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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

Consider the following gases, all at STP: \(\mathrm{Ne}, \mathrm{SF}_{6}, \mathrm{~N}_{2}, \mathrm{CH}_{4}\). (a) Which gas is most likely to depart from the assumption of the kinetic- molecular theory that says there are no attractive or repulsive forces between molecules? (b) Which one is closest to an ideal gas in its behavior? (c) Which one has the highest root-mean-square molecular speed at a given temperature? (d) Which one has the highest total molecular volume relative to the space occupied by the gas? (e) Which has the highest average kinetic- molecular energy? (f) Which one would effuse more rapidly than \(\mathrm{N}_{2} ?\) (g) Which one would have the largest van der Waals \(b\) parameter?

Table 10.3 shows that the van der Waals \(b\) parameter has units of \(\mathrm{L} / \mathrm{mol}\). This implies that we can calculate the size of atoms or molecules from \(b\). Using the value of \(b\) for \(\mathrm{Xe},\) calculate the radius of a Xe atom and compare it to the value found in Figure \(7.7,\) that is, \(140 \mathrm{pm}\). Recall that the volume of a sphere is \((4 / 3) \pi r^{3}\).

A mixture containing \(0.50 \mathrm{~mol} \mathrm{H}_{2}(g), 1.00 \mathrm{~mol} \mathrm{O}_{2}(g)\), and 3.50 \(\mathrm{mol} \mathrm{N}_{2}(g)\) is confined in a 25.0-L vessel at \(25^{\circ} \mathrm{C}\). (a) Calculate the total pressure of the mixture. (b) Calculate the partial pressure of each of the gases in the mixture.

A neon sign is made of glass tubing whose inside diameter is \(3.0 \mathrm{~cm}\) and length is \(10.0 \mathrm{~m}\). If the sign contains neon at a pressure of \(265 \mathrm{~Pa}\) at \(30^{\circ} \mathrm{C}\), how many grams of neon are in the sign? (The volume of a cylinder is \(\pi r^{2} h\).)

(a) Place the following gases in order of increasing average molecular speed at \(25^{\circ} \mathrm{C}: \mathrm{O}_{2}, \mathrm{Ar}, \mathrm{CO}, \mathrm{HCl}, \mathrm{CH}_{4} \cdot(\mathbf{b})\) Calculate the rms speed of CO molecules at \(25^{\circ} \mathrm{C} .(\mathbf{c})\) Calculate the most probable speed of an argon atom in the stratosphere, where the temperature is \(0^{\circ} \mathrm{C}\).

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