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Chapter 5: Problem 8

Convert the following: (a) 0.745 atm to \(\mathrm{mmHg}\) (b) 992 torr to bar (c) \(365 \mathrm{kPa}\) to atm (d) \(804 \mathrm{mmHg}\) to \(\mathrm{kPa}\)

Short Answer

Expert verified
(a) 565.4 mmHg (b) 1.322 bar (c) 3.602 atm (d) 107.19 kPa

Step by step solution

01

Convert 0.745 atm to mmHg

Use the conversion factor between atmospheres and millimeters of mercury: 1 atm = 760 mmHg. Multiply the given value in atm by this conversion factor. \[ 0.745 \text{ atm} \times 760 \text{ mmHg/atm} = 565.4 \text{ mmHg} \]
02

Convert 992 torr to bar

Use the conversion factor between torr and bar: 1 torr = 0.00133322 bar. Multiply the given value in torr by this conversion factor. \[ 992 \text{ torr} \times 0.00133322 \text{ bar/torr} = 1.3222184 \text{ bar} \]
03

Convert 365 kPa to atm

Use the conversion factor between kilopascals and atmospheres: 1 atm = 101.325 kPa. To convert kPa to atm, divide the given value in kPa by this conversion factor. \[\frac{365 \text{ kPa}}{101.325 \text{ kPa/atm}} = 3.602 \text{ atm} \]
04

Convert 804 mmHg to kPa

Use the conversion factor between millimeters of mercury and kilopascals: 1 mmHg = 0.133322 kPa. Multiply the given value in mmHg by this conversion factor. \[ 804 \text{ mmHg} \times 0.133322 \text{ kPa/mmHg} = 107.19 \text{ kPa} \]

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Key Concepts

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

atm to mmHg
When converting atmospheres (atm) to millimeters of mercury (mmHg), it's essential to know the conversion factor. 1 atm is equal to 760 mmHg. This means that for every 1 atm, you get 760 mmHg. To convert, simply multiply the atm value by 760.
As an example, if you have 0.745 atm, you multiply it by 760 to get:
0.745 atm 脳 760 mmHg/atm = 565.4 mmHg.
This approach makes it easy to manage pressure conversions between these two units.
Remembering these conversion factors can save time and reduce errors in calculations.
torr to bar
Converting from torr to bar involves another useful conversion factor: 1 torr is equal to 0.00133322 bar. This means, for every torr value, you can multiply it by 0.00133322 to get the value in bar.
Take the example from the exercise: 992 torr. Multiply it by 0.00133322:
992 torr 脳 0.00133322 bar/torr = 1.3222184 bar.
It鈥檚 straightforward and helps you convert pressure units effortlessly.
Memorizing this factor comes in handy for quick conversions.
kPa to atm
To convert kilopascals (kPa) to atmospheres (atm), use the conversion factor where 1 atm equals 101.325 kPa. If you have a pressure value in kPa and want to figure out how many atm it is, divide the kPa value by 101.325.
For example, consider 365 kPa:
365 kPa 梅 101.325 kPa/atm = 3.602 atm.
This process makes it simple to switch from kilopascals to atmospheres in any calculation.
Keep this conversion factor in mind for quick and accurate pressure conversions.
mmHg to kPa
Converting mmHg (millimeters of mercury) to kPa (kilopascals) involves the conversion factor where 1 mmHg equals 0.133322 kPa. You multiply the mmHg value by 0.133322 to get the value in kPa.
Using the exercise example, take 804 mmHg and multiply it by 0.133322:
804 mmHg 脳 0.133322 kPa/mmHg = 107.19 kPa.
This method ensures you can easily convert pressure measurements from mmHg to kPa.
Remember this conversion factor for efficient and accurate pressure unit conversions.

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Most popular questions from this chapter

A large portion of metabolic energy arises from the biological combustion of glucose: $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ (a) If this reaction is carried out in an expandable container at \(37^{\circ} \mathrm{C}\) and \(780 .\) torr, what volume of \(\mathrm{CO}_{2}\) is produced from \(20.0 \mathrm{~g}\) of glucose and excess \(\mathrm{O}_{2} ?\) (b) If the reaction is carried out at the same conditions with the stoichiometric amount of \(\mathrm{O}_{2}\), what is the partial pressure of each gas when the reaction is \(50 \%\) complete \((10.0 \mathrm{~g}\) of glucose remains)?

Allotropes are different molecular forms of an element, such as dioxygen \(\left(\mathrm{O}_{2}\right)\) and ozone \(\left(\mathrm{O}_{3}\right) .\) (a) What is the density of each oxygen allotrope at \(0^{\circ} \mathrm{C}\) and 760 torr? (b) Calculate the ratio of densities, \(d_{\mathrm{O}_{3}} / d_{\mathrm{O}_{2}}\) and explain the significance of this number.

A gas-filled weather balloon with a volume of \(65.0 \mathrm{~L}\) is released at sea-level conditions of 745 torr and \(25^{\circ} \mathrm{C}\). The balloon can expand to a maximum volume of \(835 \mathrm{~L}\). When the balloon rises to an altitude at which the temperature is \(-5^{\circ} \mathrm{C}\) and the pressure is 0.066 atm, will it have expanded to its maximum volume?

An "empty" gasoline can with dimensions \(15.0 \mathrm{~cm}\) by \(40.0 \mathrm{~cm}\) by \(12.5 \mathrm{~cm}\) is attached to a vacuum pump and evacuated. If the atmospheric pressure is \(14.7 \mathrm{lb} / \mathrm{in}^{2},\) what is the total force (in pounds) on the outside of the can?

Convert each of the pressures described below to atm: (a) At the peak of Mt. Everest, atmospheric pressure is only \(2.75 \times 10^{2} \mathrm{mmHg}\) (b) A cyclist fills her bike tires to 86 psi. (c) The surface of Venus has an atmospheric pressure of \(9.15 \times 10^{6} \mathrm{~Pa}\) (d) At \(100 \mathrm{ft}\) below sea level, a scuba diver experiences a pressure of \(2.54 \times 10^{4}\) torr.
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