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The temperature of a \(5.00-\mathrm{L}\) container of \(\mathrm{N}_{2}\) gas is increased from \(20^{\circ} \mathrm{C}\) to \(250^{\circ} \mathrm{C}\) . If the volume is held constant, predict qualitatively how this change affects the following: (a) the average kinetic energy of the molecules; (b) the root-mean- square speed of the molecules; (c) the strength of the impact of an average molecule with the container walls; (d) the total number of collisions of molecules with walls per second.

Short Answer

Expert verified
Increasing the temperature from \(20^{\circ}C\) to \(250^{\circ}C\) while keeping the volume constant results in an increase in: (a) the average kinetic energy of the molecules, because it is directly proportional to temperature; (b) the root-mean-square speed of the molecules, as it depends on the square root of temperature; (c) the strength of the impact of an average molecule with the container walls, as the momentum of the molecules increases with increasing temperature; and (d) the total number of collisions of molecules with walls per second, as the collision frequency increases with higher root-mean-square speed.

Step by step solution

01

(a) Average Kinetic Energy

The average kinetic energy of the gas molecules is directly proportional to the temperature (in Kelvin) of the gas. Mathematically, we have: \( \bar{K.E.} = \frac{3}{2}kT \) Where \( \bar{K.E.} \) is the average kinetic energy, \(k\) is Boltzmann's constant, \(T\) is the temperature in Kelvin. Since the temperature is increased, the average kinetic energy of the molecules will increase as well.
02

(b) Root-mean-square speed

The root-mean-square speed of the molecules is given by: \(v_{rms} = \sqrt{\frac{3kT}{m}} \) Where \(v_{rms}\) is the root-mean-square speed, \(m\) is the mass of a single molecule, and other variables are as defined earlier. As the temperature increases, the root-mean-square speed will also increase, because the temperature term is in the numerator.
03

(c) Strength of the impact with the container walls

The strength of the impact will be related to the momentum of the colliding particles. Momentum is given by: \(p = mv \) Where \(p\) is the momentum, \(m\) is the mass of a single molecule, and \(v\) is the molecule's speed. Since the root-mean-square speed of the molecules increases with temperature, the strength of the impacts with the container walls will also increase.
04

(d) Number of collisions per second

To find the total number of collisions, we need to consider the collision frequency, which is given by: \(Z = \frac{N}{V}v_{rms}\sigma \) Where \(Z\) is the number of collisions per second, \(N\) is the number of molecules, \(V\) is the volume of the container, \(\sigma\) is the collision cross-section. As the root-mean-square speed increases with increasing temperature (as described in part b), there will be an increase in the number of collisions per second with the container walls. In conclusion, an increase in temperature will lead to an increase in the average kinetic energy of the molecules, root-mean-square speed, strength of the impact with container walls, and the total number of collisions per second.

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Key Concepts

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

Kinetic Molecular Theory
The kinetic molecular theory is a fundamental concept that explains the behavior of gases. It is based on several key assumptions that help to describe how gas molecules move and interact.
\(\qquad\)
  • Gas molecules are in constant, random motion, colliding with the walls of the container and each other.
  • The volume of individual gas molecules is negligible compared to the total volume of the container.
  • There are no forces of attraction or repulsion between the molecules in an ideal gas.
  • The average kinetic energy of gas molecules is directly proportional to the temperature of the gas in Kelvin.
Understanding these principles helps us explain why changes in temperature affect gases in the ways they do.
For example, if you increase the temperature of a gas held at constant volume, the molecules move faster — hence the average kinetic energy increases. This supports the observations that temperature changes affect other properties of the gas, such as speed and collision frequency.
Root-mean-square Speed
Root-mean-square speed is a measure of the speed of particles in a gas and is particularly useful when considering their kinetic energy. It is defined by the equation:
\(\qquad\)\[v_{rms} = \sqrt{\frac{3kT}{m}}\]
\(\qquad\)Here, \(v_{rms}\) is the root-mean-square speed, \(k\) is Boltzmann’s constant, \(T\) is the temperature in Kelvin, and \(m\) is the mass of a single molecule.
As temperature rises, \(v_{rms}\) increases because temperature is in the numerator of the equation. This means the particles' speeds increase on average and they move faster. Even though individual molecules may move faster or slower, root-mean-square speed provides a useful average speed.
This concept is crucial when understanding the behavior of gases as they warm. As temperature affects \(v_{rms}\), it influences other factors like kinetic energy and collision frequency.
Collision Frequency
Collision frequency refers to the number of collisions a gas molecule makes with the walls of its container per second. It is an indicator of how often gases interact with their surroundings.
\(\qquad\)The expression for collision frequency \(Z\) is given by:
\(\qquad\)\[Z = \frac{N}{V}v_{rms}\sigma\]
\(\qquad\)Where \(N\) is the number of molecules, \(V\) is the volume of the container, \(v_{rms}\) is the root-mean-square speed, and \(\sigma\) is the collision cross-section. As the temperature increases, the \(v_{rms}\) also increases, which results in more frequent collisions.
Frequent collisions are essential for maintaining the physical properties of gases such as pressure. An increase in collision frequency due to an increase in temperature raises the pressure exerted by the gas if the volume is kept constant.
Boltzmann's Constant
Boltzmann's constant is a fundamental constant in physics denoted as \(k\). Its value is approximately \(1.38 \times 10^{-23} \, \mathrm{J/K}\) and it provides a bridge between macroscopic and microscopic physics.
\(\qquad\)Boltzmann's constant is a crucial part of various equations concerning kinetic molecular theory, such as the average kinetic energy equation:
\(\qquad\)\[\bar{K.E.} = \frac{3}{2}kT\]
\(\qquad\)The constant links temperature with energy at the molecular level, making it essential for calculations of thermal dynamics, such as changes in average kinetic energy when temperature changes.
Boltzmann’s role in these equations helps us understand phenomena observed in gases as part of everyday occurrences, like how faster moving particles at higher temperatures lead to things like cooking faster or engines running more efficiently.

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Most popular questions from this chapter

The metabolic oxidation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) in our bodies produces \(\mathrm{CO}_{2},\) which is expelled from our lungs as a gas: $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$$ (a) Calculate the volume of dry \(\mathrm{CO}_{2}\) produced at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and 0.970 atm when 24.5 \(\mathrm{g}\) of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at 1.00 \(\mathrm{atm}\) and \(298 \mathrm{K},\) to completely oxidize 50.0 \(\mathrm{g}\) of glucose.

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