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Chapter 10: Problem 110

It turns out that the van der Waals constant \(b\) equals four times the total volume actually occupied by the molecules of a mole of gas. Using this figure, calculate the fraction of the volume in a container actually occupied by Ar atoms (a) at STP, (b) at 100 atm pressure and \(0{ }^{\circ} \mathrm{C}\). (Assume for simplicity that the ideal-gas equation still holds.)

Short Answer

Expert verified
The fraction of volume occupied by Ar atoms is 0.3336 (33.36%) at STP and 3.336 (333.6%) at 100 atm pressure and 0掳C.

Step by step solution

01

Find the van der Waals constant b for Argon atoms

To find the van der Waals constant b for Argon, we can use the Ideal Gas Law equation and the given information. The Ideal Gas Law equation is given by: \(PV=nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature. We know that one mole of Ar occupies 22.4 L at STP. Standard Temperature and Pressure (STP) refers to a temperature of 273.15 K (0掳C) and a pressure of 1 atm (101.3 kPa). The conversion factor between atm and L atm/mol K is given by R = 0.0821 L atm/mol K. Since we know that b is equal to four times the volume occupied by a mole of gas, we can rewrite the Ideal Gas Law equation for one mole of Ar as follows: \(V - 4V = 0.0821 * 273.15\) Now, we can solve for V:
02

Solve for V

To solve for V, we simplify the equation: \(V - 4V = 0.0821 * 273.15\) \(-3V = 22.41\) \(V = \frac{-22.41}{-3}\) \(V = 7.47 L\) We found that the volume actually occupied by one mole of Ar gas is 7.47 L.
03

Calculate the fraction of volume occupied by Ar atoms at STP

To calculate the occupied volume fraction at STP, we can use the relationship between b and the volume occupied by one mole of Ar. We know that b = 4V. Since we found V to be 7.47 L in the previous step, we now have all the necessary values to find the fraction at STP: Fraction occupied at STP = \( \frac{V}{V_{total}} = \frac{7.47 L}{22.4 L} \) Now, just simplify the fraction. Fraction occupied at STP = \( \frac{7.47}{22.4} \) Fraction occupied at STP = 0.3336 So, the fraction of volume occupied by Ar atoms at STP is 0.3336 (33.36%).
04

Calculate the fraction of volume occupied at 100 atm pressure and 0掳C

We are given the condition of 100 atm pressure and 0掳C temperature (273.15 K), and we want to find the fraction of volume occupied by Ar atoms at this condition. We will use the Ideal Gas Law equation to get the volume at this state: \(PV = nRT \Rightarrow V = \frac{nRT}{P}\) We know that n=1, R=0.0821 L atm/mol K, T=273.15 K, and P=100 atm. Plugging in the values, we have: \(V = \frac{1 * 0.0821 * 273.15}{100}\) \(V = 2.24 L\) Now, we calculate the fraction of volume occupied by the Ar atoms: Fraction occupied at 100 atm and 0掳 C = \( \frac{V}{V_{total}} = \frac{7.47 L}{2.24 L} \) Simplify the fraction: Fraction occupied at 100 atm and 0掳 C = \( \frac{7.47}{2.24} \) Fraction occupied at 100 atm and 0掳 C = 3.336 So, the fraction of volume occupied by Ar atoms at 100 atm pressure and 0掳C is 3.336 (333.6%).

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation that relates four important properties of gases: pressure (P), volume (V), number of moles (n), and temperature (T). It is expressed as the equation \(PV = nRT\). In this equation, \(R\) is the universal gas constant that has a value of 0.0821 L atm/mol K. This law assumes that gases behave ideally, meaning that gas molecules do not interact with each other and occupy no volume. This simplification works well under many conditions, but it is important to note that real gases, like Argon, may deviate from this behavior under high pressures or low temperatures. In the exercise, the Ideal Gas Law helps calculate the volume occupied by a mole of Argon gas under different conditions.
Argon
Argon is a noble gas, represented by the chemical symbol \(\text{Ar}\). It is colorless, odorless, and tasteless, making up about 0.93% of the Earth's atmosphere. As a noble gas, Argon is generally inert, meaning it does not easily undergo chemical reactions. This makes it useful for various applications where a non-reactive atmosphere is required, such as preserving historical documents or in welding. Because of its stability, Argon serves as a good example in studying gas laws like the van der Waals equation and the Ideal Gas Law, which help describe the behavior of gases under different conditions.
molecular volume
Molecular volume refers to the volume that the molecules of a gas actually occupy. For ideal gases, we often assume that the molecules themselves do not take up any space, which simplifies calculations and is acceptable under many conditions. However, real gases do occupy some volume. In the context of the van der Waals equation, the volume parameter \(b\) represents the molecular volume, acknowledging that gas molecules do indeed take up space. It is particularly important to consider molecular volume at high pressures, where the space between molecules reduces, making their volume more significant in calculations.
Standard Temperature and Pressure
Standard Temperature and Pressure, often abbreviated as STP, is a set of conditions used as a common reference point for scientific calculations. At STP, the temperature is set at 0掳C (273.15 K), and the pressure is 1 atm (101.3 kPa). These conditions allow scientists to compare experimental results and theoretical predictions consistently. For gases like Argon, knowing the properties at STP is valuable because it supports calculations involving gas laws. For example, using the Ideal Gas Law under STP conditions, one mole of an ideal gas can be expected to occupy 22.4 liters of volume. This serves as a baseline for further investigations, such as calculating the fraction of volume occupied by Argon atoms in a given container.

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Most popular questions from this chapter

Assume that a single cylinder of an automobile engine has a volume of \(524 \mathrm{~cm}^{3}\). (a) If the cylinder is full of air at \(74{ }^{\circ} \mathrm{C}\) and \(0.980 \mathrm{~atm}\), how many moles of \(\mathrm{O}_{2}\) are present? (The mole fraction of \(\mathrm{O}_{2}\) in dry air is \(0.2095 .\) ) (b) How many grams of \(\mathrm{C}_{8} \mathrm{H}_{18}\) could be combusted by this quantity of \(\mathrm{O}_{2}\), assuming complete combustion with formation of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ?

Suppose the mercury used to make a barometer has a few small droplets of water trapped in it that rise to the top of the mercury in the tube. Will the barometer show the correct atmospheric pressure? Explain

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