Certainly! Let's address each part of the problem step-by-step, using the Maxwell velocity distribution for a gas of molecules in thermal equilibrium.
### Part (a) [tex]\(\overline{v_x^2}\)[/tex]
The Maxwell velocity distribution for a single component [tex]\(v_x\)[/tex] is given by:
[tex]\[ f(v_x) = \sqrt{\frac{m}{2\pi kT}} \exp\left(-\frac{mv_x^2}{2kT}\right) \][/tex]
We need to find [tex]\(\overline{v_x^2}\)[/tex]:
[tex]\[ \overline{v_x^2} = \int_{-\infty}^\infty v_x^2 f(v_x) \, dv_x \][/tex]
Substitute [tex]\( f(v_x) \)[/tex] into the integral:
[tex]\[ \overline{v_x^2} = \int_{-\infty}^\infty v_x^2 \sqrt{\frac{m}{2\pi kT}} \exp\left(-\frac{mv_x^2}{2kT}\right) dv_x \][/tex]
Let's simplify by using a substitution. Let [tex]\( u = \sqrt{\frac{m}{2kT}} v_x \)[/tex], hence [tex]\( dv_x = \sqrt{\frac{2kT}{m}} du \)[/tex]:
[tex]\[ \overline{v_x^2} = \int_{-\infty}^\infty \left( \frac{2kT}{m} u^2 \right) \frac{1}{\sqrt{\pi}} \exp(-u^2) \sqrt{\frac{2kT}{m}} du \][/tex]
[tex]\[ \overline{v_x^2} = \frac{2kT}{m} \int_{-\infty}^\infty \frac{u^2}{\sqrt{\pi}} \exp(-u^2) du \][/tex]
The integral [tex]\(\int_{-\infty}^\infty \frac{u^2}{\sqrt{\pi}} \exp(-u^2) du\)[/tex] is a known result that evaluates to [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ \overline{v_x^2} = \frac{2kT}{m} \cdot \frac{1}{2} = \frac{kT}{m} \][/tex]
### Part (b) [tex]\(\overline{v^2 v_x}\)[/tex]
Here, [tex]\( v \)[/tex] is the magnitude of the velocity vector, given by [tex]\( v = \sqrt{v_x^2 + v_y^2 + v_z^2} \)[/tex]. Using the same distribution, we express the distribution in three dimensions:
[tex]\[ f(v_x, v_y, v_z) = \left(\frac{m}{2\pi kT}\right)^{3/2} \exp\left(-\frac{m(v_x^2 + v_y^2 + v_z^2)}{2kT}\right) \][/tex]
To find [tex]\(\overline{v^2 v_x}\)[/tex]:
[tex]\[ \overline{v^2 v_x} = \int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty v^2 v_x f(v_x, v_y, v_z) dv_x dv_y dv_z \][/tex]
Let's use spherical coordinates ([tex]\(v, \theta, \phi\)[/tex]) where [tex]\(v_x = v \cos \theta\)[/tex]:
[tex]\[ v^2 = v^2 \text{ and } dv_x dv_y dv_z = v^2 \sin \theta dv d\theta d\phi \][/tex]
[tex]\[ \overline{v^2 v_x} = \int_{0}^\infty \int_{0}^\pi \int_{0}^{2\pi} v^2 (v \cos\theta) f(v) v^2 \sin \theta d\phi d\theta dv \][/tex]
Integrating with respect to [tex]\(\phi\)[/tex] and normalizing factors:
[tex]\[ \overline{v^2 v_x} = \int_{0}^\infty \int_{0}^\pi v^3 \cos \theta \left( v^2 \sin \theta \right) \left( \frac{m}{2\pi kT} \right)^{3/2} \exp \left( - \frac{m v^2}{2 k T} \right) d\theta dv \][/tex]
The [tex]\(\phi\)[/tex] integral yields [tex]\(2\pi\)[/tex]. The [tex]\(\cos\theta\cdot\sin\theta\)[/tex] integral will yield zero because cosine is an odd function integrated over a symmetric interval around zero. Hence,
[tex]\[ \overline{v^2 v_x} = 0 \][/tex]
### Part (c) [tex]\(\overline{v_x^2 v_y^2}\)[/tex]
Using the symmetry in the Maxwell-Boltzmann distribution and recognizing independence of the velocity components:
[tex]\[ \overline{v_x^2 v_y^2} \][/tex]
The calculation involves the product of two independent distributions:
[tex]\[ \overline{v_x^2 v_y^2} = \left( \overline{v_x^2} \right) \left( \overline{v_y^2} \right) \][/tex]
Given that both [tex]\( \overline{v_x^2} \)[/tex] and [tex]\( \overline{v_y^2} \)[/tex] equal [tex]\(\frac{kT}{m}\)[/tex]:
[tex]\[ \overline{v_x^2 v_y^2} = \left( \frac{kT}{m} \right) \left( \frac{kT}{m} \right) = \left( \frac{kT}{m} \right)^2 \][/tex]
In summary:
1. [tex]\(\overline{v_x^2} = \frac{kT}{m}\)[/tex]
2. [tex]\(\overline{v^2 v_x} = 0 \)[/tex]
3. [tex]\(\overline{v_x^2 v_y^2} = \left( \frac{kT}{m} \right)^2\)[/tex]
