## Saturday, March 05, 2011

### Temperature and Pressure Gradient in Ideal Gas

Introduction

Consider a narrow horizontal tube. The curved surface made with perfect insulator. The flat faces made with perfect heat conductor. One end of the tube is maintained at 100 oC and the other end is maintained at 0 oC. Once the steady state is attained, heat continue to flow from hot end to cold end. A temperature gradient is attained in the gas as the heat flows through it.
Q1. Is the pressure in the gas in tube uniform?
Q2. Is the density in the gas in tube uniform? Else how does it vary?

Derivation

Consider a cross section of the tube a position z=z0.
Assume at z = z0 mean free path of molecules is l

At z = z(0-l):
Density is ρ1 and
Average molecular velocity is c1
Temperature is T1

At z = z(0+l):
Density is ρ2 and
Molecular velocity is c2
Temperature is T2
The number of molecules crossing a unit area in the cross section at z=z0 from left side is:
1/4ρ1c1
The number of molecules crossing from right side is:
1/4ρ2c2
The number molecules crossing from both side has to be equal to maintain the conservation of mass.
Hence ρ1c1 = ρ2c2
That is pressure * molecular velocity is constant
ρc = A
T is proportional to the square of c. Hence:
ρ1sqrt(T1) = ρ2sqrt(T2)
Above equation is equivalent to equation for Thermal transpiration. However from the above argument, it can found that this is equation is valid for any steady state system with temperature or pressure gradient.

Other derivations
ρ1/ρ2 = sqrt(T2/T1) OR
ρ/ρ0 = sqrt(T0/T)
pressure is proportional to ρc^2
p ~ ρc^2
Applying equation (ρc = A) in above equation:
pressure is proportional to c
p ~ c

p1/p2 = c1/c2
c is proportional to square root of temperature
p1/p2 = sqrt(T1/T2) OR
p/p0 = sqrt(T/T0)

Implications
Hot air balloon
Why a hot air balloon is stiff? (see the photograph; courtesy skywaltz)

It is stiff not because of the low density inside. Had it been the case if we fill the balloon with hydrogen (with opening as for hot air balloon), it would have been stiff (Any verification on this?).
My friend Prem suggested that the stiffness may be due to the buoyancy. The upper part of the balloon can be stiff due the buoyancy. However stiffness of all wiered shaped balloons ( http://www.xarj.net/wp-content/uploads/2009/03/crazy-hot-air-balloon14.jpg ) cannot be explained by buoyancy.
It is stiff because of the high pressure inside. Pressure is high due to the high temperature inside.
However is the pressure inside high enough according to the equation (4)? Not really!
Assume the inside temperature is 100 oC (373 K). Then:
Assume outside temperature is 20 oC (293 K) and pressure is 1 kg/cm^2
Inside pressure p = sqrt(373/293) * 1
p = 1.13 kg/cm^2
That is a pressure difference of 0.13 kg/cm^2
That means a force of 1300 kg per square meter of balloon cloth. No balloon cloth can withstand that kind of force considering the area of balloon is many square meters.
Why the pressure inside lower than expected?
I have two hypothesis
(a) Heat convection - Our assumption was heat transfer is entirely through conduction. If heat transfer via convection instead of conduction, then the gas inside the balloon instead of having continuous steady state temperature gradient, there will be macro level big pieces of air cloud with different temperature and density. Heat transfer shall be through movement of such air clouds rather than through conduction. in such case there will not be any pressure build up. We can consider the air inside the balloon as a collection of air clouds (like pebbles in a bag).
In really both conduction and convection contributes and hence there will be some pressure difference (which is much less than otherwise it would have been).
(b) Quantum mechanical non local potential energy reduction - If you consider glass tube in our discussion, the pressure inside is only a local optimization. By re-ordering the molecules, another state with same temperature gradient but with no pressure gradient is possible. However there is no incremental ways to achieve that state. If we consider molecular distribution of molecules as a superimposition of all possible distributions, by a non local state reduction of molecules, pressure reduction is possible. Hence the actual pressure difference shall be less than predicted one and it will be depends on the amount of quantum mechanical state reductions.
Is both (a) and (b) are related? How is heat convection explained in kinetic theory?
Also I think as the opening at the bottom of the balloon is narrow, the pressure difference shall be high (for the same temperature inside) (Needs verification). Narrow opening will prevent heat convection and non local state reductions.
Experimental verifications
To be done.
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