Chapter 10, Problem 67GP

To determine

To Calculate:The approximate difference in air pressure.

Answer to Problem 67GP

The approximate difference in air pressure is $0.047\text{atm}$ .

Explanation of Solution

Given:

Height of the empire state building $=380\text{m}$ .

Formula used:

Pressure difference is given by,

  $\Delta P=\rho gh$

(density of atmosphere $=$$1.2{\text{kg/m}}^{\text{3}}$ )

And,

Standard pressure in pascal is $1.013\times {10}^5\text{Pa}$ .

Calculation:

  $\begin{array}{l}\frac{\Delta P}{P_0}=\frac{\left(\text{1}\text{.29}\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.\right)\text{×}\left(\text{9}\text{.8}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.\right)\text{×380m}}{{\text{P}}_{\text{0}}}\\ \text{=}\frac{\left(\text{1}\text{.29}\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.\right)\text{×}\left(\text{9}\text{.8}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.\right)\text{×380m}}{\text{1}{\text{.013×10}}^{\text{5}}\text{Pa}}\\ =0.047\end{array}$

Conclusion:

The change in pressure is, $\Delta P=0.047\text{atm}$