Deriving hydrostatic pressure equation for an ideal gas

  • Thread starter jdawg
  • Start date
  • #1
367
2
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??
 

Answers and Replies

  • #2
21,406
4,816
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.
 
  • #3
367
2
THANK YOU, can't believe I forgot to do that!
 

Related Threads on Deriving hydrostatic pressure equation for an ideal gas

Replies
11
Views
1K
Replies
1
Views
618
  • Last Post
Replies
1
Views
496
  • Last Post
Replies
6
Views
810
  • Last Post
Replies
3
Views
958
  • Last Post
Replies
7
Views
921
Replies
13
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
752
D
Top