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Chapter 5: Problem 2

Construct a table similar to Table \(5.3\) for a diatomic gas for which \(\gamma=1.4\).

Short Answer

Expert verified
Question: Construct a table for a diatomic gas with an adiabatic index of 1.4, containing columns for internal energy (U), temperature (T), pressure (P), and volume (V). Answer: Here is the table for a diatomic gas with an adiabatic index of 1.4: ``` | Index | U | T | P | V | |-------|---------------|------|------------|-----| | 1 | \(\frac{5}{2} nR(300)\) | 300K | \(nR(300)\) | 1 m³ | | 2 | \(\frac{5}{2} nR(400)\) | 400K | \(nR(400)\) | 1 m³ | | 3 | \(\frac{5}{2} nR(500)\) | 500K | \(nR(500)\) | 1 m³ | ``` Note: The table is based on the assumption of 1 mole of diatomic gas with a fixed volume of 1 m³. You can change the values of moles (n) and volume (V) as well as the gas constant (R) for other scenarios.

Step by step solution

01

Set up the table template

First, we need to set up the table template. The table should have 5 columns in total - one for indexing and 4 for the required variables \(U\), \(T\), \(P\), and \(V\). We'll add the headings for these columns and leave the cells empty for now. ``` | Index | U | T | P | V | |-------|--------|------|-----|-----| | 1 | | | | | | 2 | | | | | ... ```
02

Fill in data for the table

For this table, we need to gather or calculate sample data points for each cell in the table. To do this, let's use the equations that relate internal energy, temperature, pressure, and volume for a diatomic gas. 1. Relationship between internal energy (\(U\)) and temperature (\(T\)): \(U = \frac{5}{2} nRT\) 2. Ideal Gas Law relating temperature, pressure and volume: \(PV = nRT\) Here,\(n\) is the number of moles of the diatomic gas, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. Now let's decide which values to use as data points for each variable. Start by picking a few temperatures for column T. For example, let's use these temperatures: \(300 K\), \(400 K\) and \(500 K\) as our data points. Then, using the ideal gas law and the relationship between internal energy and temperature, we can calculate the corresponding values of U, P, and V for each data point. We assume a fixed volume of \(1 m^3\) and \(1\) mole of the diatomic gas. Keep in mind that these constants can be adjusted in case we want other data points or larger tables.
03

Calculation and populating the table

Now, let's do the calculations and populate the table using the equations mentioned earlier. - For T = 300 K: \(U = \frac{5}{2}nRT = \frac{5}{2}\times 1 \times R \times 300\) \(PV = nRT \Rightarrow P = \frac{nRT}{V} = \frac{1 \times R \times 300}{1}\) - For T = 400 K: \(U = \frac{5}{2} \times 1 \times R \times 400\) \(P = \frac{1 \times R \times 400}{1}\) - For T = 500 K: \(U = \frac{5}{2} \times 1 \times R \times 500\) \(P = \frac{1 \times R \times 500}{1}\) Now let's put the values in the table (with given volume \(V=1m^3\)): ``` | Index | U | T | P | V | |-------|---------------|------|------------|-----| | 1 | \(\frac{5}{2} nR(300)\) | 300K | \(nR(300)\) | 1 m³ | | 2 | \(\frac{5}{2} nR(400)\) | 400K | \(nR(400)\) | 1 m³ | | 3 | \(\frac{5}{2} nR(500)\) | 500K | \(nR(500)\) | 1 m³ | ``` These are the final expressions for internal energy and pressure. If you have a specific value of R and n, you can replace it and get the final numeric values for the table.

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Key Concepts

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

Understanding the Ideal Gas Law
The ideal gas law is a fundamental relationship in chemistry and physics that relates the pressure (P), volume (V), temperature (T), and number of moles (n) of an ideal gas. Expressed in the equation form, it is written as:
\( PV = nRT \).
Here, R is the gas constant, which has a value of 8.314 J/(mol K).

This law assumes that the gas molecules are point particles that interact with their surroundings but not with each other, essentially having no volume of their own and no intermolecular forces. It's useful for predicting the behavior of gases under various conditions. When we manipulate this equation, we can solve for one of the variables if the other three are known.

In the context of the textbook exercise, we used the ideal gas law to relate temperature with pressure and volume, allowing us to calculate the pressure of a diatomic gas at different temperatures when the volume is held constant.
Exploring Internal Energy
Internal energy (U) in thermodynamics refers to the total energy contained within a substance, primarily composed of the kinetic and potential energy of the particles. For an ideal gas, and particularly for a diatomic gas like in our exercise, the internal energy is related to temperature.

The equation linking internal energy to temperature for a diatomic gas is:
\( U = \frac{5}{2} nRT \).
The factor of \(\frac{5}{2}\) comes from the degrees of freedom of a diatomic molecule, which accounts for its translational and rotational motions.

In our example, we use this equation to calculate the internal energy of the gas at various temperatures, which is essential for understanding how much energy is stored at a molecular level. This understanding is fundamental in the fields of thermodynamics and statistical mechanics.
The Thermodynamic Properties of a Gas
Thermodynamic properties are characteristics that describe the state and energy of a system. In gases, these properties include pressure, volume, temperature, and internal energy.

Pressure (P) is an indication of the force exerted by the gas particles when they collide with the walls of their container. Volume (V) is the space occupied by the gas. Temperature (T) is related to the average kinetic energy of the particles, and as mentioned earlier, internal energy (U) encompasses the overall energy of the particles.

With the ideal gas law and internal energy equation, we relate these properties to describe the state of a diatomic gas. By populating a table with values of P, V, T, and U, like in the exercise, we can visualize the behavior of the gas under changing conditions, such as different temperatures while keeping the volume constant, and predict how these properties would change in response to alterations in conditions.
Heat Capacity at Constant Volume
Heat capacity at constant volume (\(C_v\)) is the amount of heat required to raise the temperature of a given quantity of a substance by one degree Celsius while maintaining constant volume.

For a diatomic gas, which has both translational and rotational degrees of freedom, the molar heat capacity at constant volume is given by:
\( C_v = \frac{5}{2} R \).
This value arises due to the kinetic theory of gases, where it is assumed that each degree of freedom contributes \(\frac{1}{2} R\) to the heat capacity.

Knowing \(C_v\) is crucial for understanding how a gas absorbs heat under constant volume conditions. This is directly related to the internal energy change when a gas is heated at constant volume, as there isn't any work done on or by the gas, so all the added heat contributes to raising its internal energy.

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Most popular questions from this chapter

Each possible normal mode of oscillation of a distributed system constitutes a mechanical degree of freedom of the system. According to the classical equipartition-of-energy theorem, each oscillatory degree of freedom is to be assigned an average thermal energy of \(k T\), where \(k\) is Boltzmann's constant and \(T\) is the absolute temperature. Hence the internal thermal energy per unit volume of a monatomic gas should be \(E_{v}=n k T\), with \(n\) given by (5.7.11). For a mole of monatomic gas, kinetic theory tells us that the internal energy is \(E_{\mu}=\frac{3}{2} R T\), where \(R\) is the gas constant per mole. We can bring the two viewpoints into agreement by choosing a suitable value of \(\omega_{\max }\). Compute this value and show that the minimum wavelength is of the order of the mean spacing of the atoms in the gas.

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