Chapter 2, Problem 2.3P

Interpretation Introduction

Interpretation: Assuming the ideal gas behavior, the height at a particular pressure is to be determined by applying the hydrostatic equilibrium equation and thus finding the error from previous linked question.

Concept introduction: The hydrostatic equilibrium equation can be applied in order to determine the change in pressure for the change in height for a given fluid.

The hydrostatic equilibrium equation can be given as,

  ${\text{P}}_{\text{a}}{\text{-P}}_{\text{b}}{\text{= ρg(Z}}_{\text{a}}{\text{-Z}}_{\text{b}}\text{)}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}$ ……. (1)

  $\begin{array}{l}{\text{P}}_{\text{a}}\text{}\text{= }\text{}\text{Pressure }\text{}\text{at }\text{}\text{the }\text{}\text{given point A}\\ {\text{P}}_{\text{b}}\text{}\text{= }\text{}\text{Pressure }\text{}\text{at }\text{}\text{the }\text{}\text{given point B}\\ \text{g = Acceleration due to gravity = 9}{\text{.8 m/s}}^{\text{2}}\\ {\text{Z}}_{\text{a}}\text{ =Elevation }\text{}\text{at }\text{}\text{point A}\\ {\text{Z}}_{\text{b}}\text{ =Elevation }\text{}\text{at }\text{}\text{point B}\end{array}$

The density, $\text{ρ}$ in equation (1) can be written as,

  $\text{ρ = }\frac{{\text{P}}_{\text{avg}}\text{M}}{\text{RT}}$ ……. (2)

  $\begin{array}{l}{\text{P}}_{\text{avg}}\text{=Average pressure of air}\\ \text{M=Molecular weight of air}\\ \text{R=Universal gas constant }\text{}\text{= 8}\text{.314 J/mol}\text{.K}\\ \text{T=Temperature of earth's atmosphere}\\ \end{array}$