# Ideal gas temperature and pressure gradients?

Consider an ideal gas in a $d\times d\times L$ box with the $L$ dimension in the $x$-direction. Suppose that the opposite $d\times d$ sides of the box are held at temperatures $T_1$ and $T_2$ with $T_2>T_2$ and that the system reaches a steady state. 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 $x$-direction is $$T(x) = \left[T_1^{3/2}+(T_2^{3/2}-T_1^{3/2})\frac{x}{L}\right]^{2/3}$$ What is the corresponding pressure gradient $P(x)$ in the steady state?

-

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.

-
Wow why didn't I think of this? Thanks Mark. –  joshphysics Feb 6 '13 at 1:08