Question 788870
{{{P[h] = P[0]*e^"(-mgh/RT)"}}}
{{{P[h]/P[0]=e^"(-mgh/RT)"}}} or {{{P[0]/P[h]=e^"(mgh/RT)"}}}
{{{ln(P[h]/P[0])=-mgh/RT}}} or {{{ln(P[0]/P[h])=mgh/RT}}}
{{{ln(1013.25/900)=mgh/RT}}} --> {{{h=(RT/mg)*ln(1013.25/900)}}}
Assuming that {{{T}}} is the temperature at sea level, or it does not change much from 0 to h meters above sea level,
{{{h=(8.314462*6.5/(0.0289644*9.80665))*ln(1013.25/900)}}}  gives us the height in meters (because those would be the units of {{{RT/mg}}}} with, R, T, m, and g in the units given).
I do not have a temperature, so I will assume {{{20^oC}}} (293 kelvins)
{{{h=(8.314462*293/(0.0289644*9.80665))*ln(1013.25/900)=1017)}}}
So the height is about 1,000 meters.