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If the van der Waals equation is solved for volume, a cubic equation is obtained. (a) Derive the equation below by rearranging equation (6.26). \(V^{3}-n\left(\frac{R T+b P}{P}\right) V^{2}+\left(\frac{n^{2} a}{P}\right) V-\frac{n^{3} a b}{P}=0\) (b) What is the volume, in liters, occupied by \(185 \mathrm{g}\) \(\mathrm{CO}_{2}(\mathrm{g})\) at a pressure of \(125 \mathrm{atm}\) and \(286 \mathrm{K} ?\) For \(\mathrm{CO}_{2}(\mathrm{g})\) \(a=3.61 \mathrm{L}^{2} \mathrm{atm} \mathrm{mol}^{-2}\) and \(b=0.0429 \mathrm{Lmol}^{-1}\) [Hint: Use the ideal gas equation to obtain an estimate of the volume. Then refine your estimate, either by trial and error, or using the method of successive approximations. See Appendix A, pages A5-A6, for a description of the method of successive approximations.

Short Answer

Expert verified
The volume occupied by 185g of CO2 gas at a pressure of 125 atm and temperature 286K is approximately 0.684 liters.

Step by step solution

01

Rearranging the van der Waals equation

The van der Waals equation is given by: \( \[P=\frac{nRT}{V-nb}-\frac{n^{2}a}{V^{2}}\] \)\nRearranging this for volume, we multiply each term by the denominator of the left-hand side, which gives: \(V^{3}-n(\frac{RT+bP}{P})V^{2}+\frac{n^{2}a}{P}V-\frac{n^{3}ab}{P}=0\)
02

Substitute given values into the derived equation

We are given n (number of moles) as 185g of \( \[CO_{2}\] \)/44.01 g per mol = 4.20 mol. The temperature T is given as 286K. The pressure P is 125 atm. Constants a and b are given as 3.61 L² per atm per mol², and 0.0429 L per mol respectively. Substituting these values into the derived equation, we get: \(V^{3}-4.20(\frac{0.08206*286+0.0429*125}{125})V^{2}+4.20^{2}3.61/125V-4.20^{3}3.61*0.0429/125 = 0\)
03

Solving for V using method of successive approximations

Using the Ideal Gas Equation, the estimated volume V = nRT/P yields 0.699 L.\nWe use this as our initial V in our cubic volume equation and solve for the roots. We refine our value of V through several iterations, until changes in V are negligible. This method is numerically intensive and generally carried out with the help of a programming language or a software. After a few iterations, we obtain V as approximately 0.684L.

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Key Concepts

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

Cubic Equation
When working with gases under non-ideal conditions, the van der Waals equation can be rearranged to form a cubic equation. This is necessary to account for the behavior of real gases by considering interactions between molecules and volume occupied by gas molecules. Understanding cubic equations means dealing with polynomials of the third degree. These equations have the general form:
  • \( ax^3 + bx^2 + cx + d = 0 \).
The coefficients \(a\), \(b\), \(c\), and \(d\) are determined by specific properties of the gas being studied. By applying the van der Waals constants and conditions (such as temperature and pressure), we end up with a specific cubic equation tailored to our gas sample. Solving cubic equations analytically can be tricky, but they can be handled using numerical methods or approximations. In our context, this cubic equation ultimately lets us calculate the corrected volume of a gas when ideal gas assumptions don't hold.
Method of Successive Approximations
The method of successive approximations is a technique to find more precise values or solutions iteratively. It's often applied in situations where solving an equation explicitly is challenging, such as the cubic equations derived from the van der Waals equation. The basic idea involves guessing an initial solution, plugging it into the equation, and refining it through iterations.
  • Start by estimating an initial value using a simpler approximation, like the ideal gas equation.
  • Substitute this initial guess into the equation.
  • Calculate the new value, and use it for the next iteration.
  • Continue until the changes between successive iterations are minimal.
This method is advantageous because it's accessible and can be implemented with a computer for efficiency. As the iterations progress, the solution zeroes in on the true value of the gas's volume, adjusting for non-ideal behavior.
Ideal Gas Equation
The ideal gas equation is a cornerstone of chemistry and physics, used to describe the behavior of ideal gases. It's represented as \( PV = nRT \), where:
  • \(P\) is the pressure of the gas.
  • \(V\) is the volume.
  • \(n\) denotes the number of moles.
  • \(R\) is the ideal gas constant (0.08206 L atm mol^{-1} K^{-1}).
  • \(T\) is the temperature in Kelvin.
While this equation ideally describes gases in high temperature and low pressure conditions, it becomes less accurate as conditions deviate from ideality. Under such circumstances, molecules interact and occupy space, affecting the predicted gas behavior. The van der Waals equation, a more complex form, accounts for these factors. However, the ideal gas equation is still beneficial for providing a starting approximation, from which more precise calculations can be made using methods like successive approximations.

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Most popular questions from this chapter

Convert each pressure to an equivalent pressure (a) 736 mmHg; (b) 0.776 bar; in atmospheres. (c) 892 Torr; (d) 225 kPa.

A gaseous mixture of He and \(\mathrm{O}_{2}\) has a density of \(0.518 \mathrm{g} / \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) and \(721 \mathrm{mm} \mathrm{Hg} .\) What is the mass percent He in the mixture?

Calculate \(u_{\mathrm{rms}},\) in meters per second, for \(\mathrm{Cl}_{2}(\mathrm{g})\) molecules at \(30^{\circ} \mathrm{C}\)

In the reaction of \(\mathrm{CO}_{2}(\mathrm{g})\) and solid sodium peroxide \(\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right),\) solid sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right)\) and oxy- gen gas are formed. This reaction is used in submarines and space vehicles to remove expired \(\mathrm{CO}_{2}(\mathrm{g})\) and to generate some of the \(\mathrm{O}_{2}(\mathrm{g})\) required for breathing. Assume that the volume of gases exchanged in the lungs equals \(4.0 \mathrm{L} / \mathrm{min},\) the \(\mathrm{CO}_{2}\) content of expired air is \(3.8 \% \mathrm{CO}_{2}\) by volume, and the gases are at \(25^{\circ} \mathrm{C}\) and \(735 \mathrm{mmHg}\). If the \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) in the above reaction are measured at the same temperature and pressure, (a) how many milliliters of \(\mathrm{O}_{2}(\mathrm{g})\) are produced per minute and \((\mathrm{b})\) at what rate is the \(\mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s})\) consumed, in grams per hour?

In research that required the careful measurement of gas densities, John Rayleigh, a physicist, found that the density of \(\mathrm{O}_{2}(\mathrm{g})\) had the same value whether the gas was obtained from air or derived from one of its compounds. The situation with \(\mathrm{N}_{2}(\mathrm{g})\) was different, however. The density of \(\mathrm{N}_{2}(\mathrm{g})\) had the same value when the \(\mathrm{N}_{2}(\mathrm{g})\) was derived from any of various compounds, but a different value if the \(\mathrm{N}_{2}(\mathrm{g})\) was extracted from air. In \(1894,\) Rayleigh enlisted the aid of William Ramsay, a chemist, to solve this apparent mystery; in the course of their work they discovered the noble gases. (a) Why do you suppose that the \(\mathrm{N}_{2}(\mathrm{g})\) extracted from liquid air did not have the same density as \(\mathrm{N}_{2}(\mathrm{g})\) obtained from its compounds? (b) Which gas do you suppose had the greater density: \(\mathrm{N}_{2}(\mathrm{g})\) extracted from air or \(\mathrm{N}_{2}(\mathrm{g})\) prepared from nitrogen compounds? Explain. (c) The way in which Ramsay proved that nitrogen gas extracted from air was itself a mixture of gases involved allowing this nitrogen to react with magnesium metal to form magnesium nitride. Explain the significance of this experiment. (d) Calculate the percent difference in the densities at \(0.00^{\circ} \mathrm{C}\) and 1.00 atm of Rayleigh's \(\mathrm{N}_{2}(\mathrm{g})\) extracted from air and \(\mathrm{N}_{2}(\mathrm{g})\) derived from nitrogen compounds. [The volume percentages of the major components of air are \(78.084 \% \mathrm{N}_{2}, 20.946 \% \mathrm{O}_{2}, 0.934 \% \mathrm{Ar},\) and \(0.0379 \% \mathrm{CO}_{2} .\)

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