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

Short Answer

Expert verified
The atmospheric pressure on top of Mt. Everest is approximately \(0.348 \mathrm{atm}\), \(265 \mathrm{mmHg}\), \(35261 \mathrm{Pa}\), and \(0.353 \mathrm{bars}\).

Step by step solution

01

(a) Convert the pressure to atm

First, let's convert the pressure in torr to atm. We can use the conversion factor \(\frac{1 \mathrm{atm}}{760 \mathrm{torr}}\). \(265 \mathrm{torr} \times \frac{1 \mathrm{atm}}{760 \mathrm{torr}} = a \mathrm{atm}\) Calculate the value of a: \(a = \frac{265}{760} \approx 0.348 \mathrm{atm}\)
02

(b) Convert the pressure to mmHg

Since 1 torr is equal to 1 mmHg, the pressure value is already in mmHg. \(265 \mathrm{torr} = 265 \mathrm{mmHg}\)
03

(c) Convert the pressure to pascals

To convert the pressure to pascals, we first convert it to atm (as we have already done in part (a)) and then use the conversion factor \(\frac{101325 \mathrm{Pa}}{1 \mathrm{atm}}\). \(0.348 \mathrm{atm} \times \frac{101325 \mathrm{Pa}}{1 \mathrm{atm}} = \mathrm{pascals}\) Calculate the value in pascals: \(0.348 \times 101325 \approx 35261 \mathrm{Pa}\)
04

(d) Convert the pressure to bars

To convert the pressure to bars, we use the value we obtained in pascals and use the conversion factor \(\frac{1 \mathrm{bar}}{100000 \mathrm{Pa}}\). \(35261 \mathrm{Pa} \times \frac{1 \mathrm{bar}}{100000 \mathrm{Pa}} = \mathrm{bars}\) Calculate the value in bars: \(\frac{35261}{100000} \approx 0.353 \mathrm{bars}\) The pressure on top of Mt. Everest is approximately \(0.348 \mathrm{atm}\), \(265 \mathrm{mmHg}\), \(35261 \mathrm{Pa}\), and \(0.353 \mathrm{bars}\).

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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 weight of the air above us in the Earth's atmosphere, exerted on Earth's surface. This concept is crucial because it helps us understand how pressure changes based on altitude and weather conditions. At sea level, the standard atmospheric pressure is 1 atm, equivalent to 101,325 pascals or 760 torr. This value serves as a baseline used in various pressure conversion calculations. However, as you ascend in altitude, like climbing Mt. Everest, the atmospheric pressure decreases because there are fewer air molecules above exerting pressure. Understanding atmospheric pressure is important in numerous science and engineering fields, such as meteorology and aeronautics.
unit conversion
Unit conversion is a fundamental concept that involves changing a measurement from one unit to another. This skill is essential across scientific and engineering disciplines, where precise measurements matter. To perform unit conversions:
  • Identify the original and desired units.
  • Use a conversion factor, which is a numerical ratio or formula that equates one unit to another.
  • Multiply the original measurement by the conversion factor to find the value in the new units.
Effective unit conversion often requires a reference for common conversion factors, such as 1 atm = 760 torr or 1 atm = 101,325 pascals. This process allows for consistent and accurate interpretation of measurements across different contexts, like converting the pressure on Mt. Everest from torr to atm.
Mt. Everest pressure
Mt. Everest, the highest peak on Earth, presents unique conditions including extremely low pressures compared to sea level. The reduced atmospheric pressure of 265 torr on top of Mt. Everest challenges both humans and equipment. For comparison, the sea level pressure is about 760 torr; thus, the pressure on Everest is significantly lower, providing less oxygen per breath. This factor is pivotal for climbers who must acclimate to avoid altitude sickness. Knowing and converting this pressure into units like atm or pascals assists in planning and managing the challenges of extreme altitude environments. For example, the 265 torr converts to approximately 0.348 atm, which is a useful measurement for understanding the thin air conditions at the summit.
torr to atm conversion
Converting from torr to atm is a straightforward yet essential calculation in understanding pressure changes, like those experienced on Mt. Everest. The key conversion factor here is that 1 atm is equal to 760 torr. Here's how you can do it:
  • Start with the pressure value in torr.
  • Apply the conversion factor: \(\frac{1 \, atm}{760 \, torr}\).
  • Calculate to find the pressure in atm. For example, 265 torr converts to approximately 0.348 atm.
This conversion is vital in various fields such as aviation and medicine, where pressures are often described in more than one unit, making it necessary to switch between them for practical applications.

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Most popular questions from this chapter

Assume that an exhaled breath of air consists of \(74.8 \% \mathrm{~N}_{2}\), \(15.3 \% \mathrm{O}_{2}, 3.7 \% \mathrm{CO}_{2}\), and \(6.2 \%\) water vapor. (a) If the total pressure of the gases is \(0.980 \mathrm{~atm}\), calculate the partial pressure of each component of the mixture. (b) If the volume of the exhaled gas is \(455 \mathrm{~mL}\) and its temperature is \(37^{\circ} \mathrm{C}\), calculate the number of moles of \(\mathrm{CO}_{2}\) exhaled. (c) How many grams of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) would need to be metabolized to produce this quantity of \(\mathrm{CO}_{2} ?\) (The chemical reaction is the same as that for combustion of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\). See Section 3.2.)

Consider a mixture of two gases, \(A\) and \(B\), confined in a closed vessel. A quantity of a third gas, \(C\), is added to the same vessel at the same temperature. How does the addition of gas \(C\) affect the following: (a) the partial pressure of gas \(A,(b)\) the total pressure in the vessel, (c) the mole fraction of gas \(\mathrm{B}\) ?

Large amounts of nitrogen gas are used in the manufacture of ammonia, principally for use in fertilizers. Suppose \(120.00 \mathrm{~kg}\) of \(\mathrm{N}_{2}(g)\) is stored in a 1100.0-L metal cylinder at \(280^{\circ} \mathrm{C}\). (a) Calculate the pressure of the gas, assuming ideal-gas behavior. (b) By using data in Table 10.3, calculate the pressure of the gas according to the van der Waals equation. (c) Under the conditions of this problem, which correction dominates, the one for finite volume of gas molecules or the one for attractive interactions?

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 average 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 per second.

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 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?

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