Chapter 3: Problem 49
The RMS speed of oxygen molecules in a gas is \(V .\) If the temperature is doubled and the oxygen molecules dissociated into oxygen atoms, the RMS speed will become (a) \(\bar{V}\) (b) \(\sqrt{2} V\) (c) \(2 V\) (d) \(4 V\)
Short Answer
Expert verified
\(2V\)
Step by step solution
01
Understand RMS Speed and Temperature Relationship
The root mean square (RMS) speed of molecules in a gas is related to the temperature through the equation \(V_{rms} = \sqrt{\frac{3kT}{m}}\), where \(k\) is the Boltzmann constant, \(T\) is the temperature, and \(m\) is the mass of a single molecule of the gas.
02
Equate Initial and Final RMS Speeds
When the temperature is doubled \((2T)\), and the molecules dissociate into atoms, the mass \(m\) of the particles in the gas becomes half \((m/2)\) because an oxygen molecule \(O_2\) consists of two oxygen atoms \(O\). So, the final RMS speed can be found using the modified equation \(V_{final} = \sqrt{\frac{3k(2T)}{m/2}}\).
03
Solve for the Final RMS Speed
Plugging in the values into the equation, you can simplify to \(V_{final} = \sqrt{4 \cdot \frac{3kT}{m}} = \sqrt{4} \cdot \sqrt{\frac{3kT}{m}} = 2V_{rms}\). This shows that after doubling the temperature and dissociating the molecule into atoms, the new RMS speed is \(2V_{rms}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Root Mean Square Velocity
Understanding the root mean square (RMS) speed, or RMS velocity, is essential when studying the movement of gas particles. RMS speed is a measure of the speed of particles in a gas that represents the square root of the average squared speeds of the particles. In physics, it’s a way to quantify the velocity of particles in thermal equilibrium.
The formula for RMS speed is given by \[\begin{equation}V_{rms} = \sqrt{\frac{3kT}{m}}\end{equation}\],
where k is the Boltzmann constant, T is the absolute temperature of the gas, and m is the mass of a single particle. The key to this formula is not just the presence of these variables, but also their relationships. For instance, the RMS speed is directly proportional to the square root of the temperature. This means that if the temperature increases, the RMS speed increases as well.
In the exercise at hand, the RMS speed plays a critical role in understanding how the speed of gas particles changes when the temperature is altered and when oxygen molecules (O_2) dissociate into individual oxygen atoms (O).
The formula for RMS speed is given by \[\begin{equation}V_{rms} = \sqrt{\frac{3kT}{m}}\end{equation}\],
where k is the Boltzmann constant, T is the absolute temperature of the gas, and m is the mass of a single particle. The key to this formula is not just the presence of these variables, but also their relationships. For instance, the RMS speed is directly proportional to the square root of the temperature. This means that if the temperature increases, the RMS speed increases as well.
In the exercise at hand, the RMS speed plays a critical role in understanding how the speed of gas particles changes when the temperature is altered and when oxygen molecules (O_2) dissociate into individual oxygen atoms (O).
Temperature and Kinetic Energy Relationship
The temperature of a gas is intimately related to the kinetic energy of its particles. This is a fundamental concept in thermodynamics and physical chemistry. In a gas, the temperature increase leads to an increase in the kinetic energy of the molecules, which in turn affects their speed.
The kinetic energy (KE) of a particle is given by the equation\[\begin{equation}KE = \frac{1}{2}mv^2\end{equation}\],
which states that the kinetic energy is proportional to the mass (m) of the particle and the square of its speed (v). However, when dealing with a collection of gas particles, we consider the average kinetic energy, which relates to the temperature (T) as seen in the formula for RMS speed.
If the temperature of a gas is doubled, as posed in our example exercise, the average kinetic energy of the particles also essentially doubles, leading to a change in speed, particularly the RMS speed. This direct dependency of RMS speed on temperature is key to solving the exercise.
The kinetic energy (KE) of a particle is given by the equation\[\begin{equation}KE = \frac{1}{2}mv^2\end{equation}\],
which states that the kinetic energy is proportional to the mass (m) of the particle and the square of its speed (v). However, when dealing with a collection of gas particles, we consider the average kinetic energy, which relates to the temperature (T) as seen in the formula for RMS speed.
If the temperature of a gas is doubled, as posed in our example exercise, the average kinetic energy of the particles also essentially doubles, leading to a change in speed, particularly the RMS speed. This direct dependency of RMS speed on temperature is key to solving the exercise.
Gas Molecule Dissociation
Gas molecule dissociation refers to the process by which molecules consisting of two or more atoms split into smaller species, usually individual atoms or smaller molecules. In our exercise, oxygen molecules (O_2), which normally consist of two oxygen atoms, are dissociating into separate oxygen atoms (O).
Dissociation affects the mass (m) of the gas particles involved in calculations of RMS speed. As stated earlier, the mass in the RMS speed formula is that of a single particle. When dealing with molecules, this is the molecular mass, but upon dissociation, it becomes the atomic mass. Since the molecular mass of an O_2 molecule is essentially double the atomic mass of an oxygen atom, dissociation results in halving the mass used in our RMS speed calculation.
Therefore, the dissociation of oxygen molecules into atoms is a critical consideration in the step-by-step solution as it alters the particle mass and thus the final RMS speed.
Dissociation affects the mass (m) of the gas particles involved in calculations of RMS speed. As stated earlier, the mass in the RMS speed formula is that of a single particle. When dealing with molecules, this is the molecular mass, but upon dissociation, it becomes the atomic mass. Since the molecular mass of an O_2 molecule is essentially double the atomic mass of an oxygen atom, dissociation results in halving the mass used in our RMS speed calculation.
Therefore, the dissociation of oxygen molecules into atoms is a critical consideration in the step-by-step solution as it alters the particle mass and thus the final RMS speed.
Physical Chemistry Problems
Physical chemistry problems involve the application of physical principles to chemically significant scenarios. The exercise we explored is an excellent example of a classical physical chemistry problem where knowledge of molecular behavior and mathematical relationships are used to predict changes in a system's properties.
Solving such problems generally requires an understanding of various principles, such as the kinetic theory of gases, thermodynamics, and the laws governing energy and mass. The step-by-step approach, as shown in the exercise, is particularly effective as it allows students to reason through each variable's impact and how changes in these variables affect the overall system.
Students are often advised to break down complex problems into smaller, manageable steps, carefully considering how alterations in conditions—such as temperature changes or molecular dissociation—impact the outcome. This systematic method is not only key to solving physical chemistry problems but also to deeply understand the underlying concepts at play.
Solving such problems generally requires an understanding of various principles, such as the kinetic theory of gases, thermodynamics, and the laws governing energy and mass. The step-by-step approach, as shown in the exercise, is particularly effective as it allows students to reason through each variable's impact and how changes in these variables affect the overall system.
Students are often advised to break down complex problems into smaller, manageable steps, carefully considering how alterations in conditions—such as temperature changes or molecular dissociation—impact the outcome. This systematic method is not only key to solving physical chemistry problems but also to deeply understand the underlying concepts at play.