Answer:
The value of [tex]v_{x}^2[/tex] is [tex]\dfrac{KT}{M}[/tex].
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.
[tex]\bar{v}_{x}^2=\bar{v}_{y}^{2}=\bar{v}_{z}^{2}[/tex]
[tex]v_{rms}^2=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}[/tex]
[tex]v_{rms}^2=3v_{x}^2[/tex]
[tex]v_{x}^{2}=\dfrac{v_{rms}^2}{3}[/tex]
Equation of ideal gas
[tex]PV=RT[/tex]
[tex]P=\dfrac{nKT}{V}[/tex]
Here, R = nK
We know that,
[tex]\dfrac{nKT}{V}=\dfrac{nM}{3V}v_{rms}^2[/tex]
[tex]v_{rms}^2=\dfrac{3KT}{M}[/tex]....(I)
Put the value of [tex]v_{rms}^2[/tex] in the equation (I)
[tex]v_{x}^2=\dfrac{1}{3}\times\dfrac{3KT}{M}[/tex]
[tex]v_{x}^2=\dfrac{KT}{M}[/tex]
Hence, The value of [tex]v_{x}^2[/tex] is [tex]\dfrac{KT}{M}[/tex].
