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The pressure recorded for an argon gas cylinder is \(44.0 \mathrm{lb} \cdot\) inch \(^{-2}\). Convert this pressure into (a) kPa; (b) Torr; (c) bar; (d) atm.

Short Answer

Expert verified
The pressure of the gas cylinder is (a) 303.369 kPa; (b) 2277.57 Torr; (c) 3.0337 bar; (d) 2.9924 atm.

Step by step solution

01

Conversion to Pascals

First, convert the pressure from pounds per square inch (psi) to Pascals (Pa). Use the conversion factor: 1 psi = 6894.757 Pa.
02

Calculate Pressure in Pascals

Multiply the given pressure by the conversion factor: \(44.0 \text{ psi} \times 6894.757 \text{ Pa/psi} = 303369.268 \text{ Pa} \).
03

Convert Pascals to Kilopascals

To convert from Pa to kPa, divide the pressure in Pa by 1000: \(303369.268 \text{ Pa} \div 1000 = 303.369 \text{ kPa}\).
04

Convert Pascals to Torr

Use the conversion factor 1 Pa = 0.00750062 Torr. Multiply the pressure in Pascals by this factor: \(303369.268 \text{ Pa} \times 0.00750062 \text{ Torr/Pa} = 2277.57 \text{ Torr}\).
05

Convert Pascals to bar

To convert Pa to bar, use the conversion factor 1 Pa = 1e-5 bar. Multiply the pressure in Pascals by this factor: \(303369.268 \text{ Pa} \times 1e-5 \text{ bar/Pa} = 3.0337 \text{ bar}\).
06

Convert Pascals to Atmospheres

For the conversion to atmospheres, use the conversion factor 1 atm = 101325 Pa. Divide the pressure in Pascals by this factor: \(303369.268 \text{ Pa} \div 101325 \text{ Pa/atm} = 2.9924 \text{ atm}\).

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Key Concepts

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

Unit Conversion
When working with measurements, it's common to encounter the need to switch between different units to match the context of a problem or the requirements of a system. Unit conversion is a fundamental skill in science and engineering disciplines, as it ensures consistency and understanding across various fields and applications. It involves using conversion factors, which are fixed constants that represent how numerically one unit of measurement relates to another.

To convert units, you simply multiply the numerical value you have by the appropriate conversion factor. It is crucial to arrange the conversion factor so that units cancel out, leaving you with the desired unit. A chain of conversions may be necessary when no direct conversion factor is available, such as converting pounds per square inch (psi) to Pascals and then to kilopascals (kPa). Mastery of unit conversion ensures accuracy in scientific calculations and clear communication of measurements.
Pascals
The Pascal (Pa) is the SI (International System of Units) unit of pressure and named after the French mathematician Blaise Pascal. It measures the force of one newton exerted over an area of one square meter. Due to its small size, the Pascal is more commonly used in scientific calculations rather than everyday applications. When dealing with real-world pressures, such as that in a gas cylinder as in our exercise, larger units like kilopascals or atmospheres are typically more convenient.

The conversion from psi to Pascals is pivotal because Pa acts a starting point from which we can convert to various other units used in industry and science, such as kilopascals (kPa), Torr, or atmospheres (atm), thus illustrating the interconnectivity and versatility of the Pascal in pressure measurement.
Kilopascals

Understanding Kilopascals

The kilopascal (kPa) is a multiple of the Pascal and is widely used for expressing pressures, especially in meteorology and engineering. One kilopascal is equal to one thousand Pascals. The convenience of kilopascals stems from their magnitude, which makes them more manageable for use in everyday situations than the smaller Pascal.

In the given exercise, converting Pascals to kilopascals simplifies the pressure reading, making it easier to understand and relate to. It’s a straightforward conversion by simply dividing the pressure in Pascals by 1,000, resulting in a more compact numerical value without changing the essence of the measurement.
Torr

Exploring the Unit of Torr

The unit Torr was named after the Italian mathematician Evangelista Torricelli, and it's another way to measure pressure. It is approximately equal to the pressure exerted by a millimeter of mercury (mmHg). The relationship between Pascals and Torr is such that 1 Pa is equivalent to 0.00750062 Torr. This unit is often used in vacuum physics and barometry where precise measurements of pressure are crucial.

In the exercise's context, converting Pascals to Torr is significant for fields that require a more practical approach to pressure measurement, like blood pressure monitors and aircraft altimeters, which historically used millimeters of mercury as a reference.
Bar

The Metric Unit: Bar

The bar is a metric unit of pressure, but not an SI unit. It is defined as 100,000 Pascals, making it very closely aligned with the atmospheric pressure at sea level, which is approximately 1 bar. For many practical purposes, it is used interchangeably with atmospheres due to their near-equivalency.

In industrial applications and weather reporting, bar is favored because it provides a convenient scale for reading and comparing pressures without resorting to large numbers. The conversion from Pascals to bar in the exercise demonstrates the unit's suitability for higher pressure levels typically found in the operation of machinery and equipment.
Atmospheres

Atmospheric Pressure Unit

An atmosphere (atm) is a unit of measurement that represents the average pressure at sea level on Earth. It is a traditional unit still widely used in many fields, including meteorology and aviation. The standard atmosphere is defined as exactly 101,325 Pascals.

This unit provides an intuitive understanding of pressure levels relative to the ambient air pressure we experience every day. Converting Pascals to atmospheres, as seen in the exercise, is necessary for applications such as scuba diving and pneumatic systems where referencing pressure to atmospheric levels is essential.

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Most popular questions from this chapter

When Robert Boyle conducted his experiments, he measured pressure in inches of mercury (in \(\mathrm{Hg}\) ). On a day when the atmospheric pressure was \(29.85 \mathrm{inHg}\), he trapped some air in the tip of a J-tube (1) and measured the difference in height of the mercury in the two arms of the tube \((b)\). When \(h=12.0\) inches, the height of the gas in the tip of the tube was \(32.0\) in. Boyle then added additional mercury and the level rose in both arms of the tube so that \(h=30.0\) inches \((2)\). (a) What was the height of the air space (in inches) in the tip of the tube in \((2)\) ? (b) What was the pressure of the gas in the tube in (1) and in (2) in inHg?

(a) A \(125-\mathrm{mL}\) flask contains argon at \(1.30\) atm and \(77^{\circ} \mathrm{C}\). What amount of Ar is present (in moles)? (b) A \(120 .-\mathrm{mL}\) flask contains \(2.7 \mu \mathrm{g}\) of \(\mathrm{O}_{2}\) at \(17^{\circ} \mathrm{C}\). What is the pressure (in Torr)? (c) A 20.0-L flask at \(215 \mathrm{~K}\) and 20. Torr contains nitrogen. What mass of nitrogen is present (in grams)? (d) A 16.7-g sample of krypton exerts a pressure of \(1.00 \times 10^{2} \mathrm{~m}\) Torr at \(44^{\circ} \mathrm{C}\). What is the volume of the container (in liters)? (e) A \(2.6-\mu \mathrm{L}\) ampoule of xenon has a pressure of \(2.00\) Torr at \(15^{\circ} \mathrm{C}\). How many Xe atoms are present?

Calculate the molar kinetic energy (in joules per mole) of a sample of \(\mathrm{Ne}(\mathrm{g})\) at (a) \(25.00^{\circ} \mathrm{C}\) and (b) \(26.00^{\circ} \mathrm{C}\). (c) The difference between the answers to parts (a) and (b) is the energy per mole that it takes to raise the temperature of \(\mathrm{Ne}(\mathrm{g})\) by \(1^{\circ} \mathrm{C}\). The quantity is known as the molar heat capacity. What is its value?

A flask of volume \(5.00 \mathrm{~L}\) is evacuated and \(43.78 \mathrm{~g}\) of solid dinitrogen tetroxide, \(\mathrm{N}_{2} \mathrm{O}_{4}\), is introduced at \(-196^{\circ} \mathrm{C}\). The sample is then warmed to \(25^{\circ} \mathrm{C}\), during which time the \(\mathrm{N}_{2} \mathrm{O}_{4}\) vaporizes and some of it dissociates to form brown \(\mathrm{NO}_{2}\) gas. The pressure slowly increases until it stabilizes at \(2.96\) atm. (a) Write a balanced equation for the reaction. (b) If the gas in the flask at \(25^{\circ} \mathrm{C}\) were all \(\mathrm{N}_{2} \mathrm{O}_{4}\), what would the pressure be? (c) If all the gas in the flask converted into \(\mathrm{NO}_{2}\), what would the pressure be? (d) What are the mole fractions of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) once the pressure stabilizes at \(2.96 \mathrm{~atm}\) ?

(a) The van der Waals parameters for helium are \(a=\) \(3.41 \times 10^{-2} \mathrm{~L}^{2} \cdot\) atm \(\cdot \mathrm{mol}{ }^{-2}\) and \(b=2.38 \times 10^{-2} \mathrm{~L}^{-\mathrm{mol}^{-1}}\). Calculate the apparent volume (in \(\mathrm{pm}^{3}\) ) and radius (in pm) of a helium atom as determined from the van der Waals parameters. (b) Estimate the volume of a helium atom on the basis of its atomic radius. (c) How do these quantities compare? Should they be same? Discuss.

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