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A gauge on a compressed gas cylinder reads 2200 psi (pounds per square inch; 1 atm \(=14.7\) psi). Express this pressure in each of the following units. a. standard atmospheres b. mega pascals (MPa) c. torr

Short Answer

Expert verified
The pressure can be expressed in the following units: a. \(149.660\) atm, b. \(15.158\) MPa, and c. \(113661.6\) torr.

Step by step solution

01

Convert psi to atm

Given pressure = 2200 psi We know that 1 atm = 14.7 psi Therefore, to convert 2200 psi to atm, we'll use the following formula: Pressure in atm = (Pressure in psi) / (Conversion factor) Pressure in atm = 2200 psi / 14.7 psi/atm = 149.660 atm a. The pressure in standard atmospheres is approximately 149.660 atm.
02

Convert atm to MPa

Now, we will convert the pressure from atm to MPa. We know that: 1 atm = 101.325 kPa, and 1 MPa = 1000 kPa Therefore, to convert the pressure from atm to MPa, we'll follow these steps: 1. Convert atm to kPa Pressure in kPa = (Pressure in atm) × (Conversion factor) Pressure in kPa = 149.660 atm × 101.325 kPa/atm ≈ 15158.31 kPa 2. Convert kPa to MPa Pressure in MPa = (Pressure in kPa) / 1000 Pressure in MPa = 15158.31 kPa / 1000 = 15.158 MPa b. The pressure in mega pascals is approximately 15.158 MPa.
03

Convert atm to torr

Lastly, we will convert the pressure from atm to torr. We know that: 1 atm = 760 torr To convert the pressure from atm to torr, we'll use the following formula: Pressure in torr = (Pressure in atm) × (Conversion factor) Pressure in torr = 149.660 atm × 760 torr/atm ≈ 113661.6 torr c. The pressure in torr is approximately 113661.6 torr.

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Key Concepts

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

Standard Atmospheres
When we talk about pressure in terms of standard atmospheres (atm), we are referring to a unit of pressure that is universally used because of its simplicity and direct relation to the pressure exerted by the Earth's atmosphere at sea level. 1 atm is defined as exactly 101,325 pascals (Pa) and is also equivalent to about 14.7 pounds per square inch (psi). This makes it easier to compare different pressure readings from various units.To convert pressure measured in psi to atm, you divide the pressure value by 14.7. For instance, if you have a pressure reading of 2200 psi, converting this to standard atmospheres involves the calculation: Pressure in atm = \( \frac{2200\, \text{psi}}{14.7\, \text{psi/atm}} \) = 149.660 atm.Knowing how to convert between psi and atm can help you understand many real-world scenarios, such as tire pressure and weather systems.
Mega Pascals
Mega pascals (MPa) are part of the SI (International System of Units) family of pressure units. They are especially useful in scientific and engineering contexts due to their ability to represent large pressure values simply. 1 MPa is equal to 1,000,000 pascals (Pa), which is great for scenarios where the pressures are high, such as stresses in construction materials or pressures in industrial machinery.To convert from atmospheres (atm) to MPa, you'll perform two steps:
  • First, convert the atm value to kilopascals (kPa), since 1 atm = 101.325 kPa.
  • Then, convert kPa to MPa by dividing by 1000, because 1 MPa = 1000 kPa.
So, if you have a pressure of 149.660 atm, you convert this to MPa as follows:Pressure in MPa = \( \left(149.660\, \text{atm} \times 101.325\, \text{kPa/atm}\right) \div 1000 \) = 15.158 MPa.Understanding this conversion is key in fields like civil engineering and mechanical engineering, where such pressures are commonly encountered.
Torr
"Torr" is a pressure unit named after the Italian physicist Evangelista Torricelli, who is credited with inventing the barometer. It's a crucial unit for scientists, especially in the field of chemistry, where laboratory pressures are often measured. 1 torr is defined as 1/760 of an atmosphere, making it a smaller and more precise unit ideal for low-pressure environments. Since 1 atm is equal to 760 torr, converting from atm to torr involves a simple multiplication. For a pressure reading of 149.660 atm, the conversion is straightforward: Pressure in torr = 149.660 atm × 760 torr/atm ≈ 113661.6 torr. Using torr is common in laboratory settings where precise pressure measurements are needed, such as in vacuum systems and gas reactions. Having this knowledge at your fingertips enables you to seamlessly switch between different units of pressure.

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Most popular questions from this chapter

A chemist weighed out 5.14 g of a mixture containing unknown amounts of \(\mathrm{BaO}(s)\) and \(\mathrm{CaO}(s)\) and placed the sample in a \(1.50-\mathrm{L}\) flask containing \(\mathrm{CO}_{2}(g)\) at \(30.0^{\circ} \mathrm{C}\) and 750 . torr. After the reaction to form BaCO_ \(_{3}(s)\) and \(\mathrm{CaCO}_{3}(s)\) was completed, the pressure of \(\mathrm{CO}_{2}(g)\) remaining was 230 . torr. Calculate the mass percentages of \(\mathrm{CaO}(s)\) and \(\mathrm{BaO}(s)\) in the mixture.

The rate of effusion of a particular gas was measured and found to be 24.0 mL/min. Under the same conditions, the rate of effusion of pure methane \(\left(\mathrm{CH}_{4}\right)\) gas is 47.8 mL/min. What is the molar mass of the unknown gas?

Calculate the root mean square velocities of \(\mathrm{CH}_{4}(g)\) and \(\mathrm{N}_{2}(g)\) molecules at 273 \(\mathrm{K}\) and 546 \(\mathrm{K} .\)

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of \(200 . \mathrm{L} / \mathrm{min}\) at 1.50 \(\mathrm{atm}\) and ambient temperature. Air is added to the chamber at 1.00 \(\mathrm{atm}\) and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g),\) three times as much oxygen as is necessary is reacted. Assuming air is 21 mole percent \(\mathrm{O}_{2}\) and 79 \(\mathrm{mole}\) percent \(\mathrm{N}_{2},\) calculate the flow rate of air necessary to deliver the required amount of oxygen. b. Under the conditions in part a, combustion of methane was not complete as a mixture of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) was produced. It was determined that 95.0\(\%\) of the carbon in the exhaust gas was present in \(\mathrm{CO}_{2}\) . The remainder was present as carbon in CO. Calculate the composition of the exhaust gas in terms of mole fraction of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) . Assume \(\mathrm{CH}_{4}\) is completely reacted and \(\mathrm{N}_{2}\) is unreacted.

Consider three identical flasks filled with different gases. Flask \(\mathrm{A} : \mathrm{CO}\) at 760 torr and \(0^{\circ} \mathrm{C}\) Flask \(\mathrm{B} : \mathrm{N}_{2}\) at 250 torr and \(0^{\circ} \mathrm{C}\) Flask \(\mathrm{C} : \mathrm{H}_{2}\) at 100 torr and \(0^{\circ} \mathrm{C}\) a. In which flask will the molecules have the greatest average kinetic energy? b. In which flask will the molecules have the greatest average velocity?

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