Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Question:**
What pressure (in atm) is exerted by a column of toluene (C₇H₈) 87 m high? The density of toluene is 0.867 g/cm³.
**Explanation:**
To solve this problem, the pressure exerted by a liquid column can be calculated using the formula:
\[ P = h \cdot \rho \cdot g \]
Where:
- \( P \) is the pressure,
- \( h \) is the height of the liquid column (in meters),
- \( \rho \) is the density of the liquid (in kg/m³),
- \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).
First, convert the density from \(\text{g/cm}^3\) to \(\text{kg/m}^3\):
\[
0.867 \, \text{g/cm}^3 = 867 \, \text{kg/m}^3
\]
Next, use the formula to calculate the pressure:
\[
P = 87 \, \text{m} \times 867 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2
\]
Finally, convert the pressure from pascals (Pa) to atmospheres (atm), using the conversion factor \( 1 \, \text{atm} = 101,325 \, \text{Pa} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffb787237-32bd-4647-baa5-49b05320a55c%2F15d4cf97-7eac-4b0b-999d-dc5b71f2c036%2Fw6zosr_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
What pressure (in atm) is exerted by a column of toluene (C₇H₈) 87 m high? The density of toluene is 0.867 g/cm³.
**Explanation:**
To solve this problem, the pressure exerted by a liquid column can be calculated using the formula:
\[ P = h \cdot \rho \cdot g \]
Where:
- \( P \) is the pressure,
- \( h \) is the height of the liquid column (in meters),
- \( \rho \) is the density of the liquid (in kg/m³),
- \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).
First, convert the density from \(\text{g/cm}^3\) to \(\text{kg/m}^3\):
\[
0.867 \, \text{g/cm}^3 = 867 \, \text{kg/m}^3
\]
Next, use the formula to calculate the pressure:
\[
P = 87 \, \text{m} \times 867 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2
\]
Finally, convert the pressure from pascals (Pa) to atmospheres (atm), using the conversion factor \( 1 \, \text{atm} = 101,325 \, \text{Pa} \).
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