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

Short Answer

Expert verified
The short answers for the conversions are: (a) \(692.12 \mathrm{~torr}\), (b) \(68.5 \mathrm{~kPa}\), (c) \(0.8618 \mathrm{~atm}\), (d) \(1.305 \mathrm{~atm}\), and (e) \(36.74 \mathrm{~psi}\).

Step by step solution

01

Identify the conversion factor

In this case, we need the conversion factor for atm to torr, which is 1 atm = 760 torr.
02

Perform the conversion

To convert 0.912 atm to torr, we simply multiply by the conversion factor: \(0.912 \mathrm{~atm} \times \frac{760 \mathrm{~torr}}{1 \mathrm{~atm}} = 692.12 \mathrm{~torr}\) #b) Convert 0.685 bar to kilopascals#
03

Identify the conversion factors

In this case, we need the conversion factors for bar to Pascal (Pa) and Pascal to kilopascals (kPa), which are 1 bar = 10^5 Pa and 1 kPa = 10^3 Pa.
04

Perform the conversion

To convert 0.685 bar to kPa, we first convert bar to Pa, and then convert Pa to kPa: \(0.685 \mathrm{~bar} \times \frac{10^5 \mathrm{~Pa}}{1 \mathrm{~bar}} \times \frac{1 \mathrm{~kPa}}{10^3 \mathrm{~Pa}} = 68.5 \mathrm{~kPa}\) #c) Convert 655 mm Hg to atmospheres#
05

Identify the conversion factor

In this case, we need the conversion factor for mm Hg to atm, which is 1 atm = 760 mm Hg.
06

Perform the conversion

To convert 655 mm Hg to atm, we simply divide by the conversion factor: \(\frac{655 \mathrm{~mm} \mathrm{Hg}}{760 \mathrm{~mm} \mathrm{Hg} \cdot 1 \mathrm{~atm}} = 0.8618 \mathrm{~atm}\) #d) Convert 1.323 x 10^5 Pa to atmospheres#
07

Identify the conversion factor

In this case, we need the conversion factor for Pa to atm, which is 1 atm = 1.01325 x 10^5 Pa.
08

Perform the conversion

To convert 1.323 x 10^5 Pa to atm, we divide by the conversion factor: \(\frac{1.323 \times 10^{5} \mathrm{~Pa}}{1.01325 \times 10^{5} \mathrm{~Pa} \cdot 1 \mathrm{~atm}} = 1.305 \mathrm{~atm}\) #e) Convert 2.50 atm to psi#
09

Identify the conversion factor

In this case, we need the conversion factors for atm to Pa and Pa to psi, which are 1 atm = 1.01325 x 10^5 Pa and 1 psi = 6894.76 Pa.
10

Perform the conversion

To convert 2.50 atm to psi, we first convert atm to Pa, and then convert Pa to psi: \(2.50 \mathrm{~atm} \times \frac{1.01325 \times 10^{5} \mathrm{~Pa}}{1 \mathrm{~atm}} \times \frac{1 \mathrm{~psi}}{6894.76 \mathrm{~Pa}} = 36.74 \mathrm{~psi}\)

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Key Concepts

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

atm to torr
Converting atm (atmospheres) to torr is a straightforward process once you understand the relationship between these two units. Atmospheric pressure (atm) is a unit of pressure used to measure the force exerted by the atmosphere. Torr, named after the Italian physicist Evangelista Torricelli, is another unit of pressure often used in scientific contexts. The conversion factor you need here is wisely standardized: **1 atm = 760 torr**.

To convert atmospheric pressure from atm to torr, you simply multiply the value in atm by 760. For example, converting 0.912 atm to torr involves multiplying 0.912 by 760, resulting in approximately 692.12 torr. This process highlights the simplicity and directness of the conversion, making it easy to switch between these units when necessary.
bar to kilopascals
The conversion from bars to kilopascals is important in various scientific and engineering applications. Here's why: The **bar** is a metric unit of pressure, and it is commonly used in meteorology and oceanography. On the other hand, the **kilopascal (kPa)** is a SI unit favored in many scientific fields.

The conversion is based on the well-known factors: **1 bar = 100,000 Pa** (Pascals) and **1 kPa = 1,000 Pa**. First, convert the bar figure into Pascals by multiplying by 100,000. Then, convert Pascals to kilopascals by dividing by 1,000. Therefore, converting 0.685 bar to kilopascals involves calculating 0.685 imes 100,000 / 1,000 = 68.5 kPa.
  • Step 1: Bar to Pascals: multiply by 100,000.
  • Step 2: Pascals to kilopascals: divide by 1,000.
mm Hg to atm
Millimeters of Mercury (mm Hg) is a unit of pressure that originated from the measurement of atmospheric pressure via a mercury column. It's especially common in medical settings. Converting mm Hg to atm involves understanding the simple relationship: **1 atm = 760 mm Hg**.

To convert mm Hg to atm, you divide the mm Hg value by 760. For instance, transforming 655 mm Hg into atmospheres is resolved by calculating 655 divided by 760, leading to 0.8618 atm.
  • This process shows how efficiently you can convert pressure units by using division.
Remember, opting for atm or mm Hg often depends on the context, such as local barometric readings or medical purposes.
Pa to atm
Converting pascals (Pa) to atmospheres (atm) is a useful skill in fields like physics and engineering. The pascal is the SI unit of pressure, whereas the atmosphere is a more traditional metric that often relates to Earth's atmospheric pressure at sea level.

The conversion factor to remember is: **1 atm = 1.01325 x 10^5 Pa**. You determine the atm value by dividing the pressure in pascals by this conversion factor. For example, changing 1.323 x 10^5 Pa to atm involves calculating 1.323 x 10^5 divided by 1.01325 x 10^5, which yields approximately 1.305 atm. This gives you a clear way to transition between scientific contexts that may use either unit readily.
atm to psi
Converting atm to psi (pounds per square inch) is often required in industries like automotive and aeronautics where psi is more prevalent. The conversion involves a two-step process.

The factors at play are: **1 atm = 1.01325 x 10^5 Pa** and **1 psi = 6894.76 Pa**. First, convert the atm value to pascals by multiplying by 1.01325 x 10^5. Then, convert to psi by dividing by 6894.76.

For instance, converting 2.50 atm to psi means you take 2.50 x 1.01325 x 10^5 and divide by 6894.76, which results in approximately 36.74 psi. This step-by-step process ensures precise calculations, necessary for accurate pressure measurements in practical applications like tire pressure.

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Most popular questions from this chapter

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.

Chlorine dioxide gas \(\left(\mathrm{ClO}_{2}\right)\) is used as a commercial bleaching agent. It bleaches materials by oxidizing them. In the course of these reactions, the \(\mathrm{ClO}_{2}\) is itself reduced. (a) What is the Lewis structure for \(\mathrm{ClO}_{2} ?\) (b) Why do you think that \(\mathrm{ClO}_{2}\) is reduced so readily? (c) When a \(\mathrm{ClO}_{2}\), molecule gains an electron, the chlorite ion, \(\mathrm{ClO}_{2}^{-},\) forms. Draw the Lewis structure for \(\mathrm{ClO}_{2}^{-}\). (d) Predict the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in the \(\mathrm{ClO}_{2}^{-}\) ion. (e) One method of preparing \(\mathrm{ClO}_{2}\) is by the reaction of chlorine and sodium chlorite: $$\mathrm{Cl}_{2}(g)+2 \mathrm{NaClO}_{2}(s) \longrightarrow 2 \mathrm{ClO}_{2}(g)+2 \mathrm{NaCl}(s)$$ If you allow \(15.0 \mathrm{~g}\) of \(\mathrm{NaClO}_{2}\) to react with \(2.00 \mathrm{~L}\) of chlorine gas at a pressure of \(152.0 \mathrm{kPa}\) at \(21^{\circ} \mathrm{C},\) how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

A \(50.0 \mathrm{~g}\) sample of solid \(\mathrm{CO}_{2}\) (dry ice) is added at \(-100^{\circ} \mathrm{C}\) to an evacuated (all of the gas removed) container with a volume of \(5.0 \mathrm{~L}\). If the container is sealed and then allowed to warm to room temperature \(\left(25^{\circ} \mathrm{C}\right)\) so that the entire solid \(\mathrm{CO}_{2}\) is converted to a gas, what is the pressure inside the container?

Assume that an exhaled breath of air consists of \(74.8 \% \mathrm{~N}_{2}, 15.3 \% \mathrm{O}_{2}, 3.7 \% \mathrm{CO}_{2},\) and \(6.2 \%\) water vapor. (a) If the total pressure of the gases is \(99.8 \mathrm{kPa}\), calculate the partial pressure of each component of the mixture. (b) If the volume of the exhaled gas is \(455 \mathrm{~mL}\) and its temperature is \(37^{\circ} \mathrm{C},\) calculate the number of moles of \(\mathrm{CO}_{2}\) exhaled. (c) How many grams of glucose ( \(\left.\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) would need to be metabolized to produce this quantity of \(\mathrm{CO}_{2} ?\) (The chemical reaction is the same as that for combustion of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\). See Section 3.2 and Problem \(10.57 .\) )

Does the effect of intermolecular attraction on the properties of a gas become more significant or less significant if (a) the gas is compressed to a smaller volume at constant temperature; \((\mathbf{b})\) the temperature of the gas is increased at constant volume?

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