The typical atmospheric pressure on top of Mt. Everest \((29,028 \mathrm{ft})\) is about 265 torr. Convert this pressure to (a) atm, b) \(\mathrm{mm} \mathrm{Hg},\) (c) pascals, (d) bars, (e) psi.

Short Answer

Expert verified
The atmospheric pressure on top of Mt. Everest is approximately (a) 0.348684 atm, (b) 265 mm Hg, (c) 35347.8716 Pa, (d) 0.353452 bars, and (e) 5.121727 psi.

Step by step solution

01

Write down the conversion factors

For this exercise, we will need the following conversion factors: 1 atm = 760 torr 1 torr = 1 mm Hg 1 atm = 101325 Pa 1 atm = 1.01325 bar 1 atm = 14.696 psi
02

Convert pressure to atm

We are given the pressure on top of Mt. Everest in torr (265 torr) and are asked to convert this to atm. We will use the conversion factor to do this: \(P_\text{Everest (atm)} = \dfrac{P_\text{Everest (torr)}}{\text{Conversion factor}}\) \(P_\text{Everest (atm)} = \dfrac{265 \,\text{torr}}{760 \,\text{torr/atm}} = 0.348684 \, \text{atm}\) (a) The pressure on top of Mt. Everest is about 0.348684 atm.
03

Convert pressure to mm Hg

Since 1 torr is equal to 1 mm Hg, the pressure in mm Hg is the same as the given pressure in torr: (b) The pressure on top of Mt. Everest is 265 mm Hg.
04

Convert pressure to pascals

Next, we will convert the pressure in atm to pascals using the conversion factor: \(P_\text{Everest (Pa)} = P_\text{Everest (atm)}\times \text{Conversion factor}\) \(P_\text{Everest (Pa)} = 0.348684 \,\text{atm} \times 101325 \,\text{Pa/atm} = 35347.8716 \, \mathrm{Pa}\) (c) The pressure on top of Mt. Everest is approximately 35347.8716 Pa.
05

Convert pressure to bars

We will now convert the pressure in atm to bars: \(P_\text{Everest (bars)} = P_\text{Everest (atm)}\times \text{Conversion factor}\) \(P_\text{Everest (bars)} = 0.348684 \,\text{atm} \times 1.01325 \,\text{bars/atm} = 0.353452 \, \mathrm{bars}\) (d) The pressure on top of Mt. Everest is approximately 0.353452 bars.
06

Convert pressure to psi

Finally, we will convert the pressure in atm to psi: \(P_\text{Everest (psi)} = P_\text{Everest (atm)}\times \text{Conversion factor}\) \(P_\text{Everest (psi)} = 0.348684 \, \text{atm} \times 14.696 \, \text{psi/atm} = 5.121727 \, \mathrm{psi}\) (e) The pressure on top of Mt. Everest is approximately 5.121727 psi.

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Key Concepts

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

Atmospheric Pressure
Atmospheric pressure is the force exerted by the weight of the air in the Earth's atmosphere. It is an important factor in meteorology, aviation, and the behavior of gases. At sea level, the standard atmospheric pressure is 101,325 Pascals (Pa), which is equivalent to 1 atmosphere (atm). However, atmospheric pressure decreases with altitude, which is why the pressure on top of mountains such as Mt. Everest is significantly lower. Understanding atmospheric pressure is crucial for activities ranging from weather forecasting to calculating the boiling point of water at different elevations.

When working with atmospheric pressure, it's important to realize that it can be expressed in multiple units including torr, millimeters of mercury (mm Hg), and atmospheres (atm). These units are often converted from one to another in scientific calculations.
Torr to Atm Conversion
Converting pressure from torr to atm involves using the defined relationship between these two units of pressure. Specifically, 1 atm is defined as being equal to 760 torr. Thus, to convert torr to atm, you would divide the number of torr by 760. It's a straightforward calculation, but attention to precision is important.

The conversion from torr to atm is a basic principle in the study of pressure and is essential for understanding various scientific scenarios, such as measuring pressure in chemistry or in meteorology.
Pressure Units
Pressure units are varied and serve to quantify the force applied by a substance (typically a gas or liquid) per unit area. Common pressure units include atmospheres (atm), torr, millimeters of mercury (mm Hg), Pascals (Pa), bars, and pounds per square inch (psi). Each of these units is used in different contexts: Pascals are standard in the metric system and used widely in science; bars are similar to atmospheres and convenient for atmospheric pressure readings; psi is common in mechanical and tire pressure measurements.

Understanding the relationships and conversion factors between these units is important in fields such as physics, engineering, and meteorology. It is also crucial for ensuring accuracy when comparing measurements taken in different units.
Scientific Measurement
Scientific measurement is a foundational aspect of empirical science; it enables researchers and professionals to obtain, compare, and communicate data with precision and repeatability. Integral to this process is the use of standardized units, like those for measuring pressure, and the understanding of how to convert between these units. Whether it's converting torr to atm or Pascals to bars, mastering these conversions is essential for data integrity and clear scientific communication.

The step-by-step approach shown in the conversion of atmospheric pressure on top of Mt. Everest to various pressure units exemplifies the methodical nature of scientific measurement. Science relies on such methodologies to quantify observations and support hypothesis testing, making measurement skills critical for students and practitioners alike.

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Most popular questions from this chapter

A gaseous mixture of \(\mathrm{O}_{2}\) and \(\mathrm{Kr}\) has a density of \(1.104 \mathrm{~g} / \mathrm{L}\) at 355 torr and \(400 \mathrm{~K}\). What is the mole percent \(\mathrm{O}_{2}\) in the mixture?

A 1.42-g sample of helium and an unknown mass of \(\mathrm{O}_{2}\) are mixed in a flask at room temperature. The partial pressure of the helium is 42.5 torr, and that of the oxygen is 158 torr. What is the mass of the oxygen?

Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. \((\mathrm{a})\) If \(1.56 \mathrm{~g}\) of cyclopropane has a volume of \(1.00 \mathrm{~L}\) at 0.984 atm and \(50.0^{\circ} \mathrm{C}\), what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, \(\mathrm{CH}_{4} ?\)

In Sample Exercise 10.16 , we found that one mole of \(\mathrm{Cl}_{2}\) confined to \(22.41 \mathrm{~L}\) at \(0{ }^{\circ} \mathrm{C}\) deviated slightly from ideal behavior. Calculate the pressure exerted by \(1.00 \mathrm{~mol} \mathrm{Cl}_{2}\) confined to a smaller volume, \(5.00 \mathrm{~L}\), at \(25^{\circ} \mathrm{C} .\) (a) First use the ideal-gas equation and (b) then use the van der Waals equation for your calculation. (Values for the van der Waals constants are given in Table \(10.3 .)\) (c) Why is the difference between the result for an ideal gas and that calculated using the van der Waals equation greater when the gas is confined to \(5.00 \mathrm{~L}\) compared to \(22.4 \mathrm{~L} ?\)

A sample of \(5.00 \mathrm{~mL}\) of diethylether \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5},\right.\) density \(=\) \(0.7134 \mathrm{~g} / \mathrm{mL}\) ) is introduced into a 6.00 - \(\mathrm{L}\) vessel that already contains a mixture of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), whose partial pressures are \(P_{\mathrm{N}_{2}}=0.751 \mathrm{~atm}\) and \(P_{\mathrm{O}_{2}}=0.208 \mathrm{~atm} .\) The temperature is held at \(35.0^{\circ} \mathrm{C}\), and the diethylether totally evaporates. (a) Calculate the partial pressure of the diethylether. (b) Calculate the total pressure in the container.

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