/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 The typical atmospheric pressure... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 21

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When a large evacuated flask is filled with argon gas, its mass increases by \(3.224 \mathrm{~g}\). When the same flask is again evacuated and then filled with a gas of unknown molar mass, the mass increase is 8.102 g. (a) Based on the molar mass of argon, estimate the molar mass of the unknown gas. (b) What assumptions did you make in arriving at your answer?

The temperature of a 5.00-L container of \(\mathrm{N}_{2}\) gas is increased from \(20^{\circ} \mathrm{C}\) to \(250^{\circ} \mathrm{C}\). If the volume is held constant, predict qualitatively how this change affects the following: (a) the average kinetic energy of the molecules; (b) the root-mean-square speed of the molecules; (c) the strength of the impact of an average molecule with the container walls; (d) the total number of collisions of molecules with walls ner second.

After the large eruption of Mount St. Helens in 1980 , gas samples from the volcano were taken by sampling the downwind gas plume. The unfiltered gas samples were passed over a goldcoated wire coil to absorb mercury (Hg) present in the gas. The mercury was recovered from the coil by heating it and then analyzed. In one particular set of experiments scientists found a mercury vapor level of \(1800 \mathrm{ng}\) of Hg per cubic meter in the plume at a gas temperature of \(10^{\circ} \mathrm{C}\). Calculate (a) the partial pressure of Hg vapor in the plume, (b) the number of \(\mathrm{Hg}\) atoms per cubic meter in the gas, \((\mathrm{c})\) the total mass of Hg emitted per day by the volcano if the daily plume volume was \(1600 \mathrm{~km}^{3}\).

At constant pressure, the mean free path \((\lambda)\) of a gas molecule is directly proportional to temperature. At constant temperature, \(\lambda\) is inversely proportional to pressure. If you compare two different gas molecules at the same temperature and pressure, \(\lambda\) is inversely proportional to the square of the diameter of the gas molecules. Put these facts together to create a formula for the mean free path of a gas molecule with a proportionality constant (call it \(R_{\mathrm{mfp}}\), like the ideal-gas constant) and define units for \(R_{\mathrm{mfp}}\).

An aerosol spray can with a volume of \(250 \mathrm{~mL}\) contains \(2.30 \mathrm{~g}\) of propane gas \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) as a propellant. (a) If the can is at \(23{ }^{\circ} \mathrm{C}\), what is the pressure in the can? (b) What volume would the propane occupy at STP? (c) The can's label says that exposure to temperatures above \(130^{\circ} \mathrm{F}\) may cause the can to burst. What is the pressure in the can at this temperature?
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.