Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What pressure (in atm) is exerted by a column of toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right) 87 \mathrm{~m}\) high? The density of toluene is \(0.867 \mathrm{~g} / \mathrm{cm}^{3}\)

Short Answer

Expert verified
The pressure exerted by the toluene column is approximately 7.32 atm.

Step by step solution

01

Convert Density to kg/m³

The given density of toluene is 0.867 g/cm³. We need to convert this to kg/m³. Since 1 g/cm³ = 1000 kg/m³, we have: \(0.867 \, \text{g/cm}^3 = 867 \, \text{kg/m}^3\).
02

Calculate the Hydrostatic Pressure

The pressure exerted by a liquid column is given by the formula: \(P = \rho gh\), where \(\rho\) is the density in kg/m³, \(g\) is the acceleration due to gravity (9.81 m/s²), and \(h\) is the height of the column in meters. Substitute \(\rho = 867\), \(g = 9.81\), and \(h = 87\) into the formula: \[ P = 867 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 87 \, \text{m} = 741099.57 \, \text{Pa} \].
03

Convert Pressure from Pa to atm

1 atmosphere (atm) is equivalent to 101325 Pascal (Pa). To convert the pressure from Pascal to atm, use the formula: \( P_{\text{atm}} = \frac{P}{101325} \). Hence, \( P_{\text{atm}} = \frac{741099.57}{101325} \approx 7.32 \text{ atm} \).

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.

Toluene Density
Toluene is a common organic solvent with interesting properties, one of which is its density. Knowing the density of a substance allows you to determine how much mass is in a given volume, which is crucial for calculating pressures in fluids. The density of toluene is given as 0.867 g/cm³. However, for calculations involving pressure over large areas or heights, the density is often converted to kg/m³.
To convert from g/cm³ to kg/m³, you use the conversion factor: 1 g/cm³ is equal to 1000 kg/m³. Thus, for toluene, this calculation looks like:
  • 0.867 g/cm³ × 1000 kg/m³ = 867 kg/m³
This conversion is essential for using the density in formulas that require SI units, like the hydrostatic pressure formula, which is expressed in terms of kg/m³, m/s², and meters. Remember, converting units accurately is a foundational skill in physics and helps ensure precise calculations.
Pressure Conversion
Pressure is a measure of force applied over an area. In certain calculations, especially those involving fluids like toluene, it’s important to work with pressure units that are most applicable to the situation. The term ‘hydrostatic pressure’ refers to the pressure exerted by a static fluid due to the force of gravity. The hydrostatic pressure (Default: P = ρgh),where
  • \( P \) is the pressure
  • \( \rho \) is the fluid density
  • \( g \) is the acceleration due to gravity
  • \( h \) is the height of the fluid column
is crucial in determining how much pressure a liquid exerts at a given depth.
For our scenario with toluene, the hydrostatic pressure was calculated:
\[ 867 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 87 \, \text{m} = 741099.57 \, \text{Pa} \]
An understanding of how different units interact is necessary to accurately convert and compare these values.
Pascal to Atmosphere Conversion
After calculating the pressure in units of Pascals from the hydrostatic column of toluene, it may be necessary to convert this to atmospheres for practical applications. Pascal (Pa) and atmosphere (atm) are both units of pressure, but they belong to different systems. One atmosphere is defined as 101325 Pascals.
To convert a pressure from Pascals to atmospheres, use the formula:
  • \( P_{\text{atm}} = \frac{P_{\text{Pa}}}{101325} \)
For instance, converting 741099.57 Pa:
  • \( P_{\text{atm}} = \frac{741099.57}{101325} \approx 7.32 \, \text{atm} \)
Understanding this conversion is handy when you need to interpret pressures in a context more familiar to everyday activities, like atmospheric pressure at sea level, making the information more relatable and easier to understand. Employing such conversions is a practical skill for scientists and engineers who often need to switch between different units to communicate their results effectively.

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

An unknown gas evolved from the fermentation of glucose is found to effuse through a porous barrier in 15.0 min. Under the same conditions of temperature and pressure, it takes an equal volume of \(\mathrm{N}_{2} 12.0 \mathrm{~min}\) to effuse through the same barrier. Calculate the molar mass of the unknown gas, and suggest what the gas might be.

The \({ }^{235} \mathrm{U}\) isotope undergoes fission when bombarded with neutrons. However, its natural abundance is only 0.72 percent. To separate it from the more abundant \({ }^{238} \mathrm{U}\) isotope, uranium is first converted to \(\mathrm{UF}_{6},\) which is easily vaporized above room temperature. The mixture of the \({ }^{235} \mathrm{UF}_{6}\) and \({ }^{238} \mathrm{UF}_{6}\) gases is then subjected to many stages of effusion. Calculate how much faster \({ }^{235} \mathrm{UF}_{6}\) effuses than \({ }^{238} \mathrm{UF}_{6}\)

Use the kinetic theory of gases to explain why hot air rises.

How does the kinetic molecular theory explain Boyle's law, Charles's law, Avogadro's law, and Dalton's law of partial pressures?

A student tries to determine the volume of a bulb like the one shown in Figure \(10.12 .\) These are her results: mass of the bulb filled with dry air at \(23^{\circ} \mathrm{C}\) and \(744 \mathrm{mmHg}=91.6843 \mathrm{~g} ;\) mass of evacuated bulb \(=\) \(91.4715 \mathrm{~g}\). Assume the composition of air is 78 percent \(\mathrm{N}_{2}, 21\) percent \(\mathrm{O}_{2},\) and 1 percent argon by volume. What is the volume (in mL) of the bulb? (Hint: First calculate the average molar mass of air, as shown in Problem \(3.129 .)\)

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