Average velocity of particles and maxwell distribution

[KFC]

Consider many same particles are moving randomly in a constant pressure and constant temperature box. The average speed $$\overline{v}$$ of the particle can be calculated by maxwell distribution. Now assume the density of particles in the box is $$n$$, so in unit time, total number of particle hitting unit area of a wall is given by

$$\frac{1}{4}n\overline{v}$$

I am thinking this problem, actually, we are considering a small volume in front of the wall being hit. The cross area of region is unit while the length of that region is $$d$$, so in time $$t$$

$$\frac{d}{t} \approx \overline{v}$$

Since not all particle inside that region will hit the wall, only those moving towards the wall will do. Well if we consider each particle in that region might have velocity along x, -x, y, -y, z, -z direction. So only one-eighth of them will along the direction towards the wall, so the expression we get should be

$$\frac{1}{8}n\overline{v}$$

but why in many book, it is 1/4 instead?

 

[paweld]

I am thinking this problem, actually, we are considering a small volume in front of the wall being hit. The cross area of region is unit while the length of that region is $$d$$, so in time $$t$$

$$\frac{d}{t} \approx \overline{v}$$

Since not all particle inside that region will hit the wall, only those moving towards the wall will do. Well if we consider each particle in that region might have velocity along x, -x, y, -y, z, -z direction. So only one-eighth of them will along the direction towards the wall, so the expression we get should be

$$\frac{1}{8}n\overline{v}$$

but why in many book, it is 1/4 instead?

Your reasoning isn't proper. Let z be direction normal to the surface element. On average $$(S \overline{v_z} t) n$$ particles hits the surface S (it's the number of particles in a cuboid with base surface S and height $$\overline{v_z} t$$; $$\overline{v_z}$$ is average veliocity in z direction of particles moving towards considered surface element; components in other direction are unimportant because on average as much particles leave cuboid as come into throught its side faces; it's not true that only one-eighth is moving towards surface). So $$\overline{v_z} n$$ particles hits unit surface element per unit time.

All we need to find now is the dependence between $$\overline{v_z}$$ and $$\overline{v}$$. In order to do it we use spherical coordinates. Let $$\theta$$ be the angle between z axis and $$\vec{v}$$ and $$\phi$$ be the angle between the projection of $$\vec{v}$$ on xy plane and x axis. Because all direction are equally probable we have:
$$\overline{v_z}=\frac{\int_0^{\pi/2} \int_0^{2\pi} \overline{v} \cos\theta (\sin\theta d\theta d\phi)} {\int_0^\pi \int_0^{2\pi} \sin\theta d\theta d\phi}= \overline{v}\frac{2 \pi \int_0^{\pi/2} \sin\theta d(\sin\theta)}{4 \pi}=\frac{1}{4} \overline{v}$$
In first integral we integrate $$\theta$$ form $$0$$ to $$\pi/2$$ because
we consider only particles moving towards our surface.
$$(\sin\theta d\theta d\phi)$$ is an infitesimal solid angle.

 

[KFC]

Your reasoning isn't proper. Let z be direction normal to the surface element. On average $$(S \overline{v_z} t) n$$ particles hits the surface S (it's the number of particles in a cuboid with base surface S and height $$\overline{v_z} t$$; $$\overline{v_z}$$ is average veliocity in z direction of particles moving towards considered surface element; components in other direction are unimportant because on average as much particles leave cuboid as come into throught its side faces; it's not true that only one-eighth is moving towards surface). So $$\overline{v_z} n$$ particles hits unit surface element per unit time.

All we need to find now is the dependence between $$\overline{v_z}$$ and $$\overline{v}$$. In order to do it we use spherical coordinates. Let $$\theta$$ be the angle between z axis and $$\vec{v}$$ and $$\phi$$ be the angle between the projection of $$\vec{v}$$ on xy plane and x axis. Because all direction are equally probable we have:
$$\overline{v_z}=\frac{\int_0^{\pi/2} \int_0^{2\pi} \overline{v} \cos\theta (\sin\theta d\theta d\phi)} {\int_0^\pi \int_0^{2\pi} \sin\theta d\theta d\phi}= \overline{v}\frac{2 \pi \int_0^{\pi/2} \sin\theta d(\sin\theta)}{4 \pi}=\frac{1}{4} \overline{v}$$
In first integral we integrate $$\theta$$ form $$0$$ to $$\pi/2$$ because
we consider only particles moving towards our surface.
$$(\sin\theta d\theta d\phi)$$ is an infitesimal solid angle.

Thanks for your reply. I didn't see the reply until now, sorry :(

You reasoning is quite mathematics. I am not completely understand how do you get obtain the integral and why do you need to divide it by another integral?