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Chapter 5: Problem 78

The van der Waals equation of state (Equation \(5.3-7\) ) is to be used to estimate the specific molar volume \(\hat{V}(\mathrm{L} / \mathrm{mol})\) of air at specified values of \(T(\mathrm{K})\) and \(P(\mathrm{atm}) .\) The van der Waals constants for air are \(a=1.33 \mathrm{atm} \cdot \mathrm{L}^{2} / \mathrm{mol}^{2}\) and \(b=0.0366 \mathrm{L} / \mathrm{mol}\) (a) Show why the van der Waals equation is classified as a cubic equation of state by expressing it in the form $$f(\hat{V})=c_{3} \hat{V}^{3}+c_{2} \hat{V}^{2}+c_{1} \hat{V}+c_{0}=0$$ where the coefficients \(c_{3}, c_{2}, c_{1},\) and \(c_{0}\) involve \(P, R, T, a,\) and \(b .\) Calculate the values of these coefficients for air at \(223 \mathrm{K}\) and 50.0 atm. (Include the units when giving the values.) (b) What would the value of \(\hat{V}\) be if the ideal-gas equation of state were used for the calculation? Use this value as an initial estimate of \(\tilde{V}\) for air at \(223 \mathrm{K}\) and 50.0 atm and solve the van der Waals equation using Goal Seek or Solver in Excel. What percentage error results from the use of the ideal-gas equation of state, taking the van der Waals estimate to be correct? (c) Set up a spreadsheet to carry out the calculations of Part (b) for air at \(223 \mathrm{K}\) and several pressures. The spreadsheet should appear as follows: The polynomial expression for \(\hat{V}\left(f=c_{3} \hat{V}^{3}+c_{2} \hat{V}^{2}+\cdots\right)\) should be entered in the \(f(V)\) column, and the value in the \(V\) column should be determined using Goal Seek or Solver in Excel.

Short Answer

Expert verified
First, rearrange the van Der Waals equation to its cubic form \( \hat{V}^{3} - (b+\frac{RT}{P})\hat{V}^{2} + \frac{a}{P}\hat{V} - \frac{ab}{P} = 0\), which allows to identify the coefficients in terms of P, R, T, a, and b. Then, calculate the coefficients using the given values. Afterwards, calculate \(\hat{V}\) using ideal-gas equation, and use it as the estimate to solve van der Waals equation. Finally, calculate the percentage error between the solutions given by ideal-gas equation and van der Waals equation, and create an Excel spreadsheet to carry out calculations for different pressure values.

Step by step solution

01

- Write Down the van der Waals equation:

The van der Waals equation is given by \( (P + \frac{a}{\hat{V}^2})(\hat{V} - b) = RT \). It is required to express this in the cubic form \(f(\hat{V})=c_{3} \hat{V}^{3}+c_{2} \hat{V}^{2}+c_{1} \hat{V}+c_{0}=0\).
02

- Rearrange the equation to its cubic form:

The cubic form of the van der Waals equation is obtained by expanding and rearranging the terms, and it becomes \( \hat{V}^{3} - (b+\frac{RT}{P})\hat{V}^{2} + \frac{a}{P}\hat{V} - \frac{ab}{P} = 0\). This gives \(c_{3} = 1\), \(c_{2} = -(b+\frac{RT}{P})\), \(c_{1} = \frac{a}{P}\), and \(c_{0} = -\frac{ab}{P}\).
03

- Calculate the coefficients for the given conditions:

Substitute the given values \(T = 223K\), \(P = 50.0atm\), \(R=0.0821L.atm/K.mol\) (universal gas constant), \(a=1.33atm.L^2/mol^2\) and \(b=0.0366L/mol\) into \(c_2, c_1, c_0\) equations to find their numerical values.
04

- Compare with the Ideal-Gas Equation estimate:

Calculate the molar volume \(\hat{V}\) using the ideal-gas equation \(PV = RT\). Use the obtained \(\hat{V}\) value as an initial estimate and solve the cubic equation (van der Waals equation) using numerical methods. Also calculate the percentage error taking the van der Waals estimate to be correct.
05

- Excel Spreadsheet Formation:

Lastly, create a spreadsheet to carry out the calculations for air at 223K and several pressures. The polynomial expression for \(\hat{V}\) should be in the \(f(V)\) column, and the \(V\) value has to be determined via the Excel Solver tool.

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Key Concepts

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

Cubic Equation of State
When exploring the states and properties of gases, particularly those close to the boundaries where they might liquefy, it's essential to have a model that accounts for interactions not present in an ideal gas. The van der Waals equation is one such model, enhancing the classic ideal-gas law by factoring in the volume occupied by gas molecules and the attractive forces between them.

Classified as a cubic equation of state, the van der Waals equation is written as \( f(\hat{V})=\hat{V}^3 - (b + \frac{RT}{P})\hat{V}^2 + \frac{a}{P}\hat{V} - \frac{ab}{P} = 0 \), where \( \hat{V} \) signifies the specific molar volume, \( P \) is the pressure, \( T \) is the temperature, and \( R \) is the ideal gas constant. The coefficients \( c_3, c_2, c_1, c_0 \) are related to these variables and the substance-specific van der Waals constants \( a \) and \( b \). These adjustments reflect real gas behavior, with the cubic nature of the equation deriving from the equation being set to zero after multiplication and expansion of the contributing factors.

This cubic form is powerful as it can yield up to three real volume solutions corresponding to various possible states of the gas under given conditions. For example, in a system where phase transitions occur, one solution may correlate with the gaseous state, another with the liquid state, and a third being unphysical or representing a metastable state. Typically, numerical methods or graphical techniques are employed to solve the cubic equation and find the specific molar volume of the gas.
Specific Molar Volume
The term 'specific molar volume', denoted as \( \hat{V} \), is intrinsic to understanding gases' behavior under different pressures and temperatures. It represents the volume that one mole of gas occupies under specified conditions. This concept is crucial when dealing with equations of state because it directly relates to how tightly molecules are packed and how they interact in a gas sample.

Using the van der Waals equation incorporates specific molar volume into a more accurate framework than the ideal-gas equation by including finite volume and intermolecular forces. For an ideal gas, these interactions and volumes are assumed nonexistent, leading to the ideal-gas law \( PV = nRT \), with \( V \) being the volume of \( n \) moles of gas. However, gases exhibit deviations from the ideal behavior, especially under high pressure and low temperature, making the van der Waals equation, where \( \hat{V} \) is a central parameter, a more realistic model.

As a practical application, by knowing the van der Waals constants \( a \) and \( b \), and the conditions of temperature and pressure for a gas like air, one can compute the specific molar volume using the rearranged cubic form of the van der Waals equation, which provides insights into the actual volume occupied by air molecules in different states鈥攁 valuable calculation in fields like thermodynamics and chemical engineering.
Ideal-Gas Equation
The ideal-gas equation, \( PV = nRT \), is a cornerstone in the study of thermodynamics and plays a vital role in the initial estimation of gas volume within the van der Waals framework. In this equation, \( P \) represents the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature. This formula presupposes that the gas molecules do not interact and that they occupy no space on their own; an approximation that holds true under low-pressure and high-temperature conditions.

In cases where a gas behaves nearly ideally, the ideal-gas equation serves as a reliable model for calculating its volume. However, in situations involving high pressure or low temperature, where gases behave non-ideally, the ideal-gas equation can still provide a preliminary volume estimate which is then refined using more complex models like the van der Waals equation. This step is evident in exercises where the initial molar volume obtained from the ideal-gas equation is used as a starting point for iterative methods that solve the van der Waals cubic equation.

Moreover, comparing the volume calculated using the ideal-gas equation with that obtained via the van der Waals equation offers insight into the magnitude of deviation from ideality. This comparison is often expressed as a percentage error, which quantifies the discrepancy between the ideal and real behavior of gases, highlighting the importance and impact of molecular volume and intermolecular forces in predicting gas behavior.

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Most popular questions from this chapter

Ethane at \(25^{\circ} \mathrm{C}\) and 1.1 atm (abs) flowing at a rate of \(100 \mathrm{mol} / \mathrm{s}\) is burned with \(20 \%\) excess oxygen at \(175^{\circ} \mathrm{C}\) and 1.1 atm \((\text { abs }) .\) The combustion products leave the furnace at \(800^{\circ} \mathrm{C}\) and 1 atm. (a) What is the volumetric flow rate of oxygen (L/s) fed to the furnace? (b) What should the volumetric flow rate of the combustion products be? State all assumptions you make. (c) The volumetric flow rate of the combustion products is measured and found to be different from the value calculated in Part (b). Assuming that no mistakes were made in the calculation, what could be going on that could lead to the discrepancy? Consider assumptions made in the calculations and things that can go wrong in a real system.

The volume of a dry box (a closed chamber with dry nitrogen flowing through it) is \(2.0 \mathrm{m}^{3}\). The dry box is maintained at a slight positive gauge pressure of \(10 \mathrm{cm} \mathrm{H}_{2} \mathrm{O}\) and room temperature \(\left(25^{\circ} \mathrm{C}\right) .\) If the contents of the box are to be replaced every five minutes, calculate the required mass flow rate of nitrogen in \(g / \min\) by (a) direct solution of the ideal-gas equation of state and (b) conversion from standard conditions. You may assume the gas in the dry box is well mixed.

A balloon \(20 \mathrm{m}\) in diameter is filled with helium at a gauge pressure of 2.0 atm. A man is standing in a basket suspended from the bottom of the balloon. A restraining cable attached to the basket kecps the balloon from rising. The balloon (not including the gas it contains), the basket, and the man have a combined mass of \(150 \mathrm{kg}\). The temperature is \(24^{\circ} \mathrm{C}\) that day, and the barometer reads \(760 \mathrm{mm} \mathrm{Hg}\) (a) Calculate the mass (kg) and weight (N) of the helium in the balloon. (b) How much force is exerted on the balloon by the restraining cable? (Recall: The buoyant force on a submerged object equals the weight of the fluid- -in this case, the air- -displaced by the object. Neglect the volume of the basket and its contents.) (c) Calculate the initial acceleration of the balloon when the restraining cable is released. (d) Why does the balloon eventually stop rising? What would you need to know to calculate the altitude at which it stops? (e) Suppose at its point of suspension in midair the balloon is heated, raising the temperature of the helium. What happens and why?

Many references give the specific gravity of gases with reference to air. For example, the specific gravity of carbon dioxide is 1.53 relative to air at the same temperature and pressure. Show that this value is correct as long as the ideal-gas equation of state applies.

During your summer vacation, you plan an epic adventure trip to scale Mt. Kilimanjaro in Tanzania. Dehydration is a great danger on such a climb, and it is essential to drink enough water to make up for the amount you lose by breathing. (a) During your pre-trip physical, your physician measured the average flow rate and composition of the gas you exhaled (expired air) while performing light activity. The results were \(11.36 \mathrm{L} / \mathrm{min}\) at body temperature (37^) C) and 1 atm, 17.08 mole\% oxygen, 3.25\% carbon dioxide, 6.12 mole\% \(\mathrm{H}_{2} \mathrm{O},\) and the balance nitrogen. The ambient (inspired) air contained 1.67 mole\% water and a negligible amount of carbon dioxide. Calculate the rate of mass lost through the breathing process (kg/day) and the volume of water in liters you would have to drink per day just to replace the water lost in respiration. Consider your lungs to be a continuous steady-state system, with input streams being inspired air and water and \(\mathrm{CO}_{2}\) transferred from the blood and output streams being expired air and \(\mathrm{O}_{2}\) transferred to the blood. Assume no nitrogen is transferred to or from the blood. (b) You made the trip to Tanzania and completed the climb to Uhuru Peak, the summit of Kilimanjaro, at an altitude of 5895 meters above sea level. The ambient temperature and pressure there averaged \(-9.4^{\circ} \mathrm{C}\) and \(360 \mathrm{mm} \mathrm{Hg},\) and the air contained \(0.46 \mathrm{mole} \%\) water. The molar flow rate of your expired air was roughly the same as it had been at sea level, and the expired air contained \(14.86 \% \mathrm{O}_{2}\) \(3.80 \% \mathrm{CO}_{2},\) and \(13.20 \% \mathrm{H}_{2} \mathrm{O} .\) Calculate the rate of mass lost (g/day) through breathing and water you would have to drink (L/day) just to replace the water lost in respiration. (c) The equality of the molar flow rates of expired air at sea level and at Uhuru Peak is due to a cancellation of effects, one of which would tend to increase the rate at higher altitudes and the other to decrease it. What are those effects? (Hint: Use the ideal-gas equation of state in your solution, and think about how the oxygen concentration at a high altitude would likely affect your breathing rate.)
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