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In the United States, barometric pressures are generally reported in inches of mercury (in. \(\mathrm{Hg}\) ). On a beautiful summer day in Chicago, the barometric pressure is \(30.45 \mathrm{in} . \mathrm{Hg}\). (a) Convert this pressure to torr. (b) Convert this pressure to atm.

Short Answer

Expert verified
The barometric pressure of \( 30.45 \; \mathrm{in. \; Hg} \) is equal to \( 772.29 \; \mathrm{torr} \) and \( 1.0162 \; \mathrm{atm} \).

Step by step solution

01

Convert in. Hg to mm Hg

To convert the given pressure from inches of mercury to millimeters of mercury, we can use the conversion factor 1 in. Hg = 25.4 mm Hg. So, we multiply the given pressure by this factor: \( 30.45 \; \mathrm{in. \; Hg} \times \frac{25.4 \; \mathrm{mm \; Hg}}{1 \; \mathrm{in. \; Hg}} \) #Step 2: Calculate the pressure in torr#
02

Convert mm Hg to torr

Now that we have the pressure in mm Hg, we can easily convert it to torr, since 1 mm Hg = 1 torr. So the pressure in torr is equal to: \( 30.45 \times 25.4 = 772.29 \; \mathrm{torr} \) #Step 3: Convert the pressure to atmospheres#
03

Convert torr to atm

Finally, we'll convert the pressure from torr to atmospheres using the conversion factor 1 atm = 760 torr. So, we divide the pressure in torr by this factor: \( \frac{772.29 \; \mathrm{torr}}{760 \; \mathrm{torr/atm}} \) #Step 4: Calculate the pressure in atmospheres#
04

Find the pressure in atm

Now, we can calculate the pressure in atmospheres: \( \frac{772.29}{760} = 1.0162 \; \mathrm{atm} \) In conclusion, the barometric pressure of 30.45 in. Hg is equal to 772.29 torr and 1.0162 atm.

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Key Concepts

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

Pressure Unit Conversion
Understanding pressure unit conversions is essential for students in fields such as meteorology, physics, and engineering. Pressure, which is force per unit area, can be reported in a variety of units including pounds per square inch (psi), pascals (Pa), bar, atmospheres (atm), and millimeters of mercury (mm Hg or torr).

To convert between these units, one must use established conversion factors, which are ratios that express how much of one unit is equivalent to another. For example, the conversion from inches of mercury to millimeters of mercury is based on the direct relationship that 1 in. Hg equals exactly 25.4 mm Hg. Therefore, to convert from in. Hg to mm Hg (or torr), you simply multiply by 25.4.

Being proficient with these conversions allows for seamless communication and comparison of data across different scientific and engineering disciplines.
Atmospheric Pressure
The term atmospheric pressure refers to the force exerted by the Earth's atmosphere at a given point. It is a crucial factor in meteorology, affecting weather patterns, and is also important in various scientific applications, such as determining the boiling point of liquids.

At sea level, the standard atmospheric pressure is defined as 101,325 pascals (Pa) or equivalently 1 atmosphere (atm). Different units can be used to express this value, such as torr or inches of mercury. Atmospheric pressure varies with altitude and weather conditions, hence why understanding the concept and the ability to convert between units is important for accurately interpreting weather reports and performing scientific calculations.
Torr
The unit torr is named after Evangelista Torricelli, an Italian physicist and mathematician who is credited with inventing the barometer. One torr is defined as precisely \( \frac{1}{760} \) of a standard atmosphere.

It's important to note that the torr is not part of the International System of Units (SI), but is still widely used in certain fields, particularly in the measurement of vacuum pressures and in meteorology. Since the relationship between mm Hg and torr is 1:1, converting from in. Hg to torr simply involves multiplying the inch measurement by 25.4. This relationship highlights the historical use of mercurial barometers, where atmospheric pressure was determined by the height of a mercury column.
Atmospheres
An atmosphere (atm) is a unit of pressure based on the average atmospheric pressure at sea level. It was originally intended to represent the typical pressure that the Earth's atmosphere exerts at sea level and is defined as 101,325 pascals (Pa) or 760 millimeters of mercury (760 mm Hg or 760 torr).

In practical applications, the atm is convenient for high-pressure systems, such as those found in the fields of oceanography and hydraulics. Converting from torr to atm is a straightforward division, with the understanding that the standard atmosphere is equivalent to 760 torr. Subsequently, dividing a pressure value in torr by 760 yields the pressure in atm, making it easier for students and professionals alike to interpret and utilize pressure data.

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Most popular questions from this chapter

An herbicide is found to contain only \(\mathrm{C}, \mathrm{H}, \mathrm{N}\), and \(\mathrm{Cl}\). The complete combustion of a \(100.0\)-mg sample of the herbicide in excess oxygen produces \(83.16 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) and \(73.30 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) vapor at STP. A separate analysis shows that the sample also contains \(16.44 \mathrm{mg}\) of \(\mathrm{Cl}\). (a) Determine the percentage of the composition of the substance. (b) Calculate its empirical formula. (c) What other information would you need to know about this compound to calculate its true molecular formula?

Chlorine dioxide gas \(\left(\mathrm{ClO}_{2}\right)\) is used as a commercial bleaching agent. It bleaches materials by oxidizing them. In the course of these reactions, the \(\mathrm{ClO}_{2}\) is itself reduced. (a) What is the Lewis structure for \(\mathrm{ClO}_{2}\) ? (b) Why do you think that \(\mathrm{ClO}_{2}\) is reduced so readily? (c) When a \(\mathrm{ClO}_{2}\) molecule gains an electron, the chlorite ion, \(\mathrm{ClO}_{2}^{-}\), forms. Draw the Lewis structure for \(\mathrm{ClO}_{2}^{-}\). (d) Predict the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in the \(\mathrm{ClO}_{2}^{-}\)ion. (e) One method of preparing \(\mathrm{ClO}_{2}\) is by the reaction of chlorine and sodium chlorite: $$ \mathrm{Cl}_{2}(g)+2 \mathrm{NaClO}_{2}(s) \longrightarrow 2 \mathrm{ClO}_{2}(g)+2 \mathrm{NaCl}(s) $$ If you allow \(15.0 \mathrm{~g}\) of \(\mathrm{NaClO}_{2}\) to react with \(2.00 \mathrm{~L}\) of chlorine gas at a pressure of \(1.50\) atm at \(21^{\circ} \mathrm{C}\), how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

(a) How high in meters must a column of water be to exert a pressure equal to that of a \(760-\mathrm{mm}\) column of mercury? The density of water is \(1.0 \mathrm{~g} / \mathrm{mL}\), whereas that of mercury is \(13.6 \mathrm{~g} / \mathrm{mL}\). (b) What is the pressure, in atmospheres, on the body of a diver if he or she is \(39 \mathrm{ft}\) below the surface of the water when atmospheric pressure at the surface is \(0.97\) atm?

Both Jacques Charles and Joseph Louis Guy-Lussac were avid balloonists. In his original flight in 1783 , Jacques Charles used a balloon that contained approximately \(31,150 \mathrm{~L}\) of \(\mathrm{H}_{2}\). He generated the \(\mathrm{H}_{2}\) using the reaction between iron and hydrochloric acid: $$ \mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ How many kilograms of iron were needed to produce this volume of \(\mathrm{H}_{2}\) if the temperature was \(22^{\circ} \mathrm{C}\) ?

Briefly explain the significance of the constants \(a\) and \(b\) in the van der Waals equation.

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