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Chapter 10: Problem 27

(a) Amonton's law expresses the relationship between pressure and temperature. Use Charles's law and Boyle's law to derive the proportionality relationship between \(P\) and \(T .(\mathbf{b})\) If a car tire is filled to a pressure of \(220.6 \mathrm{kPa}\) measured at \(24^{\circ} \mathrm{C}\), what will be the tire pressure if the tires heat up to \(49^{\circ} \mathrm{C}\) during driving?

Short Answer

Expert verified
Using Charles's law and Boyle's law, we derive the proportionality relationship between pressure and temperature, known as Amonton's law: \(P \propto \frac{1}{T}\). Given the initial tire pressure \(P_1 = 220.6\,\text{kPa}\) at \(T_1 = 24^{\circ}\mathrm{C}\), we find the tire pressure \(P_2\) at \(T_2 = 49^{\circ}\mathrm{C}\) using the relationship \(\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\). The resulting tire pressure when the tires heat up to \(49^{\circ}\mathrm{C}\) during driving is approximately \(240.57\,\text{kPa}\).

Step by step solution

01

Reminder of the formulas of Charles's law and Boyle's law

Charles's law states the relationship between the temperature and volume of an ideal gas, keeping the pressure constant. The formula for Charles's law is: \[V \propto T\] Boyle's law states the relationship between the pressure and volume of an ideal gas when the temperature is held constant. The formula for Boyle's law is: \[P \propto \frac{1}{V}\]
02

Combine Charles's law and Boyle's law to derive the proportionality between P and T

To derive the proportionality relationship between \(P\) and \(T\), we need to combine both laws. From Charles's law, we can write: \[V = k_{1}T\] where \(k_{1}\) is some constant of proportionality. Similarly, from Boyle's law, we can write: \[P = k_{2}\frac{1}{V}\] where \(k_{2}\) is some constant of proportionality. Now, we can substitute the expression for \(V\) from Charles's law into the expression of Boyle's law: \[P = k_{2}\frac{1}{k_{1}T}\] We can rewrite the equation as: \[P = \frac{k_{2}}{k_{1}} \cdot \frac{1}{T}\] Let \(\frac{k_{2}}{k_{1}}\) be another constant \(k\), so: \[P \propto \frac{1}{T}\] This is the proportionality relationship between \(P\) and \(T\), which is also known as Amonton's law.
03

Use the derived proportionality to find the tire pressure at \(49^{\circ}\mathrm{C}\)

Given: Initial tire pressure \(P_{1} = 220.6\,\text{kPa}\) at \(T_{1}=24^{\circ}\mathrm{C}\). We need to find the tire pressure \(P_{2}\) at \(T_{2} = 49^{\circ}\mathrm{C}\). First, convert the Celsius temperatures to Kelvin: \[T_{1} = 24^{\circ}\mathrm{C} + 273.15 = 297.15 \,\text{K}\] \[T_{2} = 49^{\circ}\mathrm{C} + 273.15 = 322.15 \,\text{K}\] Now, we can use the proportionality relationship we derived, which means: \[\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\] Now, substituting the given values and solving for \(P_{2}\): \[P_{2} = \frac{T_{2}}{T_{1}} \cdot P_{1}\] \[P_{2} = \frac{322.15 \,\text{K}}{297.15 \,\text{K}} \cdot 220.6\,\text{kPa}\] Calculating the tire pressure \(P_2\): \[P_{2} = 240.57\, \text{kPa}\] Therefore, the tire pressure when the tires heat up to \(49^{\circ}\mathrm{C}\) during driving will be approximately \(240.57\, \text{kPa}\).

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Key Concepts

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

Charles's Law
Charles's Law is a fundamental principle in the study of gases. It describes how gases expand when heated and contract when cooled. This law establishes a direct relationship between the volume of a gas and its temperature, assuming the pressure remains constant. In simpler terms, if the temperature of a gas increases, the volume increases as well, and vice versa.Charles's Law can be mathematically expressed as:\[ V \propto T \]Here, \( V \) stands for the volume of the gas and \( T \) represents the temperature in Kelvin. This direct proportionality implies that when you increase the temperature, the molecules within the gas move faster, requiring more space. Thus, the volume expands to maintain equilibrium. It is crucial to use Kelvin for temperature because it starts at absolute zero, providing a clear starting point for thermodynamic calculations. This concept is fundamental in understanding how gases behave under changing thermal conditions.
Boyle's Law
Boyle's Law relays the relationship between pressure and volume in gases, maintaining constant temperature. It reveals that as the volume of a gas decreases, its pressure increases, demonstrating an inverse relationship. This means if you compress a gas into a smaller volume, its particles are more tightly packed, leading to increased pressure.Mathematically, Boyle's Law is described by the formula:\[ P \propto \frac{1}{V} \]In this equation, \( P \) denotes the pressure of the gas and \( V \) stands for the volume. This inverse relationship is key to many practical applications, such as maintaining tire pressure in cars. When driving, the volume of air in a tire remains constant, but changes in external temperature affect the internal pressure. Therefore, understanding Boyle's Law helps predict and manage these changes, ensuring safety and efficiency in various mechanical settings.
Ideal Gas Law
The Ideal Gas Law provides a comprehensive framework for understanding the behavior of gases by integrating other simpler gas laws. It combines the relationships outlined by Charles's Law, Boyle's Law, and Amonton's Law to express the connection between pressure, volume, temperature, and the number of moles of a gas.The formula for the Ideal Gas Law is:\[ PV = nRT \]Here:
  • \( P \) is the pressure,
  • \( V \) denotes the volume,
  • \( n \) represents the number of moles,
  • \( R \) is the ideal gas constant,
  • \( T \) stands for the temperature in Kelvin.
This law allows us to calculate any one of the state properties of an ideal gas if the others are known. It assumes that gas molecules do not interact except for elastic collisions, which makes it an ideal approximation under many conditions. The Ideal Gas Law is vital in fields like chemistry and engineering, providing essential insights for designing reaction chambers, forecasting weather, and more.

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Most popular questions from this chapter

An aerosol spray can with a volume of \(125 \mathrm{~mL}\) contains \(1.30 \mathrm{~g}\) of propane gas \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) as a propellant. (a) If the can is at \(25^{\circ} \mathrm{C}\), what is the pressure in the can? (b) What volume would the butane occupy at \(S T P ?\) (c) The can's label says that exposure to temperatures above \(50^{\circ} \mathrm{C}\) may cause the can to burst. What is the pressure in the can at this temperature?

A 6.53 -g sample of a mixture of magnesium carbonate and calcium carbonate is treated with excess hydrochloric acid. The resulting reaction produces \(1.72 \mathrm{~L}\) of carbon dioxide gas at \(28^{\circ} \mathrm{C}\) and \(99.06 \mathrm{kPa}\) pressure. (a) Write balanced chemical equations for the reactions that occur between hydrochloric acid and each component of the mixture. (b) Calculate the total number of moles of carbon dioxide that forms from these reactions. (c) Assuming that the reactions are complete, calculate the percentage by mass of magnesium carbonate in the mixture.

Both Jacques Charles and Joseph Louis Guy-Lussac were avid balloonists. In his original flight in 1783 , Jacques Charles used a balloon that contained approximately \(31,150 \mathrm{~L}\) of \(\mathrm{H}_{2}\). He generated the \(\mathrm{H}_{2}\) using the reaction between iron and hydrochloric acid: $$\mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ How many kilograms of iron were needed to produce this volume of \(\mathrm{H}_{2}\) if the temperature was \(22{ }^{\circ} \mathrm{C} ?\)

A rigid vessel containing a \(3: 1 \mathrm{~mol}\) ratio of carbon dioxide and water vapor is held at \(200^{\circ} \mathrm{C}\) where it has a total pressure of \(202.7 \mathrm{kPa}\). If the vessel is cooled to \(10^{\circ} \mathrm{C}\) so that all of the water vapor condenses, what is the pressure of carbon dioxide? Neglect the volume of the liquid water that forms on cooling.

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