Deriving hydrostatic pressure equation for an ideal gas

[jdawg]

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]

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!