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Perform the following conversions: (a) \(0.912\) atm to torr, (b) \(0.685\) bar to kilopascals, (c) \(655 \mathrm{~mm} \mathrm{Hg}\) to atmospheres, (d) \(1.323 \times 10^{5} \mathrm{~Pa}\) to atmospheres, (e) \(2.50 \mathrm{~atm}\) to psi.

Short Answer

Expert verified
The short answer to the given conversions is: (a) $0.912$ atm $\approx 693.12$ torr (b) $0.685$ bar $= 68.5$ kPa (c) $655$ mmHg $\approx 0.861$ atm (d) $1.323 \times 10^{5}$ Pa $\approx 1.306$ atm (e) $2.50$ atm $\approx 36.74$ psi

Step by step solution

01

(a) Conversion of \(0.912\) atm to torr

To convert from atm to torr, we need to multiply by the conversion factor: $$1 \, \mathrm{atm} = 760 \, \mathrm{torr}$$ Thus, we have: $$0.912 \, \mathrm{atm} \times \frac{760 \, \mathrm{torr}}{1 \, \mathrm{atm}} \approx 693.12 \, \mathrm{torr}$$
02

(b) Conversion of \(0.685\) bar to kilopascals

To convert from bar to kilopascals, we need to multiply by the conversion factor: $$1 \, \mathrm{bar} = 100 \, \mathrm{kPa}$$ Thus, we have: $$0.685 \, \mathrm{bar} \times \frac{100 \, \mathrm{kPa}}{1 \, \mathrm{bar}} = 68.5 \, \mathrm{kPa}$$
03

(c) Conversion of \(655 \, \mathrm{mmHg}\) to atmospheres

To convert from mmHg (torr) to atmospheres, we need to divide by the conversion factor: $$1 \, \mathrm{atm} = 760 \, \mathrm{torr}$$ Thus, we have: $$655 \, \mathrm{mmHg} \times \frac{1 \, \mathrm{atm}}{760 \, \mathrm{torr}} \approx 0.861 \, \mathrm{atm}$$
04

(d) Conversion of \(1.323 \times 10^{5} \, \mathrm{Pa}\) to atmospheres

To convert from pascals (Pa) to atmospheres, we need to divide by the conversion factor: $$1 \, \mathrm{atm} = 101,325 \, \mathrm{Pa}$$ Thus, we have: $$1.323 \times 10^{5} \, \mathrm{Pa} \times \frac{1 \, \mathrm{atm}}{101,325 \, \mathrm{Pa}} \approx 1.306 \, \mathrm{atm}$$
05

(e) Conversion of \(2.50 \, \mathrm{atm}\) to psi

To convert from atmospheres to psi, we need to multiply by the conversion factor: $$1 \, \mathrm{atm} = 14.696 \, \mathrm{psi}$$ Thus, we have: $$2.50 \, \mathrm{atm} \times \frac{14.696 \, \mathrm{psi}}{1 \, \mathrm{atm}} \approx 36.74 \, \mathrm{psi}$$

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Key Concepts

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

Atmospheres
An atmosphere (atm) is a unit of pressure that is widely used in various scientific fields. It is defined as the pressure exerted by the weight of the Earth's atmosphere at sea level. One atmosphere is roughly equivalent to 101,325 Pascals. The concept of atmospheres is crucial in understanding pressure at different altitudes and environments. It's a standard unit of measure for expressing atmospheric pressure, making it valuable for chemical experiments and processes that involve gases. To convert atmospheres into other pressure units like torr or Pascals, you simply use conversion factors. For example: - 1 atm = 760 torr - 1 atm = 101,325 Pa This means that if you have pressure in atmospheres, you can multiply it by 760 to convert it to torr, or multiply by 101,325 to find it in Pascals. Keeping track of conversion factors is key to moving smoothly between different pressure measurements.
Torr
The torr is another unit of pressure, named after the Italian physicist Evangelista Torricelli. It is frequently used in fields like medicine, vacuum measurements, and meteorology. One torr is precisely defined as \[1 \, \mathrm{torr} = \frac{1}{760} \, \mathrm{atm}\] This relationship helps because you can easily convert between atmospheres and torr. Just remember that:- 1 atm equals 760 torr- Hence, if you want to switch from atm to torr, multiply by 760. Conversely, divide by 760 to convert from torr to atm.Understanding how to navigate between these units is important, especially in scientific research that requires precise pressure control.
Pascals
Pascals (Pa) are the SI unit of pressure and are named after Blaise Pascal, a French mathematician and physicist. This unit provides a straightforward way to express pressure resulting from a force applied over an area, with one Pascal equivalent to one Newton per square meter. Pascals are commonly used in engineering and physics due to their simple relation to other fundamental SI units. The equation is: - 1 Pa = 1 N/m² Because Pascals are a smaller unit, pressures are often given in kilopascals (kPa), where: - 1 kPa = 1,000 Pa This conversion makes the figures more manageable, especially when dealing with everyday pressures such as in automotive tires or weather predictions. In practical applications, converting other units to Pascals involves using: - 1 atm = 101,325 Pa Using this factor, you can convert measurements in atm to Pascals, and vice versa, by straightforward multiplication or division.
Kilopascals
Kilopascals (kPa) offer an extended range for expressing pressure and are widely used in both everyday life and scientific domains. By converting large numbers of Pascals into kPa, it allows for more manageable and readable figures.You convert from other pressure units to kilopascals by using the conversion factor between those units and Pascals, and then dividing by 1,000. For example, with bar, a unit often used for atmospheric pressure:- 1 bar = 100,000 Pa = 100 kPaTo convert pressures from bar to kPa, simply multiply by 100:\[\text{Example:} \, 0.685 \, \text{bar} \times 100 \text{kPa/bar} = 68.5 \, \text{kPa}\]Kilopascals offer a useful bridge between smaller and larger units, making pressure data more understandable, especially in fields like meteorology and engineering.

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Most popular questions from this chapter

In an experiment reported in the scientific literature, male cockroaches were made to run at different speeds on a miniature treadmill while their oxygen consumption was measured. In \(1 \mathrm{hr}\) the average cockroach running at \(0.08 \mathrm{~km} / \mathrm{hr}\) consumed \(0.8 \mathrm{~mL}\) of \(\mathrm{O}_{2}\) at 1 atm pressure and \(24^{\circ} \mathrm{C}\) per gram of insect mass. (a) How many moles of \(\mathrm{O}_{2}\) would be consumed in \(1 \mathrm{hr}\) by a \(5.2-\mathrm{g}\) cockroach moving at this speed? (b) This same cockroach is caught by a child and placed in a 1-qt fruit jar with a tight lid. Assuming the same level of continuous activity as in the research, will the cockroach consume more than \(20 \%\) of the available \(\mathrm{O}_{2}\) in a 48-hr period? (Air is \(21 \mathrm{~mol} \% \mathrm{O}_{2}\).)

(a) If the pressure exerted by ozone, \(\mathrm{O}_{3}\), in the stratosphere is \(3.0 \times 10^{-3}\) atm and the temperature is \(250 \mathrm{~K}\), how many ozone molecules are in a liter? (b) Carbon dioxide makes up approximately \(0.04 \%\) of Earth's atmosphere. If you collect a \(2.0-\mathrm{L}\) sample from the atmosphere at sea level ( \(1.00\) atm) on a warm day \(\left(27^{\circ} \mathrm{C}\right)\), how many \(\mathrm{CO}_{2}\) molecules are in your sample?

Suppose you are given two 1-L flasks and told that one contains a gas of molar mass 30 , the other a gas of molar mass 60 , both at the same temperature. The pressure in flask A is \(\mathrm{X}\) atm, and the mass of gas in the flask is \(1.2 \mathrm{~g}\). The pressure Which flask contains the gas of molar mass 30 , and which contains the gas of molar mass 60 ?

A gas bubble with a volume of \(1.0 \mathrm{~mm}^{3}\) originates at the bottom of a lake where the pressure is \(3.0\) atm. Calculate its volume when the bubble reaches the surface of the lake where the pressure is 730 torr, assuming that the temperature doesn't change.

An open-end manometer containing mercury is connected to a container of gas, as depicted in Sample Exercise \(10.2\). What is the pressure of the enclosed gas in torr in each of the following situations? (a) The mercury in the arm attached to the gas is \(15.4 \mathrm{~mm}\) higher than in the one open to the atmosphere; atmospheric pressure is \(0.985 \mathrm{~atm}\). (b) The mercury in the arm attached to the gas is \(12.3 \mathrm{~mm}\) lower than in the one open to the atmosphere; atmospheric pressure is \(0.99\) atm.

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