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Chapter 10: Problem 97

In Sample Exercise 10.16 , we found that one mole of \(\mathrm{Cl}_{2}\) confined to \(22.41 \mathrm{~L}\) at \(0{ }^{\circ} \mathrm{C}\) deviated slightly from ideal behavior. Calculate the pressure exerted by \(1.00 \mathrm{~mol} \mathrm{Cl}_{2}\) confined to a smaller volume, \(5.00 \mathrm{~L}\), at \(25^{\circ} \mathrm{C} .\) (a) First use the ideal-gas equation and (b) then use the van der Waals equation for your calculation. (Values for the van der Waals constants are given in Table \(10.3 .)\) (c) Why is the difference between the result for an ideal gas and that calculated using the van der Waals equation greater when the gas is confined to \(5.00 \mathrm{~L}\) compared to \(22.4 \mathrm{~L} ?\)

Short Answer

Expert verified
In this exercise, we calculated the pressure exerted by 1.00 mol of Cl2 confined to a volume of 5.00 L at 25掳C using both the ideal-gas equation and the van der Waals equation. The ideal-gas equation gave a result of 49.2 atm, while the van der Waals equation gave a result of 51.9 atm. The difference between the two is greater when the gas is confined to a smaller volume (such as 5.00 L) compared to a larger volume (such as 22.4 L) because the van der Waals equation accounts for the intermolecular forces, which become significant at smaller volumes due to the molecules being closer to each other.

Step by step solution

01

(a) Find pressure using the ideal-gas equation

The ideal-gas equation is given as follows: \[PV = nRT\] where: - P is the pressure - V is the volume (force and \(V = 5.00 \mathrm{~L}\) - T is the temperature (force and \(T = 25^{\circ} \mathrm{C}\) which is equal to \(T = 298\mathrm{~K}\) when converted to Kelvin) - n is the number of moles (\(n = 1.00 \mathrm{~mol}\)) - R is the gas constant (\(R = 0.08206 \frac{\mathrm{L~atm}}{\mathrm{K~mol}}\)) We need to find the pressure, P. Rearrange the equation to solve for P: \[P = \frac{nRT}{V}\] Now, plug in the values of n, R, T, and V: \[P = \frac{(1.00 \mathrm{~mol})(0.08206 \frac{\mathrm{L~atm}}{\mathrm{K~mol}})(298\mathrm{~K})}{5.00 \mathrm{~L}}\] Calculate the result: \[P = 49.2 \mathrm{~atm}\] So, the pressure using the ideal-gas equation is 49.2 atm.
02

(b) Find pressure using the van der Waals equation

The van der Waals equation accounts for the deviations from ideal behaviour due to molecular interactions. The equation is given as follows: \[\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT\] For this exercise, we're given the van der Waals constants for Cl2: - a = 6.49 L虏atm/mol虏 - b = 0.0562 L/mol Plug in the values of n, R, T, V, a, and b: \[\left(P + \frac{(6.49 \frac{\mathrm L^2\mathrm{~atm}}{\mathrm{mol^2}})(1.00 \mathrm{~mol})^2}{(5.00 \mathrm{~L})^2}\right)\{(5.00 \mathrm{~L}) - (0.0562 \frac{\mathrm{L}}{\mathrm{mol}})(1.00 \mathrm{~mol})\} = (1.00 \mathrm{~mol})(0.08206\frac{\mathrm{L~atm}}{\mathrm{K~mol}})(298\mathrm{~K})\] Solve the equation for P: \[P = \frac{nRT + a(\frac{n}{V})^2(V - nb) - nbR}{V - nb}\] Plug in the same values: \[P = \frac{(1.00 \mathrm{~mol})(0.08206 \frac{\mathrm{L~atm}}{\mathrm{K~mol}})(298\mathrm{~K}) + (6.49 \frac{\mathrm{L^2\mathrm{~atm}}{\mathrm{mol^2}})(\frac{1.00 \mathrm{~mol}}{5.00 \mathrm{~L}})^2(5.00 \mathrm{~L} - (0.0562 \frac{\mathrm{L}}{\mathrm{mol}})(1.00 \mathrm{~mol})) - (0.0562 \frac{\mathrm{L}}{\mathrm{mol}})(1.00 \mathrm{~mol})(0.08206\frac{\mathrm{L~atm}}{\mathrm{K~mol}})}{(5.00 \mathrm{~L} - (0.0562\frac{\mathrm{L}}{\mathrm{mol}})(1.00 \mathrm{~mol}))}\] Calculate the result: \[P = 51.9 \mathrm{~atm}\] So, the pressure using the van der Waals equation is 51.9 atm.
03

(c) Explain the difference in pressures

The ideal-gas equation gave a result of 49.2 atm, while the van der Waals equation gave a result of 51.9 atm. The difference between the two is greater when the gas is confined to a smaller volume compared to a larger volume (22.4 L) due to the fact that the van der Waals equation accounts for the intermolecular forces. At a smaller volume, the interactions between gas molecules becomes significant as they are now located closer to each other. These forces make the gas deviate from ideal behaviour, as predicted by the ideal-gas law. The van der Waals equation accounts for the molecular size and attractions, therefore providing a better estimation of the pressure exerted by the gas when confined to smaller volumes.

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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 foundational concept in chemistry, providing a simple equation to relate the pressure, volume, temperature, and amount of a gas. It is given by the formula:
  • \(PV = nRT\)
Here:
  • \(P\) represents pressure,
  • \(V\) is volume,
  • \(n\) is the molar amount of gas,
  • \(R\) is the universal gas constant, and
  • \(T\) is the temperature in Kelvin.
This equation assumes that gases are composed of point particles that occupy no volume and experience no intermolecular forces.
This model works well under many common conditions like standard temperature and pressure.Under these assumptions, we performed a calculation using this law to determine the pressure exerted by one mole of chlorine gas in a 5.00 L container at 25掳C.
By converting the temperature to Kelvin and inserting the known values into the equation, we found that the pressure is approximately 49.2 atm.
However, this is an approximation, as it neglects the forces and space taken by real gas molecules, which leads to deviations from ideal behavior, especially under high pressure or low volume conditions.
Real Gases
Real gases deviate from the assumptions of the ideal-gas law because they possess finite volume and experience intermolecular forces.
These deviations become significant under conditions of high pressure or low volume, where gas particles are forced closer together.The van der Waals equation is a mathematical model that accurately describes the behavior of real gases. It modifies the ideal-gas equation by introducing two additional terms to account for:
  • the attractive forces between molecules (represented by the constant \(a\)), and
  • the finite volume of the gas particles (represented by the constant \(b\)).
The equation is expressed as:
  • \[(P + \frac{an^2}{V^2})(V - nb) = nRT\]
In our example with chlorine gas, using the van der Waals equation provided a pressure of 51.9 atm.
This is a more accurate measure than the ideal-gas law as it factors in the molecular interactions and finite size of the molecules, acknowledging their non-ideal behavior.
Intermolecular Forces
Intermolecular forces are the interactions between neighboring molecules, which can attract or repel them.
These forces are significant in real gases, particularly when they are compressed into tighter spaces, like in smaller volumes.
They include attractions like hydrogen bonds and van der Waals forces, the latter being relevant in this context. The van der Waals forces are essential in adjusting the real pressure of a gas from the idealized prediction. When gases are compressed:
  • The particles have less space to move apart, increasing the likelihood of intermolecular attractions.
  • These attractions can cause a reduction in pressure compared to what the ideal-gas law predicts.
In our example with 5.00 L of chlorine gas, these forces become important.
The van der Waals equation adjusts for them, explaining why it predicts a higher pressure than the simple ideal-gas law model.
Understanding these forces helps explain why gases behave differently from theoretical predictions, particularly under stress from compression and high pressure.

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Most popular questions from this chapter

When a large evacuated flask is filled with argon gas, its mass increases by \(3.224 \mathrm{~g}\). When the same flask is again evacuated and then filled with a gas of unknown molar mass, the mass increase is 8.102 g. (a) Based on the molar mass of argon, estimate the molar mass of the unknown gas. (b) What assumptions did you make in arriving at your answer?

A piece of dry ice (solid carbon dioxide) with a mass of \(5.50 \mathrm{~g}\) is placed in a 10.0 - \(\mathrm{L}\) vessel that already contains air at 705 torr and \(24^{\circ} \mathrm{C}\). After the carbon dioxide has totally vaporized, what is the partial pressure of carbon dioxide and the total pressure in the container at \(24{ }^{\circ} \mathrm{C} ?\)

Gaseous iodine pentafluoride, \(\mathrm{IF}_{5},\) can be prepared by the reaction of solid iodine and gaseous fluorine: $$ \mathrm{I}_{2}(s)+5 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{IF}_{5}(g) $$ A 5.00-L flask containing \(10.0 \mathrm{~g} \mathrm{I}_{2}\) is charged with \(10.0 \mathrm{~g} \mathrm{~F}_{2}\), and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is \(125^{\circ} \mathrm{C}\). (a) What is the partial pressure of \(\mathrm{IF}_{5}\) in the flask? (b) What is the mole fraction of \(\mathrm{IF}_{5}\) in the flask (c) Draw the Lewis structure of \(\mathrm{IF}_{5}\). (d) What is the total mass of reactants and products in the flask?

(a) Place the following gases in order of increasing average molecular speed at \(25^{\circ} \mathrm{C}: \mathrm{Ne}, \mathrm{HBr}, \mathrm{SO}_{2}, \mathrm{NF}_{3}, \mathrm{CO}\) (b) Calculate the rms speed of \(\mathrm{NF}_{3}\) molecules at \(25^{\circ} \mathrm{C} .\) (c) Calculate the most probable speed of an ozone molecule in the stratosphere, where the temperature is \(270 \mathrm{~K}\).

After the large eruption of Mount St. Helens in 1980 , gas samples from the volcano were taken by sampling the downwind gas plume. The unfiltered gas samples were passed over a goldcoated wire coil to absorb mercury (Hg) present in the gas. The mercury was recovered from the coil by heating it and then analyzed. In one particular set of experiments scientists found a mercury vapor level of \(1800 \mathrm{ng}\) of Hg per cubic meter in the plume at a gas temperature of \(10^{\circ} \mathrm{C}\). Calculate (a) the partial pressure of Hg vapor in the plume, (b) the number of \(\mathrm{Hg}\) atoms per cubic meter in the gas, \((\mathrm{c})\) the total mass of Hg emitted per day by the volcano if the daily plume volume was \(1600 \mathrm{~km}^{3}\).
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