Chapter 10, Problem 10.1P

Interpretation Introduction

Interpretation:

Local barometric pressure is 28.48 inches of mercury, the pressure in units of $\text{atm}$ , $\text{bar}$ and $\text{Pa}$ should be calculated.

Concept introduction:

Following conversion factors will be used to calculate the pressure in units of $\text{atm}$ , $\text{bar}$ and $\text{Pa}$.

  $\frac{\text{1 atm}}{\text{760 mm Hg}},\frac{1.01325\text{ bar}}{1\text{ atm}}\text{ and }\frac{101,325\text{ Pa}}{76\text{0 mm Hg}}$

Answer to Problem 10.1P

Solution:

28.48 inches of mercury will have $\text{0}\text{.9518 atm}$ , $\text{0}\text{.9644 bar}$ and $\text{9}{\text{.645×10}}^{\text{4}}\text{ Pa}$ pressure.

Explanation of Solution

The given units are related with each other as shown below:

Given barometric pressure is 28.48 inches of mercury (28.48 in Hg)
First calculating in units of mm Hg

  $28.48\text{ in Hg}\times \frac{2.54\text{ cm}}{1\text{ in}}\text{×}\frac{\text{10 mm}}{\text{1 cm}}=7\text{23}\text{.4 mm Hg}$

Multiply the barometric pressure $7\text{23}\text{.4 mm Hg}$ with the conversion factor $\frac{\text{1 atm}}{\text{760 mm Hg}},\frac{1.01325\text{ bar}}{1\text{ atm}}\text{ and }\frac{101,325\text{ Pa}}{76\text{0 mm Hg}}$

  $7\text{23}\text{.4 mm Hg}\times \frac{\text{1 atm}}{\text{760 mm Hg}}=\text{ 0}\text{.9518 atm}$

Therefore, barometric pressure is $\text{0}\text{.9518 atm}$

Multiplying $\text{0}\text{.9518 atm}$ with the conversion factor $\frac{1.01325\text{ bar}}{1\text{ atm}}$

  $\text{0}\text{.9518 atm }\times \frac{1.01325\text{ bar}}{1\text{ atm}}\text{= 0}\text{.9644 bar}$

Multiply the barometric pressure $7\text{23}\text{.4mm Hg}$ with the conversion factor $\frac{101,325\text{ Pa}}{76\text{0 mm Hg}}$

  $\text{723}\text{.4 mm Hg × }\frac{\text{101,325 Pa}}{\text{760 mm Hg}}\text{ = 9}{\text{.645×10}}^{\text{4}}\text{ Pa}$

Therefore, the barometric pressure is $\text{9}{\text{.645×10}}^{\text{4}}\text{ Pa}$.

Conclusion

28.48 inches of mercury will have $\text{0}\text{.9518 atm}$ , $\text{0}\text{.9644 bar}$ and $\text{9}{\text{.645×10}}^{\text{4}}\text{ Pa}$ pressure.