/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Compressed air can be pumped und... [FREE SOLUTION] | 91Ó°ÊÓ

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Compressed air can be pumped underground into huge caverns as a form of energy storage. The volume of a cavern is \(5.6 \times 10^{5} \mathrm{~m}^{3},\) and the pressure of the air in it is \(7.7 \times 10^{6} \mathrm{~Pa}\). Assume that air is a diatomic ideal gas whose internal energy \(U\) is given by \(U=\frac{5}{2} n R T\) If one home uses \(30.0 \mathrm{~kW} \cdot \mathrm{h}\) of energy per day, how many homes could this internal energy serve for one day?

Short Answer

Expert verified
The compressed air can serve 1000 homes for one day.

Step by step solution

01

Calculate the Number of Moles

First, we use the formula for the ideal gas law: \[ PV = nRT \]Given:\( V = 5.6 \times 10^{5} \, \mathrm{m}^{3} \),\( P = 7.7 \times 10^{6} \, \mathrm{Pa} \),\( R = 8.314 \, \mathrm{J/(mol \cdot K)} \).Assuming standard room temperature (approximately 298 K), we substitute to find:\[ n = \frac{PV}{RT} = \frac{(7.7 \times 10^{6})(5.6 \times 10^{5})}{8.314 \times 298} \approx 1.74 \times 10^{9} \text{ moles} \]
02

Calculate Internal Energy

Now, compute the internal energy of the gas using:\[ U = \frac{5}{2} nRT \]Substituting the calculated number of moles and assuming \(T = 298 \mathrm{~K}\):\[ U = \frac{5}{2} \times 1.74 \times 10^{9} \times 8.314 \times 298 \approx 1.08 \times 10^{12} \, \mathrm{J} \]
03

Convert Energy Usage Per Home to Joules

Convert the daily energy usage per home from kilowatt-hours to joules using:\[ 1 \, \mathrm{kWh} = 3.6 \times 10^{6} \, \mathrm{J} \]Energy per home per day:\[ 30.0 \, \mathrm{kWh} \times 3.6 \times 10^{6} \, \mathrm{J/kWh} = 1.08 \times 10^{9} \, \mathrm{J} \]
04

Calculate Number of Homes Served

Divide the total internal energy by the energy required per home:\[ \frac{1.08 \times 10^{12}}{1.08 \times 10^{9}} = 1000 \]Therefore, the internal energy can serve 1000 homes.

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Key Concepts

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

Understanding Internal Energy
Internal energy is a critical concept in thermodynamics. It represents the total energy contained within a system due to the movement and interactions of its molecules and atoms. In the context of gases, this energy is largely due to the kinetic energy of the particles.
For diatomic gases, which consist of molecules made up of two atoms, the internal energy can be expressed as: \[ U = \frac{5}{2} nRT \] where:
  • \(n\) is the number of moles of the gas
  • \(R\) is the ideal gas constant
  • \(T\) is the temperature in Kelvin
This formula accounts for the translational, rotational, and to some extent vibrational movements, assuming the gas behaves ideally.
Thus, in our exercise, knowing the internal energy helps us determine how much energy is stored in the compressed air, and therefore, how many homes it can supply energy to.
Properties of Diatomic Gas
Diatomic gases, such as oxygen and nitrogen, are common in the Earth's atmosphere. These gases are made up of molecules consisting of two atoms. This structural characteristic influences their thermodynamic behavior.
One impact is in their specific heat capacities, which reflect the energy required to change the temperature of the gas.
For diatomic ideal gases, the degrees of freedom differ compared to monatomic gases, leading to a slightly different formula for internal energy: \[ U = \frac{5}{2} nRT \] This formula reflects their enhanced internal energy contributions from rotational movements, necessary for accounting the effects of molecular rotation on the energy of the gas.
This understanding aids in analyzing the stored energy when diatomic gases are compressed in caverns.
Energy Conversion and Usage
Energy conversion is a fundamental process in harnessing and utilizing energy stored in various forms. In our scenario, compressed air stores potential energy, which can then be converted to electricity.
This involves converting the stored internal energy of the gas into mechanical work and then into electrical energy.
To calculate how many homes the stored energy in the cavern can serve, we compared the total energy stored with the energy consumption of a single home.
Given the internal energy stored in the cavern and the daily energy usage per home, this conversion calculation allowed us to determine that 1000 homes could be serviced.
This practical application of energy conversion shows the importance of understanding internal energy outputs in real-world scenarios.
Role of Thermodynamics
Thermodynamics is the scientific study of energy transactions and transformations. It addresses various laws and principles that govern the behavior of energy in any given system.
When looking at compressing air into caverns, thermodynamics helps predict and compute energy changes and transfers involved in the process.
A deep understanding of these principles allows us to effectively calculate the stored energy and potential output of compressed gases.
The ideal gas law: \[ PV = nRT \]is one of the core principles involved in thermodynamics, providing us with a way to relate pressure, volume, and temperature with the amount of gas in moles.
In our exercise, thermodynamics principles have been applied to calculate how efficiently energy can be stored and utilized, thereby boiling down complex interactions to simpler calculations.

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Most popular questions from this chapter

Multiple-Concept Example 4 reviews the principles that play roles in this problem. A primitive diving bell consists of a cylindrical tank with one end open and one end closed. The tank is lowered into a freshwater lake, open end downward. Water rises into the tank, compressing the trapped air, whose temperature remains constant during the descent. The tank is brought to a halt when the distance between the surface of the water in the tank and the surface of the lake is \(40.0 \mathrm{~m}\). Atmospheric pressure at the surface of the lake is \(1.01 \times 10^{5} \mathrm{~Pa}\). Find the fraction of the tank's volume that is filled with air.

A tank contains \(0.85 \mathrm{~mol}\) of molecular nitrogen \(\left(\mathrm{N}_{2}\right)\). Determine the mass (in grams) of nitrogen that must be removed from the tank in order to lower the pressure from 38 to 25 atm. Assume that the volume and temperature of the nitrogen in the tank do not change.

Refer to Interactive Solution \(\underline{14.21}\) at for help with problems like this one. An apartment has a living room whose dimensions are \(2.5 \mathrm{~m} \times 4.0 \mathrm{~m} \times 5.0 \mathrm{~m}\). Assume that the air in the room is composed of \(79 \%\) nitrogen \(\left(\mathrm{N}_{2}\right)\) and \(21 \%\) oxygen \(\left(\mathrm{O}_{2}\right)\). At a temperature of \(22{ }^{\circ} \mathrm{C}\) and a pressure of \(1.01 \times 10^{5} \mathrm{~Pa}\), what is the mass (in grams) of the air?

The artificial sweetener NutraSweet is a chemical called aspartame \(\left(\mathrm{C}_{14} \mathrm{H}_{18} \mathrm{~N}_{2} \mathrm{O}_{5}\right)\). What is (a) its molecular mass (in atomic mass units) and (b) the mass (in \(\mathrm{kg}\) ) of an aspartame molecule?

The drawing shows an ideal gas confined to a cylinder by a massless piston that is attached to an ideal spring. Outside the cylinder is a vacuum. The cross-sectional area of the piston is \(A\). The initial pressure, volume, and temperature of the gas are, respectively, \(P_{0}, V_{0}\), and \(T_{0}\), and the spring is initially stretched by an amount \(x_{0}\) with respect to its unstrained length. The gas is heated, so that its final pressure, volume, and temperature are \(P_{\mathrm{f}}, V_{\mathrm{f}},\) and \(T_{\mathrm{f}}\) and the spring is stretched by an amount \(x_{\mathrm{f}}\) with respect to its unstrained length. Assume that \(x_{0}\) and \(x_{\mathrm{f}}\) are positive variables. (a) What is the relation between the magnitude of the force required to stretch an ideal spring and the amount of the stretch with respect to the unstrained length of the spring? (b) What are the magnitudes, \(F_{0}\) and \(F_{\mathrm{F}}\), of the forces that the initial and final pressures apply to the piston and, hence, to the spring? Express your answers in terms of the pressures and the cross-sectional area of the piston. (c) According to the ideal gas law, how are the initial pressure, volume, and temperature related to the final pressure, volume, and temperature? (d) How is the final volume related to the initial volume, the cross-sectional area of the piston, and the initial and final amounts by which the spring is stretched? Account for your answer. Problem The initial temperature and volume of the gas described in the Concept Questions are \(273 \mathrm{~K}\) and \(6.00 \times 10^{-4} \mathrm{~m}^{3}\). The initial and final amounts by which the spring is stretched are, respectively, 0.0800 and \(0.1000 \mathrm{~m}\). The cross-sectional area of the piston is \(2.50 \times 10^{-3} \mathrm{~m}^{2}\). What is the final temperature of the gas?

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