Pressure variations in a gas​
Housam Binous and Ahmed Bellagi

Assuming ideal gas, the rate of change of air pressure, p, with elevation, z, obeys the following governing equation:​
P
z
=-ρg=
-29P
RT
g
​​where R = 8314 and g = 9.81.​Standard atmospheric pressure is taken equal to 101 300 Pa.​1/ Assuming that the temperature is independent of altitude and equal to 15 °C or T0=288.15 K, what would be the standard pressure at an altitude equal to 3000 meters expressed in kPa? and in mmH2O? ​2/ Assuming that the temperature is a function of altitude,
T(z)=
T
0
-B×z
with
T
0
=288.15
K and
B=0.00650 K/m
, what would be the standard pressure at an altitude equal to 3000 meters expressed in kPa? ​​References:​[1] F. M. White, Fluid Mechanics, 2nd edition, NY: McGraw-Hill, 1986.[2] J. O. Wilkes, Fluid Mechanics for Chemical Engineers, 2nd edition, NJ: Prentice Hall, 2006.
In[]:=
DSolve[{p'[z]==-29g/R/T0p[z],p[0]==101.300},p[z],z]
Out[]=
p[z]101.3
-
29gz
RT0


In[]:=
P=101.3
-
29gz
RT0

(*inkPa*)
Out[]=
101.3
-
29gz
RT0

In[]:=
z=3000;T0=288.15;R=8314;g=9.81;
In[]:=
P(*inkPa*)
Out[]=
70.9399
In[]:=
P1000/1000/9.811000(*inmmH2O*)
Out[]=
7231.38
In[]:=
Clear[z,T0,R,g]
In[]:=
T[z_]:=T0-Bz
In[]:=
DSolve[{p'[z]==-29g/R/T[z]p[z],p[0]==101.300},p[z],z]
Out[]=
p[z]101.3
-
29.g
BR
(RT0)
29g
BR
(RT0-BRz)

In[]:=
P=101.3
-
29.g
BR
(RT0)
29g
BR
(RT0-BRz)
Out[]=
101.3
-
29.g
BR
(RT0)
29g
BR
(RT0-BRz)
In[]:=
z=3000;T0=288.15;B=0.0065;R=8314;g=9.81;
P(*inkPa*)
Out[]=
70.0497