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Chapter 10: Problem 20

Perform the following conversions: \((\mathbf{a})\) 0.912 atm to torr, \((\mathbf{b})\) 0.685 bar to kilopascals, \((\mathbf{c})\) 655 \(\mathrm{mm}\) Hg to atmospheres, \((\mathbf{d})\) \(1.323 \times 10^{5}\) Pa to atmospheres, \((\mathbf{e})\) 2.50 atm to psi.

Short Answer

Expert verified
The short answer for the conversions is as follows: a) \(0.912\ \text{atm} = 693.12\ \text{Torr}\) b) \(0.685\ \text{bar} = 68.5\ \text{kPa}\) c) \(655\ \text{mm}\ \text{Hg} = 0.8618\ \text{atm}\) d) \(1.323 \times 10^{5}\ \text{Pa} = 1.306\ \text{atm}\) e) \(2.50\ \text{atm} = 36.74\ \text{psi}\)

Step by step solution

01

a) Convert 0.912 atm to Torr

To convert 0.912 atm to Torr, we'll use the conversion factor 1 atm = 760 Torr: \(0.912\ \text{atm}\times \dfrac{760\ \text{Torr}}{1\ \text{atm}} = 693.12\ \text{Torr}\)
02

b) Convert 0.685 bar to kilopascals

To convert 0.685 bar to kilopascals, we'll use the conversion factor 1 bar = 100 kPa: \(0.685\ \text{bar}\times \dfrac{100\ \text{kPa}}{1\ \text{bar}} = 68.5\ \text{kPa}\)
03

c) Convert 655 mm Hg to atmospheres

To convert 655 mm Hg to atmospheres, we'll use the conversion factor 1 atm = 760 Torr (or mmHg): \(655\ \text{mm}\ \text{Hg}\times \dfrac{1\ \text{atm}}{760\ \text{mm}\ \text{Hg}} = 0.8618\ \text{atm}\)
04

d) Convert 1.323 x 10^5 Pa to atmospheres

To convert \(1.323 \times 10^{5}\) Pa to atmospheres, we'll use the conversion factor 1 atm = 101325 Pa: \(1.323 \times 10^{5}\ \text{Pa}\times \dfrac{1\ \text{atm}}{101325\ \text{Pa}} = 1.306\ \text{atm}\)
05

e) Convert 2.50 atm to psi

To convert 2.50 atm to psi, we'll use the conversion factor 1 atm = 14.696 psi: \(2.50\ \text{atm}\times \dfrac{14.696\ \text{psi}}{1\ \text{atm}} = 36.74\ \text{psi}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Atmospheres
Atmospheres, often abbreviated as 'atm,' is a unit of pressure that measures how much force is exerted by the weight of air at sea level. One atmosphere is equivalent to the pressure of 101,325 pascals. This unit is commonly used when discussing weather, altitudes, and underwater depths. For example:

  • One standard atmosphere is used to define the boiling point of water, which is 100掳C (212掳F) at sea level.
  • At 10 meters underwater, the pressure is approximately 2 atm due to the weight of the water above.
Understanding atmospheres is important for its role in atmospheric science, diving, and aviation. It's also frequently used in chemistry for gas laws calculations.
Exploring Torr
Torr is another unit of pressure, named after the Italian physicist Evangelista Torricelli, who invented the barometer. One torr is equivalent to 1/760th of an atmosphere, or exactly one millimeter of mercury (mmHg). This makes conversion between torr and mmHg direct and simple.

Torr is commonly used in fields such as vacuum science and physics, and understanding it helps in comparing different pressure levels. For example:

  • The measurement of a perfect vacuum is 0 torr.
  • Standard atmospheric pressure is 760 torr.
Converting from atmospheres to torr involves multiplying by 760. For practical purposes, knowing this conversion factor can help perform quick pressure conversions.
Insights on Kilopascals
Kilopascals (kPa) are part of the metric system and represent a larger scale of pascals. One kilopascal equals 1,000 pascals. The kPa is widely accepted in scientific and engineering contexts because of its ease of communication and alignment with SI units.

  • Atmospheric pressure at sea level is approximately 101.3 kPa.
  • A human's blood pressure is measured in mmHg, but it can also be converted to kPa for those familiar with metric units.
Often in physics and engineering, kilopascals are used to describe pressure in various systems, such as tire pressures and weather stations. It's practical to convert between kilopascals and other units like bar or atmospheres, facilitating broader understanding across different domains.
Millimeters of Mercury (mmHg) Details
Millimeters of mercury, shortened to mmHg, is a unit that measures pressure employing the height of a column of mercury. It is mostly used in the medical field for blood pressure readings, and its value matches that of torr.

  • Normal blood pressure is defined as about 120/80 mmHg.
  • Weather barometers often use mmHg to measure atmospheric pressure.
Knowing that 1 atm equals 760 mmHg allows for easy conversion between these units. This is especially helpful in medicine and meteorology, where precise pressure readings are crucial for interpreting conditions and outcomes.
Psi and its Applications
Psi, standing for pounds per square inch, is a unit of pressure commonly used in the United States, especially in plumbing, car tires, and other mechanical systems. It represents the force of one pound-force applied to an area of one square inch.

  • Normal car tire pressures are often in the range of 30-35 psi.
  • Pressure washers may specify pressure outputs in psi for easy understanding of their power.
The conversion factor from atmospheres to psi is approximately 14.696 psi per atmosphere. Understanding this conversion helps in adjusting measurements taken in different units, ensuring accurate and useful data across various domestic and technical applications.

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Most popular questions from this chapter

(a) The compound 1-iodododecane is a nonvolatile liquid with a density of 1.20 \(\mathrm{g} / \mathrm{mL}\) . The density of mercury is 13.6 \(\mathrm{g} / \mathrm{mL} .\) What do you predict for the height of a barometer column based on 1 -iodododecane, when the atmospheric pressure is 749 torr? (b) What is the pressure, in atmospheres, on the body of a diver if he is 21 ft below the surface of the water when the atmospheric pressure is 742 torr?

Both Jacques Charles and Joseph Louis Guy-Lussac were avid balloonists. In his original flight in \(1783,\) Jacques Charles used a balloon that contained approximately \(31,150\) L of \(\mathrm{H}_{2}\) . He generated the \(\mathrm{H}_{2}\) using the reaction between iron and hy- drochloric acid: $$\mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ How many kilograms of iron were needed to produce this volume of \(\mathrm{H}_{2}\) if the temperature was \(22^{\circ} \mathrm{C}\) ?

The planet Jupiter has a surface temperature of 140 \(\mathrm{K}\) and a mass 318 times that of Earth. Mercury (the planet) has a surface temperature between 600 \(\mathrm{K}\) and 700 \(\mathrm{K}\) and a mass 0.05 times that of Earth. On which planet is the atmosphere more likely to obey the ideal-gas law? Explain.

Gaseous iodine pentafluoride,\(\mathrm{IF}_{5}\) can be prepared by the reaction of solid iodine and gaseous fluorine: $$\mathrm{I}_{2}(s)+5 \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{IF}_{5}(g)$$ A 5.00 -L flask containing 10.0 g of \(\mathrm{I}_{2}\) is charged with 10.0 \(\mathrm{g}\) of \(\mathrm{F}_{2},\) and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is \(125^{\circ} \mathrm{C}\) . (a) What is the partial pressure of \(\mathrm{IF}_{5}\) in the flask? (b) What is the mole fraction of \(\mathrm{IF}_{5}\) in the flask (c) Draw the Lewis structure of \(\mathrm{IF}_{5}\). (d) What is the total mass of reactants and products in the flask?

The atmospheric concentration of \(\mathrm{CO}_{2}\) gas is presently 407 \(\mathrm{ppm}(\) parts per million, by volume; that is, 407 \(\mathrm{L}\) of every \(10^{6} \mathrm{L}\) of the atmosphere are \(\mathrm{CO}_{2}\) . What is the mole fraction of \(\mathrm{CO}_{2}\) in the atmosphere?
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