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

Answer:

The value of $v_{x}^2$ is $\dfrac{KT}{M}$.

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.

$\bar{v}_{x}^2=\bar{v}_{y}^{2}=\bar{v}_{z}^{2}$

$v_{rms}^2=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}$

$v_{rms}^2=3v_{x}^2$

$v_{x}^{2}=\dfrac{v_{rms}^2}{3}$

Equation of ideal gas

$PV=RT$

$P=\dfrac{nKT}{V}$

Here, R = nK

We know that,

$\dfrac{nKT}{V}=\dfrac{nM}{3V}v_{rms}^2$

$v_{rms}^2=\dfrac{3KT}{M}$....(I)

Put the value of $v_{rms}^2$ in the equation (I)

$v_{x}^2=\dfrac{1}{3}\times\dfrac{3KT}{M}$

$v_{x}^2=\dfrac{KT}{M}$

Hence, The value of $v_{x}^2$ is $\dfrac{KT}{M}$.