Problem 113
Question
Perform the following conversions: (a) \(30.2\) in. \(\mathrm{Hg}\) to millimeters of mercury (b) \(890.0 \mathrm{~mm} \mathrm{Hg}\) to atmospheres (c) \(300.0 \mathrm{lb} / \mathrm{in} .^{2}\) to atmospheres
Step-by-Step Solution
Verified Answer
(a) 767.08 mm Hg
(b) 1.17 atm
(c) 20.42 atm
1Step 1: (a) Convert inches of mercury to millimeters of mercury
To perform this conversion, we need to know the conversion factor between inches (in) and millimeters (mm). We have the following information:
1 in = 25.4 mm
Now we can convert 30.2 in. Hg to mm Hg using the given conversion factor:
\(30.2 \, \text{in. Hg} \times \frac{25.4 \, \text{mm}}{1 \, \text{in}}\)
2Step 2: Calculate mm Hg value
Multiply the values:
\(30.2 \times 25.4 = 767.08\)
So, 30.2 in. Hg is equal to 767.08 mm Hg.
3Step 3: (b) Convert millimeters of mercury to atmospheres
To perform this conversion, we need to know the conversion factor between millimeters of mercury (mm Hg) and atmospheres (atm). We have the following information:
1 atm = 760 mm Hg
Now we can convert 890.0 mm Hg to atm using the given conversion factor:
\(890.0\, \text{mm Hg} \times \frac{1\, \text{atm}}{760\, \text{mm Hg}}\)
4Step 4: Calculate atmospheric pressure value
Divide the values:
\(\frac{890.0}{760} = 1.17\)
So, 890.0 mm Hg is equal to 1.17 atm.
5Step 5: (c) Convert pounds per square inch to atmospheres
To perform this conversion, we need to know the conversion factor between pounds per square inch (lb/in²) and atmospheres (atm). We have the following information:
1 atm = 14.696 lb/in²
Now we can convert 300.0 lb/in² to atm using the given conversion factor:
\(300.0\, \text{lb/in}^2 \times \frac{1\, \text{atm}}{14.696\, \text{lb/in}^2}\)
6Step 6: Calculate atmospheric pressure value
Divide the values:
\(\frac{300.0}{14.696} = 20.42\)
So, 300.0 lb/in² is equal to 20.42 atm.
Key Concepts
Converting Inches of Mercury to MillimetersConverting Millimeters of Mercury to AtmospheresConverting Pounds per Square Inch to Atmospheres
Converting Inches of Mercury to Millimeters
Understanding how to convert pressure measurements from inches of mercury (\textbf{in Hg}) to millimeters of mercury (\textbf{mm Hg}) is essential in the field of chemistry, particularly when dealing with various pressure units across different regions and scientific literature.
The conversion is straightforward once you know the basic relation between inches and millimeters: \textbf{1 inch is exactly 25.4 millimeters}. To convert \textbf{in Hg} to \textbf{mm Hg}, simply multiply the value in inches by 25.4.
The conversion is straightforward once you know the basic relation between inches and millimeters: \textbf{1 inch is exactly 25.4 millimeters}. To convert \textbf{in Hg} to \textbf{mm Hg}, simply multiply the value in inches by 25.4.
Step-by-Step Conversion Process
For instance, if you have a pressure reading of 30.2 in Hg, the conversion to millimeters will be: \[ 30.2 \, \text{in. Hg} \times \frac{25.4 \, \text{mm}}{1 \, \text{in}} = 767.08 \, \text{mm Hg} \] As a tip for students, always ensure to keep track of your units during conversion; it helps prevent mistakes and solidifies understanding of the concepts.Converting Millimeters of Mercury to Atmospheres
Another common unit conversion in chemistry is from \textbf{millimeters of mercury (mm Hg)} to \textbf{atmospheres (atm)}, which is a standard unit for measuring pressure in scientific disciplines.
The conversion factor to remember here is that \textbf{1 atmosphere} is equivalent to \textbf{760 millimeters of mercury}. To convert from \textbf{mm Hg} to \textbf{atm}, divide the value in millimeters by 760.
The conversion factor to remember here is that \textbf{1 atmosphere} is equivalent to \textbf{760 millimeters of mercury}. To convert from \textbf{mm Hg} to \textbf{atm}, divide the value in millimeters by 760.
Example of Conversion
As an example, to convert 890.0 mm Hg to atm: \[ 890.0 \, \text{mm Hg} \times \frac{1 \, \text{atm}}{760 \, \text{mm Hg}} = 1.17 \, \text{atm} \] It's important for students to recognize the significance of standard atmospheric pressure being 760 mm Hg as it represents the pressure exerted by a column of mercury one millimeter high at the standard acceleration of gravity. This fact roots the measurement of atmospheric pressure in physical phenomena.Converting Pounds per Square Inch to Atmospheres
The conversion from \textbf{pounds per square inch (lb/in²)} to \textbf{atmospheres (atm)} is invaluable for students working with systems involving different pressure units, such as those found in the United States and scientific international standards.
Remember, \textbf{1 atmosphere} is defined as \textbf{14.696 lb/in²}.
This tells us that 300.0 lb/in² is significantly higher than the standard atmospheric pressure. Students should note that this conversion is beneficial when understanding the pressure readings in pneumatic systems or when comparing weather-related pressure values across different systems.
Remember, \textbf{1 atmosphere} is defined as \textbf{14.696 lb/in²}.
- To convert \textbf{lb/in²} to \textbf{atm}, divide the value in pounds per square inch by 14.696.
This tells us that 300.0 lb/in² is significantly higher than the standard atmospheric pressure. Students should note that this conversion is beneficial when understanding the pressure readings in pneumatic systems or when comparing weather-related pressure values across different systems.
Other exercises in this chapter
Problem 111
A \(1.25-g\) sample of a gas of unknown identity occupies \(2.50 \mathrm{~L}\). and exerts a pressure of \(0.400 \mathrm{~atm}\) at \(0^{\circ} \mathrm{C}\). (a
View solution Problem 112
If a quantity of gas occupies \(850 \mathrm{~mL}\) at \(300 \mathrm{~K}\) and \(750 \mathrm{~mm} \mathrm{Hg}\), what volume in milliliters will the gas occupy a
View solution Problem 114
If \(8.50\) moles of He gas occupy a volume of \(25.0 \mathrm{~L}\) at \(28^{\circ} \mathrm{C}\), what pressure in atmospheres does the gas exert?
View solution Problem 115
If \(48.3 \mathrm{~g}\) of an unknown gas occupies \(10.0 \mathrm{~L}\) at \(40^{\circ} \mathrm{C}\) and \(3.10 \mathrm{~atm}\), what is the molar mass of the g
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