A pressure sensor shows tire pressure as \(31 \mathrm{lb} /\) in. \(^{2}\). Convert this pressure to (a) atm (b) torr (c) \(\mathrm{kPa}\)

Short Answer

Expert verified
The pressure in (a) atm is approximately 2.11, (b) torr is approximately 1604, and (c) kPa is approximately 214.

Step by step solution

01

- Conversion to atm

Understand that standard atmospheric pressure is equal to approximately 14.696 psi (pounds per square inch). Divide the given pressure by this value to convert the pressure to atm. Calculation: \[\begin{equation}1 \text{ atm } = 14.696 \text{ psi}\end{equation}\]\[\begin{equation}\text{Pressure in atm} = \frac{31 \text{ lb/in}^2}{14.696 \text{ lb/in}^2}\end{equation}\]
02

- Conversion to torr

Recognize that 1 atm is equivalent to 760 torr (also known as mmHg). Multiply the result from Step 1 by 760 to convert the pressure to torr. Calculation: \[\begin{equation}1 \text{ atm } = 760 \text{ torr}\end{equation}\]\[\begin{equation}\text{Pressure in torr} = \text{Pressure in atm} \times 760\end{equation}\]
03

- Conversion to kPa

Be aware that 1 atm is equivalent to 101.325 kPa. Multiply the result from Step 1 by 101.325 to get the pressure in kilopascals (kPa). Calculation:\[\begin{equation}1 \text{ atm } = 101.325 \text{ kPa}\end{equation}\]\[\begin{equation}\text{Pressure in kPa} = \text{Pressure in atm} \times 101.325\end{equation}\]

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Key Concepts

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

Atmospheric Pressure
Imagine the weight of air pressing down on you from the sky above. That weight creates what we call atmospheric pressure. It essentially measures the force exerted by the weight of the atmosphere. Standard atmospheric pressure is the pressure exerted by a 760 mm column of mercury at sea level and at a temperature of 0 degrees Celsius. It is often used as a reference point for pressure measurements and is equivalent to approximately 14.696 psi. Remember this value, as it will be essential for converting psi into other pressure units.

Because atmospheric pressure affects weather, breathing, and even the design of buildings and airplanes, understanding how to measure and convert it is crucial for many scientific and practical applications.
Torr
The term 'torr' can sound quite ancient, and it's because it is named after the Italian mathematician and physicist Evangelista Torricelli, credited with inventing the barometer in the 17th century. One torr is defined as exactly 1/760 of a standard atmosphere, which relates directly to the height of a mercury column that can be supported by atmospheric pressure. So, when converting from atmospheric pressure to torr, you would multiply by 760.

Why use torr?

Well, the torr is a unit that is commonly used in vacuum systems and barometric readings. It gives us a more specific measurement when compared to the atmospheric pressure, making it invaluable in fields such as meteorology and aviation.
Kilopascals
The kilopascal (kPa) is part of the International System of Units (SI) and is a unit of pressure. One kilopascal equals 1,000 pascals, where a pascal is defined as one newton of force acting on one square meter. To put it simply, it is a measure that tells us how much force is applied over an area.

For everyday use, kPa is particularly convenient for scientific and engineering calculations because it relates directly to the metric system. Moreover, it's commonly used to report barometric pressure outside of the United States and for stress calculations in engineering. Converting atmospheric pressure to kPa is also simple: just multiply by 101.325.
PSI to ATM
PSI, or pounds per square inch, is another unit of pressure commonly used, especially in the United States. It measures how much force is exerted on one square inch of an area. When converting psi to atm, you divide the number of PSI by 14.696, which is the standard atmospheric pressure in PSI.

Understanding how to make this conversion is particularly important in various fields such as tire manufacturing, where psi is a standard unit, and atmospheric sciences. Moreover, when working with systems that involve gases, such as HVAC systems or scuba diving equipment, being able to convert between these units is essential to ensure safety and optimal operation.

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Most popular questions from this chapter

Using the ideal gas law, \(P V=n R T\), calculate the following: (a) the volume of \(0.510 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) at \(47^{\circ} \mathrm{C}\) and \(1.6\) atm pressure (b) the number of grams in \(16.0 \mathrm{~L}\) of \(\mathrm{CH}_{4}\) at \(27^{\circ} \mathrm{C}\) and 600 . torr pressure (c) the density of \(\mathrm{CO}_{2}\) at \(4.00\) atm pressure and \(20.0^{\circ} \mathrm{C}\) (d) the molar mass of a gas having a density of \(2.58 \mathrm{~g} / \mathrm{L}\) at \(27^{\circ} \mathrm{C}\) and \(1.00\) atm pressure.

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$$ \begin{aligned} &\text { Fill in the table below with the missing information: }\\\ &\begin{array}{|c|c|c|c|} \hline & \text { torr } & \text { in. Hg } & \text { kilopascals (kPa) } \\ \hline \text { (a) } & & 30.2 & \\ \hline \text { (b) } & 752 & & \\ \hline \text { (c) } & & & 99.3 \\ \hline \end{array} \end{aligned} $$

What pressure will 800 . \(\mathrm{mL}\) of a gas at \(\mathrm{STP}\) exert when its volume is \(250 . \mathrm{mL}\) at \(30^{\circ} \mathrm{C}\) ?

A sample of \(\mathrm{CH}_{4}\) gas occupies a volume of \(575 \mathrm{~mL}\) at \(-25^{\circ} \mathrm{C}\). If pressure remains constant, what will be the new volume if temperature changes to: (a) \(298 \mathrm{~K}\) (b) \(32^{\circ} \mathrm{F}\) (c) \(45^{\circ} \mathrm{C}\)

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