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

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards

Over 22 million students worldwide already upgrade their learning with Vaia!

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, \( P_{\text{atm}} = \frac{741099.57}{101325} \approx 7.32 \text{ atm} \).

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

A sample of zinc metal reacts completely with an excess of hydrochloric acid: $$ \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ The hydrogen gas produced is collected over water at \(25.0^{\circ} \mathrm{C}\) using an arrangement similar to that shown in Figure \(10.14(\mathrm{a})\). The volume of the gas is \(7.80 \mathrm{~L},\) and the pressure is 0.980 atm. Calculate the amount of zinc metal in grams consumed in the reaction. (Vapor pressure of water at \(\left.25^{\circ} \mathrm{C}=23.8 \mathrm{mmHg} .\right)\)

Because the van der Waals constant \(b\) is the excluded volume per mole of a gas, we can use the value of \(b\) to estimate the radius of a molecule or atom. Consider a gas that consists of molecules, for which the van der Waals constant \(b\) is \(0.0315 \mathrm{~L} / \mathrm{mol}\). Estimate the molecular radius in pm. Assume that the molecules are spherical.

Under the same conditions of temperature and pressure, why does \(1 \mathrm{~L}\) of moist air weigh less than \(1 \mathrm{~L}\) of dry air?

Apply your knowledge of the kinetic theory of gases to the following situations. (a) Two flasks of volumes \(V_{1}\) and \(V_{2}\left(V_{2}>V_{1}\right)\) contain the same number of helium atoms at the same temperature. (i) Compare the rootmean-square (rms) speeds and average kinetic energies of the helium (He) atoms in the flasks. (ii) Compare the frequency and the force with which the He atoms collide with the walls of their containers. (b) Equal numbers of He atoms are placed in two flasks of the same volume at temperatures \(T_{1}\) and \(T_{2}\left(T_{2}>T_{1}\right) .\) (i) Compare the rms speeds of the atoms in the two flasks. (ii) Compare the frequency and the force with which the He atoms collide with the walls of their containers. (c) Equal numbers of He and neon (Ne) atoms are placed in two flasks of the same volume, and the temperature of both gases is \(74^{\circ} \mathrm{C}\). Comment on the validity of the following statements: (i) The rms speed of He is equal to that of Ne. (ii) The average kinetic energies of the two gases are equal. (iii) The rms speed of each He atom is \(1.47 \times 10^{3} \mathrm{~m} / \mathrm{s}\)

Some commercial drain cleaners contain a mixture of sodium hydroxide and aluminum powder. When the mixture is poured down a clogged drain, the following reaction occurs: $$ 2 \mathrm{NaOH}(a q)+2 \mathrm{Al}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaAl}(\mathrm{OH})_{4}(a q)+3 \mathrm{H}_{2}(g) $$ The heat generated in this reaction helps melt away obstructions such as grease, and the hydrogen gas released stirs up the solids clogging the drain. Calculate the volume of \(\mathrm{H}_{2}\) formed at \(23^{\circ} \mathrm{C}\) and 1.00 atm if \(3.12 \mathrm{~g}\) of \(\mathrm{Al}\) are treated with an excess of \(\mathrm{NaOH}\)

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