# Deriving hydrostatic pressure equation for an ideal gas

Assume: Hydrostatic Situation, ideal gas
Use empirical formula T = T0 - Bz

I have rechecked my work several times and can't seem to find a mistake. The answer is supposed to be:
p2 = p1((T0 - Bz)/T0)(g/Bz)

dp/dz = -ρg Now substitute ideal gas equation for ρ
dp = -(p/RT)g dz
∫(1/p) dp = (-g/R) ∫(1/(T0-Bz) dz After substituting T = T0 - Bz ... Integrate

let u = T0 - Bz
du = -B dz

ln(p2/p1) = (g/BR) ∫(1/u) du
ln(p2/p1) = (g/BR)(ln(u))

ln(p2/p1) = (g/BR)(ln(T0-Bz))
p2 = p1(T0 - Bz)(g/BR)

Where are they getting that T0 on the bottom??

Chestermiller
Mentor
Assume: Hydrostatic Situation, ideal gas
Use empirical formula T = T0 - Bz

I have rechecked my work several times and can't seem to find a mistake. The answer is supposed to be:
p2 = p1((T0 - Bz)/T0)(g/Bz)

dp/dz = -ρg Now substitute ideal gas equation for ρ
dp = -(p/RT)g dz
∫(1/p) dp = (-g/R) ∫(1/(T0-Bz) dz After substituting T = T0 - Bz ... Integrate

let u = T0 - Bz
du = -B dz

ln(p2/p1) = (g/BR) ∫(1/u) du
ln(p2/p1) = (g/BR)(ln(u))

ln(p2/p1) = (g/BR)(ln(T0-Bz))
p2 = p1(T0 - Bz)(g/BR)

Where are they getting that T0 on the bottom??
You forgot to substitute the lower limit of integration in the right hand side of the equation.

jdawg
THANK YOU, can't believe I forgot to do that!