# 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!