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Assume that in an alternate universe, the laws of physics are very different from ours and that "ideal" gases behave as follows: (i) At constant temperature, pressure is inversely proportional to the square of the volume. (ii) At constant pressure, the volume varies directly with the \(\frac{2}{3}\) power of the temperature. (iii) At \(273.15 \mathrm{~K}\) and 1.00 atm pressure, 1.00 mole of an ideal gas is found to occupy \(22.4 \mathrm{~L}\).

Short Answer

Expert verified
The alternate universe gas laws equations are: (1) \( PV^2 = 501.76 \) and (2) \( V = 0.384 \times T^{\frac{2}{3}} \).

Step by step solution

01

Understand the Conditions

The problem provides a modified set of gas laws in an alternate universe. These laws are: 1. Pressure is inversely proportional to the square of the volume at constant temperature: \( P \propto \frac{1}{V^2} \).2. Volume varies directly with the \(\frac{2}{3}\) power of the temperature at constant pressure: \( V \propto T^{\frac{2}{3}} \). We are also given that at 273.15 K and 1.00 atm, the volume is 22.4 L for 1.00 mole of gas.
02

Express the Laws Mathematically

Using the given conditions, express the laws mathematically. For law (i), we have the relation: \[ P \times V^2 = C_1 \] where \( C_1 \) is a constant. For law (ii), we express it as: \[ V = C_2 \times T^{\frac{2}{3}} \] where \( C_2 \) is another constant.
03

Determine Constants at Given Conditions

Using the information at 273.15 K, 1.00 atm and 22.4 L, determine the constants. First, for law (i): \[ 1.00 \times (22.4)^2 = C_1 \] So, \( C_1 = 501.76 \). Now, for law (ii), at \( T = 273.15 \): \[ 22.4 = C_2 \times (273.15)^{\frac{2}{3}} \] Thus, \( C_2 \approx \frac{22.4}{(273.15)^{\frac{2}{3}}} \). Calculate to find \( C_2 \).
04

Calculate Specific Values for Constants

Continuing from Step 3, calculate \( C_2 \): \[ C_2 \approx \frac{22.4}{(273.15)^{\frac{2}{3}}} \approx \frac{22.4}{58.33} \approx 0.384 \]. Therefore, we have \( C_1 = 501.76 \) and \( C_2 \approx 0.384 \).
05

Validate Understanding with Derived Equations

The specific derived gas equations for this alternate universe are:1. \( PV^2 = 501.76 \)2. \( V = 0.384 \times T^{\frac{2}{3}} \)These equations can now be used to analyze any other states or scenarios of the ideal gas in this universe.

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Key Concepts

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

Pressure-Volume Relationship
In the alternate universe described by the exercise, the pressure-volume relationship is quite different from what we know on Earth. Here, pressure is inversely proportional to the square of the volume at constant temperature. This means as the volume of the gas increases, the pressure decreases by an amount squared, and vice versa.

This relationship can be mathematically expressed as:
  • \( P \propto \frac{1}{V^2} \)
What this implies is that the product of pressure and the square of the volume remains constant, represented as:
  • \( P \times V^2 = C_1 \)
where \( C_1 \) is a constant specific to the conditions of the system.

What makes this relationship interesting is the heightened sensitivity of pressure changes to even small changes in volume. This is different from the more familiar inverse proportionality in our universe, where \( P \propto \frac{1}{V} \). Understanding this concept is crucial as it lays the foundation for how gases behave under these altered laws.
Temperature-Volume Relationship
In this fascinating alternate reality of physics, the temperature-volume relationship is unique. At a constant pressure, the volume of an ideal gas changes directly with the \(\frac{2}{3}\) power of the temperature. Essentially, as the temperature increases or decreases, the volume of the gas also changes but not linearly.

This direct proportionality can be represented mathematically as:
  • \( V \propto T^{\frac{2}{3}} \)
This expression implies that if you know the temperature, you can determine the volume using the equation:
  • \( V = C_2 \times T^{\frac{2}{3}} \)
where \( C_2 \) is another constant obtained under specific conditions.

This relationship is markedly different from the linear temperature-volume relationship described by Charles's Law in our universe, \( V \propto T \). In this scenario, the volume changes are more subtle compared to changes in temperature, providing an insight into unique thermodynamic properties under alternate gas laws.
Alternate Gas Laws
The alternate gas laws presented in this problem are new and unusual, and they deviate significantly from the classical gas laws familiar to us in our universe, such as Boyle's Law, Charles's Law, and the Ideal Gas Law.

In this alternate universe, two primary modifications create a distinctive set of rules:
  • Pressure and Volume: The inverse square relationship between pressure and volume.
  • Volume and Temperature: The direct relationship involving the \(\frac{2}{3}\) power of temperature.
These changes introduce us to a unique way thermodynamic quantities interact. Given that in our universe, these relationships are linear or involve simple inversions, it is fascinating to see how alternate assumptions about gas behavior bring about entirely different results.

The core takeaway from studying these laws is the realization that altering fundamental assumptions (like those of gas behavior) can lead us to entirely new mechanics. This highlights the vast possibilities in physics, where even subtle changes to laws can create a remarkably different universe, allowing us to imagine how gases behave under varied scenarios.

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Most popular questions from this chapter

(II) A typical scuba tank, when fully charged, contains \(12 \mathrm{~L}\) of air at 204 atm. Assume an "empty" tank contains air at 34 atm and is connected to an air compressor at sea level. The air compressor intakes air from the atmosphere, compresses it to high pressure, and then inputs this highpressure air into the scuba tank. If the (average) flow rate of air from the atmosphere into the intake port of the air compressor is \(290 \mathrm{~L} / \mathrm{min}\), how long will it take to fully charge the scuba tank? Assume the tank remains at the same temperature as the surrounding air during the filling process.

(I) Absolute zero is what temperature on the Fahrenheit scale?

(I) If \(3.80 \mathrm{~m}^{3}\) of a gas initially at STP is placed under a pressure of \(3.20 \mathrm{~atm}\), the temperature of the gas rises to \(38.0^{\circ} \mathrm{C}\). What is the volume?

(I) What are the following temperatures on the Kelvin scale: (a) \(66^{\circ} \mathrm{C},(b) 92^{\circ} \mathrm{F},(c)-55^{\circ} \mathrm{C}\) (d) \(5500^{\circ} \mathrm{C} ?\)

(II) It is observed that 55.50 \(\mathrm{mL}\) of water at \(20^{\circ} \mathrm{C}\) completely fills a container to the brim. When the container and the water are heated to \(60^{\circ} \mathrm{C}, 0.35 \mathrm{g}\) of water is lost. (a) What is the coefficient of volume expansion of the container? \((b)\) What is the most likely material of the container? Density of water at \(60^{\circ} \mathrm{C}\) is 0.98324 \(\mathrm{g} / \mathrm{mL} .\)

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