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Given that the van der Waals constant \(b\) is the excluded volume and that the excluded volume is four times the volume actually occupied by the gas molecules in a sample, determine what percentage of the container volume is actually occupied by \(\mathrm{CCl}_{4}(g)\) molecules at (a) STP, (b) \(10.0 \mathrm{~atm}\) and \(273 \mathrm{~K},\) and \((\mathrm{c}) 50.0 \mathrm{~atm}\) and \(273 \mathrm{~K}\)

Short Answer

Expert verified
At STP, 0.154%; at 10 atm, 1.54%; at 50 atm, 7.71%.

Step by step solution

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01

Understand the Problem

The problem requires us to determine the percentage of the container volume occupied by \( \text{CCl}_4(g) \) molecules under different conditions. Given that the van der Waals constant \( b \) represents an excluded volume that is four times the volume the gas actually occupies, we need to find this percentage for distinct pressures.
02

Determine the Gas Molar Volume

At standard temperature and pressure (STP, which is 1 atm and 273 K), the molar volume of an ideal gas is 22.414 L/mol.
03

Calculate Excluded Volume

The excluded volume \( b \) for \( \text{CCl}_4 \) molecules can generally be found in tables or empirical data, but let us assume it for calculation as \( b = 0.1383 \) L/mol for simplicity.
04

Find the Real Volume Occupied by Molecules at STP

The real volume occupied by molecules is \( \frac{b}{4} = \frac{0.1383}{4} \approx 0.034575 \) L/mol.
05

Calculate Percentage Volume Occupied at STP

The percentage volume occupied by the gas is given by \( \left( \frac{0.034575}{22.414} \right) \times 100 \approx 0.154\% \).
06

Calculate Volume Percentage at 10 atm and 273K

For 10 atm, the container volume is \( \frac{22.414}{10} = 2.2414 \) L/mol. The percentage is then \( \left( \frac{0.034575}{2.2414} \right) \times 100 \approx 1.54\% \).
07

Calculate Volume Percentage at 50 atm and 273K

For 50 atm, the container volume is \( \frac{22.414}{50} = 0.44828 \) L/mol. The percentage is \( \left( \frac{0.034575}{0.44828} \right) \times 100 \approx 7.71\% \).

Key Concepts

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

Excluded Volume
When discussing gases, the concept of excluded volume is valuable. Excluded volume, represented by the van der Waals constant 'b', is crucial in understanding real gas behavior. It accounts for the physical space that gas molecules take up. Unlike ideal gases, real gas molecules occupy space, preventing them from being compressed infinitely.

The excluded volume is typically about four times the actual volume of the gas molecules themselves. This means that when gas molecules are considered in calculations, their physical space must be acknowledged. This is crucial when applying the van der Waals equation as it modifies the ideal gas law to better reflect real gas situations.

Understanding excluded volume helps in various applications:
  • Calculating true gas behavior under different pressures.
  • Adjusting ideal gas assumptions to fit real-world scenarios.
  • Providing a clearer concept of molecule arrangement in gas phases.
This concept is particularly important when analyzing high pressure scenarios where the volume occupied by gas molecules significantly impacts pressure measurements.
Gas Molecules
Gas molecules are the tiny entities that make up a gas. In our discussion of van der Waals equation, each gas molecule has a finite size, which is why the concept of excluded volume is vital. Gas molecules are always in motion, colliding with the walls of a container and with each other, leading to observable pressure and volume changes.

Significantly:
  • Gas molecules interact not only with the container but also with each other, which affects their behavior and needs to be considered in real-life applications.
  • The volume actually occupied by these molecules is petite when compared to the total volume of a gas; this is why excluded volume and adjustments to ideal gas law assumptions are necessary.
  • Understanding the real volume these molecules occupy helps predict and calculate gas behavior under various conditions.
In calculations, as seen in exercises involving gases like CCl4, considering their actual occupied volume allows for accurate pressure and volume predictions in non-ideal situations.
Standard Temperature and Pressure
Standard Temperature and Pressure (STP) is a key reference point in studying gases. STP is defined by two conditions: temperature at 273K (0°C) and pressure at 1 atm. This standard is valuable for providing a common ground for comparing gas behaviors and calculations.

At STP:
  • The molar volume of an ideal gas is about 22.414 L/mol, which serves as a baseline for evaluating gas behavior under various pressure conditions.
  • Using STP allows chemists and students to predict how changes in temperature or pressure will affect gas volume and behavior.
  • It provides a reference for calculating deviations in real gases using the van der Waals equation.
Using STP as a reference can help simplify and accurately assess exercises involving gases, like determining occupied volumes or calculating the effects of different pressures.
Pressure
Pressure is a measure of the force exerted by gas molecules hitting the walls of their container. It plays a significant role in determining how gas behaves under different conditions. The van der Waals equation adjusts for pressure variations, especially important in non-ideal gas conditions where intermolecular forces and molecular volumes can't be ignored.

In practice:
  • High pressures can lead to more interactions between molecules, making the excluded volume a significant factor.
  • The relationship between pressure, volume, and temperature in gases is described by the ideal gas law, which can be modified using the van der Waals equation to account for real gas behaviors.
  • Pressure changes influence gas density and how gases interact with their environment.
Understanding pressure and its effects helps students predict how a gas will respond to tight spaces or high-energy environments, as in high pressure calculations.

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Most popular questions from this chapter

Ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) is emitted by fruits and is known to be responsible for their ripening. Based on this information, explain why a bunch of bananas ripens faster in a closed paper bag than in an open bowl.

Dry air near sea level has the following composition by volume: \(\mathrm{N}_{2}, 78.08\) percent; \(\mathrm{O}_{2}, 20.94\) percent; \(\mathrm{Ar}, 0.93\) percent; \(\mathrm{CO}_{2}, 0.05\) percent. The atmospheric pressure is 1.00 atm. Calculate (a) the partial pressure of each gas in atmospheres and (b) the concentration of each gas in \(\mathrm{mol} / \mathrm{L}\) at \(0^{\circ} \mathrm{C}\). (Hint: Because volume is proportional to the number of moles present, mole fractions of gases can be expressed as ratios of volumes at the same temperature and pressure.)

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) burns in air: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ Balance the equation and determine the volume of air in liters at \(45.0^{\circ} \mathrm{C}\) and \(793 \mathrm{mmHg}\) required to burn \(185 \mathrm{~g}\) of ethanol. Assume that air is 21.0 percent \(\mathrm{O}_{2}\) by volume.

In \(2.00 \mathrm{~min}, 29.7 \mathrm{~mL}\) of He effuses through a small hole. Under the same conditions of pressure and temperature, \(10.0 \mathrm{~mL}\) of a mixture of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) effuses through the hole in the same amount of time. Calculate the percent composition by volume of the mixture.

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 \(\mathrm{NaOH}\)

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