₂ :

`v = \sqrt{(3RT_2)/(M) }`

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

`\sqrt{(3RT_1)/(M) }=\sqrt{(3RT_2)/(M) }\\\\(T_2)/(T_1) = 1`