Question

A gas is equilibrium at T kelvin. If mass of one molecule is m and its component of velocity in x direction is v. Then mean of its v2 is 3kT 2kT 2 kT 3 (4) zero

Answer 1

The value of
`v_(x)^2` is
`(KT)/(M)`.

Step-by-step explanation:

A gas is equilibrium at T kelvin.

Mass = M

We know that,

The average square of the velocity in the x,y and z direction are equal.

`\bar{v}_(x)^2=\bar{v}_(y)^(2)=\bar{v}_(z)^(2)`

`v_(rms)^2=v_(x)^(2)+v_(y)^(2)+v_(z)^(2)`

`v_(rms)^2=3v_(x)^2`

`v_(x)^(2)=(v_(rms)^2)/(3)`

Equation of ideal gas

`PV=RT`

`P=(nKT)/(V)`

Here, R = nK

We know that,

`(nKT)/(V)=(nM)/(3V)v_(rms)^2`

`v_(rms)^2=(3KT)/(M)`....(I)

Put the value of
`v_(rms)^2` in the equation (I)

`v_(x)^2=(1)/(3)*(3KT)/(M)`

`v_(x)^2=(KT)/(M)`

Hence, The value of
`v_(x)^2` is
`(KT)/(M)`.