Edit My original answer was wrong. The pressure is in fact constant; if there were a gradient, then the layer between $z$ and $z+dz$ would have a net force on it, and the gas would not be in a steady state. This is in contrast to the fact that in a gravitational field, there must be a net force on such a layer that counteracts gravity in the steady state as you indicated. I should not get credit for this observation; see this question I just posted: Ideal gas temperature and pressure gradients?
On another note however, are you sure that the temperature gradient would be linear as you have indicated? This would be true if ideal gases had a constant thermal conductivity, but as far as I can tell according to these notes, the thermal conductivity of an idea gas scales as the square root of temperature; $k=\alpha\sqrt{T}$ in which case by Fourier's Law one gets that the temperature gradient in the $z$-direction is
$$
T(z) = \left[T_1^{3/2}+(T_2^{3/2}-T_1^{3/2})\frac{z}{L}\right]^{2/3}
$$
Moreover, now I'm curious to know where my first argument about chemical potential breaks down; I guess the assumption about diffusive equilibrium doesn't hold in this case.
Original (Incorrect) Answer
Cool question! Anytime one wants to compute concentration gradients, the first thing that comes to my mind is the chemical potential since it is associated with particle number; when diffusive equilibrium is achieved, each "infinitesimal" gas layer is in equilibrium with the next which means that the chemical potential as a function of $z$ is a constant.
$$
\mu(z) = \mu(0).
$$
On the other hand, the chemical potential of an ideal gas (according to Kittel & Kroemer) is
$$
\mu =kT\ln\left(\frac{n}{n_Q}\right)
$$
where $n$ is the concentration $n_Q$ is the so-called *quantum concentration" and is defined as
$$
n_Q = \left(\frac{m k T}{2\pi\hbar^2}\right)^{3/2}
$$
and $m$ is the molecular mass. In your setup, the temperature is a function of $z$, which makes the quantum concentration a function of $z$, and so is the concentration. By plugging the expression for the chemical potential into the condition for diffusive equilibrium, we obtain the following equation for $n(z)$:
$$
k T(z)\ln\left(\frac{n(z)}{n_Q(z)}\right) =\mu(0)
$$
Whose solution, after plugging in the explicit expressions for $n_Q(z)$ and $\mu(0)$ is
$$
\boxed{n(z) = \left(\frac{m k \,T(z)}{2\pi\hbar^2}\right)^{3/2}\exp\left[\frac{\mu(0)}{ k\,T(z)}\right]}
$$
Barring a massive conceptual error, I think this is call correct. It's also nice cause it apparently applies to any temperature gradient $T(z)$. Please tell me of any errors (conceptual or otherwise) if you think of any! Yay thermo!
Cheers!
The distribution of speeds in an ideal gas is given by the Maxwell-Boltzmann distribution. There are a variety of average speeds e.g. the most probable speed, the mean speed and the root mean square speed. Which one you use will depend on the application. The two equations you give are for the RMS speed:
$$ \sqrt{\langle v^2 \rangle} = \sqrt{\frac{3kT}{m}} $$
and the mean speed:
$$ \langle v \rangle = \sqrt{\frac{8kT}{\pi m}} $$
The question doesn't make it clear what speed is required, but I would guess that it is the mean.
Best Answer
It's a steady state. If there were a pressure gradient, there would be net force on the gas (ignoring gravity). There's no net force here because the air isn't accelerating. Thus the pressure is constant.
The number density varies across the box inversely to the temperature so the ideal gas law holds.