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(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 \(T\). (b) If a car tire is filled to a pressure of \(32.0 \mathrm{lbs} / \mathrm{in}^{2}\) (psi) measured at \(75^{\circ} \mathrm{F}\), what will be the tire pressure if the tires heat up to \(120^{\circ} \mathrm{F}\) during driving?

Short Answer

Expert verified
Using Charles's Law and Boyle's Law, the derived proportionality relationship between pressure and temperature (Amonton's Law) is given by \(P_2T_1 = P_1T_2\). When initial tire pressure is \(32.0\, \mathrm{lbs} / \mathrm{in}^2\) at \(75^{\circ} \mathrm{F}\) and the temperature increases to \(120^{\circ} \mathrm{F}\), the tire pressure becomes approximately \(34.709\, \mathrm{lbs} / \mathrm{in}^2\).

Step by step solution

01

(a) Deriving Amonton's Law using Charles's Law and Boyle's Law

First, let's write down the expressions for Charles's Law and Boyle's Law: Charles's Law: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) Boyle's Law: \(P_1V_1 = P_2V_2\) Our goal is to express \(P\) as a function of \(T\). Step 1: Replace \(V_2\) in Charles's Law with an expression from Boyle's Law From Boyle's Law, we know that: \(V_2 = \frac{P_1V_1}{P_2}\) Now we substitute \(V_2\) in Charles's Law: \(\frac{V_1}{T_1} = \frac{P_1V_1}{P_2T_2}\) Step 2: Simplify the equation Now let's simplify: \(P_2T_1 = P_1T_2\) This is the derived proportionality relationship between pressure and temperature, also known as Amontons's Law or Gay-Lussac's Law.
02

(b) Calculate the tire pressure at a higher temperature

Given: Initial Pressure, \(P_1 = 32.0 \, \mathrm{lbs}\,/\, \mathrm{in}^2\) Initial Temperature, \(T_1 = 75^{\circ}\mathrm{F}\) Final Temperature, \(T_2 = 120^{\circ}\mathrm{F}\) Step 1: Convert temperatures to Kelvin To work with the equation derived in part (a), we must convert the given temperatures to Kelvin. [°F] to [K] conversion formula: \[K = \frac{9}{5}(° \mathrm{F} - 32) + 273.15\] Converting initial temperature: \(T_1 = \frac{9}{5}(75-32) + 273.15 = 297.039\, \mathrm{K}\) Converting final temperature: \(T_2 = \frac{9}{5}(120-32) + 273.15 = 322.039\, \mathrm{K}\) Step 2: Apply Amonton's Law From part (a), we derived that: \(P_2T_1 = P_1T_2\) Now, we can solve for the final tire pressure, \(P_2\): \(P_2 = \frac{P_1T_2}{T_1}\) Substitute the given values: \(P_2 = \frac{32.0 \,\mathrm{lbs/\,in}^2 \times 322.039\, \mathrm{K}}{297.039\, \mathrm{K}}\) Step 3: Calculate the final tire pressure \(P_2 = 34.709\, \mathrm{lbs/\,in}^2\) The tire pressure after heating up to \(120^{\circ} \mathrm{F}\) will be approximately \(34.709 \mathrm{\,lbs} /\mathrm{in}^2\).

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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 thermodynamics that describes how gases expand when heated. Specifically, it states that the volume of a gas is directly proportional to its temperature when the pressure is held constant. This can be mathematically expressed as:
  • \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)

Here, \( V_1 \) and \( V_2 \) are the initial and final volumes, while \( T_1 \) and \( T_2 \) are the initial and final temperatures, respectively. It implies that if the temperature of a gas increases, the volume will also increase provided that the pressure does not change.
The law is crucial when considering experiments involving ideal gases, where it helps predict how a change in temperature can affect volume. In deriving other gas laws, like Amonton's Law, it's often used in combination with other principles such as Boyle's Law.
Boyle's Law
Boyle's Law is another essential gas law that addresses the relationship between pressure and volume of a gas at a constant temperature. According to this law, the pressure of a gas is inversely proportional to its volume when the temperature remains unchanged. Mathematically, it is represented as:
  • \( P_1V_1 = P_2V_2 \)

In this equation, \( P_1 \) and \( P_2 \) represent the initial and final pressures, while \( V_1 \) and \( V_2 \) symbolize the initial and final volumes. This means that as the volume of a gas decreases, the pressure increases, provided the temperature remains constant.
Boyle's Law is particularly useful in calculating pressure changes in enclosed systems, such as a tire or a piston, where temperature can be controlled. Understanding this law helps to appreciate how variables in gas systems are interconnected, which is a key to deriving more complex relationships like those seen in determining Amonton's Law.
Pressure-Temperature Relationship
The pressure-temperature relationship, also referred to as Amonton's Law, provides insight into how gas pressure changes with temperature when volume is held constant. This relationship is derived by combining the concepts from Charles's and Boyle's Laws. The derived equation is:
  • \( P_2T_1 = P_1T_2 \)

Here, \( P_1 \) and \( P_2 \) are the initial and final pressures, with \( T_1 \) and \( T_2 \) being the initial and final temperatures. The key takeaway from this law is that if the temperature of a gas increases, so does the pressure, assuming the volume stays the same.
Amonton's Law is especially useful in real-world situations like predicting how the pressure in a car tire might change when it heats up from friction during driving. By using this relationship, one can understand and anticipate the behavior of contained gases under different thermal conditions, making it crucial for various scientific and engineering applications.

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Most popular questions from this chapter

A piece of dry ice (solid carbon dioxide) with a mass of \(5.50 \mathrm{~g}\) is placed in a \(10.0\) - \(L\) vessel that already contains air at 705 torr and \(24^{\circ} \mathrm{C}\). After the carbon dioxide has totally sublimed, what is the partial pressure of the resultant \(\mathrm{CO}_{2}\) gas, and the total pressure in the container at \(24^{\circ} \mathrm{C}\) ?

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 \(2.00 \mathrm{~atm}\). 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.

Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. (a) If \(1.56 \mathrm{~g}\) of cyclopropane has a volume of \(1.00 \mathrm{~L}\) at \(0.984 \mathrm{~atm}\) and \(50.0^{\circ} \mathrm{C}\), what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, \(\mathrm{CH}_{4}\) ?

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 \(\mathrm{C}_{\mathrm{g}} \mathrm{H}_{18}\) could be combusted by this quantity of \(\mathrm{O}_{2}\), assuming complete combustion with formation of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ? as originally filled with gas?

Based on their respective van der Waals constants (Table 10.3), is Ar or \(\mathrm{CO}_{2}\) expected to behave more nearly like an ideal gas at high pressures? Explain.

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