Perform each conversion. (a) \(2.3 \mathrm{~atm}\) to torr (b) \(4.7 \times 10^{-2}\) atm to millimeters of mercury (c) \(24.8\) psi to millimeters of mercury (d) \(32.84 \mathrm{in}\). Hg to torr

Short Answer

Expert verified
(a) 1748 torr, (b) 35.72 mm Hg, (c) 1282.532 mm Hg, (d) 834.136 torr

Step by step solution

01

Convert Atmospheres to Torr for 2.3 atm

Use the conversion factor 1 atm = 760 torr. Multiply the given pressure in atmospheres by 760 torr to convert to torr. So, the conversion is: \(2.3 \, \text{atm} \times 760 \, \text{torr/atm} = 1748 \, \text{torr}\).
02

Convert Atmospheres to Millimeters of Mercury for 4.7 x 10^-2 atm

Use the conversion factor 1 atm = 760 mm Hg. Multiply the given pressure in atmospheres by 760 mm Hg to convert to millimeters of mercury. So, the conversion is: \(4.7 \times 10^{-2} \, \text{atm} \times 760 \, \text{mm Hg/atm} = 35.72 \, \text{mm Hg}\).
03

Convert PSI to Millimeters of Mercury for 24.8 psi

Use the conversion factor 1 psi = 51.715 mm Hg. Multiply the given pressure in psi by 51.715 mm Hg to convert to millimeters of mercury. So, the conversion is: \(24.8 \, \text{psi} \times 51.715 \, \text{mm Hg/psi} = 1282.532 \, \text{mm Hg}\).
04

Convert Inches of Mercury to Torr for 32.84 in. Hg

Use the conversion factor 1 in. Hg = 25.4 mm Hg and 1 mm Hg = 1 torr. First, convert inches of mercury to millimeters of mercury, then to torr. So, the conversion is: \(32.84 \, \text{in. Hg} \times 25.4 \, \text{mm Hg/in. Hg} = 834.136 \, \text{mm Hg} \), and since 1 mm Hg equals 1 torr, the value in torr is the same: 834.136 torr.
05

Round Off If Necessary

Round off the final values as per the required significant digits or the context of the calculation. In this case, no specific requirements for rounding off were provided, so we can present the numbers as calculated.

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 Vaia!

Key Concepts

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

Atmospheres to Torr
Understanding how to convert atmospheres to torr is crucial in various scientific disciplines, especially when dealing with pressure measurements. An atmosphere (\texttt{atm}) is a unit of pressure defined as the pressure exerted by the weight of the Earth's atmosphere at sea level, while the torr is a unit of pressure named after Evangelista Torricelli, an Italian physicist and mathematician. The relationship between these units is linear and uncomplicated: 1 \texttt{atm} is equivalent to 760 torr.

To perform a conversion, simply multiply the value in atmospheres by 760. For instance, if you have a pressure of 2.3 \texttt{atm}, the conversion to torr is calculated as follows: \(2.3 \texttt{ atm} \times 760 \texttt{ torr/atm} = 1748 \texttt{ torr}\). This straightforward calculation is essential for tasks ranging from calibrating laboratory instruments to setting correct pressures in industrial systems.
Atmospheres to Millimeters of Mercury
Converting atmospheres to millimeters of mercury (mm Hg) is a common practice in science, particularly in fields that involve fluid dynamics and weather prediction. Like the torr, the millimeters of mercury unit is derived from the height of a column of mercury that can be supported by a given pressure. The conversion factor between atmospheres and millimeters of mercury is the same as with torr, since 1 torr is equal to 1 mm Hg.

To convert a pressure in atmospheres to millimeters of mercury, you multiply by 760. For example, taking a small pressure of 0.047 \texttt{atm}: \(4.7 \times 10^{-2} \texttt{ atm} \times 760 \texttt{ mm Hg/atm} = 35.72 \texttt{ mm Hg}\).

This conversion is frequently needed for medical applications such as blood pressure measurements, as well as in the calibration of altimeters used in aviation.
PSI to Millimeters of Mercury
The pound per square inch (psi) is another widely used unit for measuring pressure, especially in the United States and in the fields of mechanics and hydraulics. In order to convert psi to millimeters of mercury, a specific conversion factor is used: 1 psi is equivalent to approximately 51.715 mm Hg.

When faced with the task of converting 24.8 psi to mm Hg, the calculation is as follows: \(24.8 \texttt{ psi} \times 51.715 \texttt{ mm Hg/psi} = 1282.532 \texttt{ mm Hg}\).

This relationship allows technicians to convert between measurements taken in different systems without losing accuracy, ensuring that systems are functioning correctly and safely within their designed parameters.
Inches of Mercury to Torr
In some instances, pressure is measured in inches of mercury (in. Hg), particularly in meteorology when reporting barometric pressures. To convert inches of mercury to torr, one must first convert in. Hg to mm Hg using the factor 1 in. Hg equals 25.4 mm Hg, as millimeters of mercury and torr are directly equivalent.

For example, the conversion for 32.84 in. Hg to torr includes two steps. First, convert inches to millimeters: \(32.84 \texttt{ in. Hg} \times 25.4 \texttt{ mm Hg/in. Hg} = 834.136 \texttt{ mm Hg}\). Then, recognize that 1 mm Hg is equal to 1 torr: 834.136 \texttt{ mm Hg} = 834.136 \texttt{ torr}.

This conversion is particularly helpful for aviation where aircraft altimeter settings are based on inches of mercury but may need conversion for compatibility with international standards.

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

Consider the reaction: $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) $$ A reaction flask initially contains \(0.10\) atm of \(\mathrm{SO}_{2}\) and \(0.10\) atm of \(\mathrm{O}_{2}\). What is the total pressure in the flask once the limiting reactant is completely consumed? Assume a constant temperature and volume and a \(100 \%\) reaction yield.

Consider the reaction for the synthesis of nitric acid: $$ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) $$ (a) If \(12.8 \mathrm{~L}\) of \(\mathrm{NO}_{2}(g)\), measured at \(\mathrm{STP}\), is allowed to react with \(14.9 \mathrm{~g}\) of water, find the limiting reagent and the theoretical yield of \(\mathrm{HNO}_{3}\) in grams. (b) If \(14.8 \mathrm{~g}\) of \(\mathrm{HNO}_{3}\) forms, what is the percent yield?

Calculate the volume of each gas sample at STP. (a) \(22.5 \mathrm{~mol} \mathrm{Cl}_{2}\) (b) \(3.6\) mol nitrogen (c) \(2.2\) mol helium (d) \(27 \mathrm{~mol} \mathrm{CH}_{4}\)

How many moles of gas must be forced into a 4.8-L tire to give it a gauge pressure of \(32.4\) psi at \(25^{\circ} \mathrm{C}\) ? The gauge pressure is relative to atmospheric pressure. Assume that atmospheric pressure is \(14.7\) psi so that the total pressure in the tire is \(47.1 \mathrm{psi}\).

A heliox deep-sea diving mixture delivers an oxygen partial pressure of \(0.30 \mathrm{~atm}\) when the total pressure is \(11.0 \mathrm{~atm}\). What is the partial pressure of helium in this mixture?

See all solutions

Recommended explanations on Chemistry Textbooks

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.

Sign-up for free