/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 Rewrite the van der Waals equati... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Rewrite the van der Waals equation using the molar volume rather than \(V\) and \(n\).

Short Answer

Expert verified
To rewrite the van der Waals equation using molar volume (\(v\)) instead of \(V\) and \(n\), we first define molar volume as \(v = \frac{V}{n}\). Then, we substitute \(V = nv\) in the van der Waals equation and simplify it. The final expression is \[\left(P + \frac{a}{v^2}\right)(v - b) = RT\].

Step by step solution

01

Introduce the molar volume

First, let's write the molar volume in terms of the volume and number of moles: \[v = \frac{V}{n}\]
02

Express V and n in terms of molar volume

Now, we want to express \(V\) and \(n\) in terms of the molar volume, \(v\). To do that, we can solve the molar volume equation for \(V\) and multiply both sides by \(n\): \[V=nv\]
03

Substitute V and n in the van der Waals equation

Next, let's substitute the expression for \(V\) we found in the van der Waals equation: \[\left(P + \frac{an^2}{(nv)^2}\right)(nv - nb) = nRT\]
04

Simplify the equation

Now, we simplify the modified van der Waals equation by cancelling out terms and rearranging: \[\left(P + \frac{a}{v^2}\right)(nv - nb) = nRT\] Since we want the equation to be in terms of molar volume (\(v\)), we need to eliminate factor \(n\). We can do this by dividing both sides of the equation by \(n\): \[\left(P + \frac{a}{v^2}\right)(v - b) = RT\]
05

Final expression

The van der Waals equation rewritten using molar volume is: \[\left(P + \frac{a}{v^2}\right)(v - b) = RT\]

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 91Ó°ÊÓ!

Key Concepts

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

Understanding Molar Volume
Molar volume is a measure of the space that one mole of a substance occupies at a given temperature and pressure. It is a significant concept in physical chemistry because it helps connect the microscopic world of molecules with macroscopic physical properties. When dealing with gases, the concept of molar volume becomes particularly valuable. The standard molar volume of an ideal gas at standard temperature and pressure (STP, 0°C and 1 atm) is approximately 22.4 liters. In the context of gas laws and equations, molar volume allows for a more refined analysis since it reflects the volume contribution of individual gas moles.

For practical applications, we use the formula:
\[v = \frac{V}{n}\]
where \(v\) represents the molar volume, \(V\) is the total volume of the gas, and \(n\) is the number of moles. When discussing real gases, deviations from ideal behavior are introduced, making the simple relationship between \(V\), \(n\), and \(v\) more complex. This complexity is where the van der Waals equation comes into play, helping us to model these real-life scenarios more accurately.
The Role of Physical Chemistry
Physical chemistry is a branch of chemistry focused on understanding the physical properties of molecules, the forces that act on them, and how these properties give rise to chemical phenomena. It involves the laws of physics to explain how chemical processes occur, from the very small (molecular) level to the macroscopic level, including gases, liquids, and solids. Key concepts such as thermodynamics, kinetics, and quantum mechanics are integral to physical chemistry and are used to interpret the behavior of matter.

Gas laws, which describe how pressure, volume, and temperature relate to each other for a gas, are prime examples of the principles that physical chemistry delves into. The van der Waals equation is one such law that goes a step beyond the ideal gas law to account for the finite size of molecules and the attractions between them. By manipulating this equation, we can explore and predict the real behavior of gases more accurately.
Exploring Gas Laws
Gas laws are fundamental to understanding how gases behave under different conditions. Initially, scientists like Boyle, Charles, and Avogadro described the relationship between pressure (P), volume (V), and temperature (T) in ideal gases—gases which are theoretical models where particles do not interact. An ideal gas perfectly follows the Ideal Gas Law:\[PV = nRT\]

However, real gases deviate from this ideal behavior due to molecular size and intermolecular forces. The van der Waals equation, '
\[\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT\]
'introduces two corrective factors to the Ideal Gas Law—the \(a\) and \(b\) constants that represent these non-ideal interactions and the finite size of molecules, respectively. As such, understanding how to manipulate the van der Waals equation—and gas laws in general—not only expands our comprehension of gas behavior but also allows us to predict the properties of gases in more real-world situations.
Equation Manipulation in Practice
Equation manipulation is a crucial skill in chemistry, enabling one to reshape formulas to either solve for unknown variables or express the relationships in a more convenient form. This process often includes algebraic steps such as solving for a variable, rearranging terms, factorizing, or cancelling out common factors. In the context of the van der Waals equation, manipulation is necessary to express it in terms of molar volume, leading to a clearer understanding of how pressure and temperature can influence the volume occupied by one mole of a gas.

Through a series of steps, we transformed the original van der Waals equation to '
\[\left(P + \frac{a}{v^2}\right)(v - b) = RT\]
'which represents a significant improvement because it allows for direct insights into how real gases differ from ideal gases on a per-mole basis. This understanding can be incredibly valuable in contexts such as chemical engineering or materials science, where precise measurements and predictions are crucial.

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

Consider a \(31.0 \mathrm{L}\) sample of moist air at \(60 .^{\circ} \mathrm{C}\) and one atm in which the partial pressure of water vapor is 0.131 atm. Assume that dry air has the composition 78.0 mole percent \(\mathrm{N}_{2}, 21.0\) mole percent \(\mathrm{O}_{2},\) and 1.00 mole percent Ar. a. What are the mole percentages of each of the gases in the sample? b. The percent relative humidity is defined as \(\% \mathrm{RH}=\) \(P_{H_{2}} o / P_{H_{2} O}^{*}\) where \(P_{H_{2}, O}\) is the partial pressure of water in the sample and \(P_{H, O}^{*}=0.197\) atm is the equilibrium vapor pressure of water at \(60 .^{\circ} \mathrm{C}\). The gas is compressed at \(60 .^{\circ} \mathrm{C}\) until the relative humidity is \(100 . \% .\) What volume does the mixture contain now? c. What fraction of the water will be condensed if the total pressure of the mixture is isothermally increased to 81.0 atm?

A mixture of \(\mathrm{H}_{2}\) and \(\mathrm{NH}_{3}\) has a volume of \(139.0 \mathrm{cm}^{3}\) at \(0.00^{\circ} \mathrm{C}\) and 1 atm. The mixture is cooled to the temperature of liquid nitrogen at which ammonia freezes out and the remaining gas is removed from the vessel. Upon warming the vessel to \(0.00^{\circ} \mathrm{C}\) and 1 atm, the volume is \(77.4 \mathrm{cm}^{3}\). Calculate the mole fraction of \(\mathrm{NH}_{3}\) in the original mixture.

A mixture of \(2.10 \times 10^{-3} \mathrm{g}\) of \(\mathrm{O}_{2}, 3.88 \times 10^{-3} \mathrm{mol}\) of \(\mathrm{N}_{2},\) and \(5.25 \times 10^{20}\) molecules of \(\mathrm{CO}\) are placed into a vessel of volume 5.25 L at \(12.5^{\circ} \mathrm{C}\) a. Calculate the total pressure in the vessel. b. Calculate the mole fractions and partial pressures of each gas.

One liter of fully oxygenated blood can carry 0.18 liters of \(\mathrm{O}_{2}\) measured at \(T=298 \mathrm{K}\) and \(P=1.00 \mathrm{atm}\) Calculate the number of moles of \(\mathrm{O}_{2}\) carried per liter of blood. Hemoglobin, the oxygen transport protein in blood has four oxygen binding sites. How many hemoglobin molecules are required to transport the \(\mathrm{O}_{2}\) in \(1.0 \mathrm{L}\) of fully oxygenated blood?

When Julius Caesar expired, his last exhalation had a volume of \(450 . \mathrm{cm}^{3}\) and contained 1.00 mole percent argon. Assume that \(T=300 . \mathrm{K}\) and \(P=1.00\) atm at the location of his demise. Assume further that \(T\) has the same value throughout Earth's atmosphere. If all of his exhaled Ar atoms are now uniformly distributed throughout the atmosphere, how many inhalations of \(450, \mathrm{cm}^{3}\) must we make to inhale one of the Ar atoms exhaled in Caesar's last breath? Assume the radius of Earth to be \(6.37 \times 10^{6} \mathrm{m}\). [Hint: Calculate the number of Ar atoms in the atmosphere in the simplified geometry of a plane of area equal to that of Earth's surface. See Problem P1.20 for the dependence of the barometric pressure and the composition of air on the height above Earth's surface.

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.