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The temperature of a 5.00-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 ner second.

Short Answer

Expert verified
In summary, when the temperature of a \(5.00-L\) container of \(\mathrm{N}_{2}\) gas is increased from \(20^{\circ} \mathrm{C}\) to \(250^{\circ} \mathrm{C}\), while holding the volume constant, the following changes will occur: (a) the average kinetic energy of the molecules will increase; (b) the root-mean-square speed of the molecules will increase; (c) the strength of impact of an average molecule with the container walls will increase; (d) the total number of collisions of molecules with walls per second will increase.

Step by step solution

01

Understanding the Concepts

To begin with, according to the kinetic theory of gases, the average kinetic energy of gas molecules is directly proportional to the temperature of the gas (measured in Kelvin). Also, the root-mean-square speed of gas molecules is dependent on the temperature and molecular mass. The strength of impact of molecules with the container walls and the total number of collisions with the walls per second are linked to the pressure exerted by the gas, which is determined by the number of molecules, temperature and volume of the container.
02

(a) Predicting the change in Average kinetic energy of molecules

As temperature increases, the average kinetic energy of the gas molecules also increases. This is because the gas molecules absorb heat energy, which gets transformed into kinetic energy causing them to move faster. So, the average kinetic energy of the molecules will increase.
03

(b) Predicting the change in root-mean-square speed of the molecules

The formula for the root-mean-square speed of gas molecules is given by \(\sqrt{3kT/m}\), where \(k\) is Boltzmann’s constant, \(T\) is the temperature in Kelvin and \(m\) is the molecular mass. As temperature increases, it would cause the root-mean-square speed of the molecules to increase. So, the root-mean-square speed of the molecules will increase.
04

(c) Predicting the change in strength of impact of an average molecule with the container walls

The strength of impact of molecules on the container walls is related to the pressure the gas exerts on the walls. As temperature increases, the molecules have more kinetic energy and move faster, striking the container walls with more energy. Because the volume is constant, this increase in molecule speed and energy results in an increase in pressure, thus increasing the strength of impact. So, the strength of impact of an average molecule with the container walls will increase.
05

(d) Predicting the change in the total number of collisions of molecules with walls per second

Again, with the increase in temperature, the speed of molecules increases. Faster molecules means they will hit the container walls more frequently, because they take less time to travel back and forth within the container. So, the total number of collisions of molecules with walls per second will increase.

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Key Concepts

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

Average Kinetic Energy
The concept of average kinetic energy in the kinetic theory of gases relates to how swiftly the molecules move inside a container. With nitrogen gas (\(\mathrm{N}_{2}\)), when the temperature rises, the molecules gain energy and move faster. This movement is directly linked to the temperature measured in Kelvin—a key fact in predicting changes in average kinetic energy.

- The formula linking average kinetic energy to temperature is: - \( KE_{\text{avg}} = \frac{3}{2}kT \) where \( k \) is Boltzmann's constant and \( T \) is the absolute temperature.
Understanding that higher temperature means higher kinetic energy helps clarify why the molecules speed up when the temperature increases from 20°C to 250°C. The extra heat energy transforms into kinetic energy, making the molecules move with more vigor.
Root-Mean-Square Speed
Root-mean-square speed is another fundamental concept in the kinetic theory of gases. It measures how fast molecules move in a gas. It’s not just the average speed, but a type of mean that gives a more accurate energy depiction of gas molecules.

- The root-mean-square speed \( v_{rms} \) of molecules can be calculated using this formula:

- \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( k \) is Boltzmann’s constant, \( T \) is the temperature in Kelvin, and \( m \) is the molecular mass.
This formula suggests that as the temperature increases, so does \( v_{rms} \) since both are directly proportional. In our scenario, as the temperature jumps from 20°C to 250°C, with a constant molecular mass, the \( v_{rms} \) becomes higher, indicating that molecules are moving faster.
Gas Pressure
Gas pressure is an important factor in the kinetic theory. It arises from the force exerted by gas molecules colliding with the walls of their container. When we look at a scenario where temperature inside a fixed volume increases, the molecules not only hit the walls more frequently but do so with more force.

- This increase in both frequency and force of impact ups the gas pressure.
If we hold the volume constant, like in our exercise, the only factor changing the impact is the speed of the molecules, which increases with temperature. Hence, higher temperatures lead to higher pressure, as molecules are energized more, striking container walls both harder and faster.
Molecular Collisions
Molecular collisions in gas theory are all about how often and how strongly molecules hit the container walls. In this constant volume scenario, an increase in temperature directly affects these dynamics.

- Faster molecules don’t only collide more frequently with the walls, they also strike with enhanced power.
These changes lead to more impactful collisions per second. Simply put, as molecules move quicker due to the temperature rise, they return to collide with walls more quickly, speeding up their overall collision rate. This increasing collision rate enhances both the strength and number of impacts, showing a direct correlation with temperature—a key takeaway in understanding molecular behavior in gases.

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Most popular questions from this chapter

The physical fitness of athletes is measured by " \(V_{\mathrm{O}_{2}}\) max," which is the maximum volume of oxygen consumed by an individual during incremental exercise (for example, on a treadmill). An average male has a \(V_{\mathrm{O}_{2}}\) max of \(45 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass \(/ \mathrm{min}\), but a world-class male athlete can have a \(V_{\mathrm{O}_{2}}\) max reading of \(88.0 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass \(/ \mathrm{min.}\) (a) Calculate the volume of oxygen, in mL, consumed in 1 hr by an average man who weighs 185 lbs and has a \(V_{\mathrm{O}_{2}}\) max reading of 47.5 \(\mathrm{mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass \(/ \mathrm{min}\) (b) If this man lost \(20 \mathrm{lb}\), exercised, and increased his \(V_{\mathrm{O}_{2}}\) max to \(65.0 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass \(/ \mathrm{min}\), how many mL of oxygen would he consume in \(1 \mathrm{hr}\) ?

(a) What are the mole fractions of each component in a mixture of \(15.08 \mathrm{~g}\) of \(\mathrm{O}_{2}, 8.17 \mathrm{~g}\) of \(\mathrm{N}_{2}\), and \(2.64 \mathrm{~g}\) of \(\mathrm{H}_{2} ?\) (b) What is the partial pressure in atm of each component of this mixture if it is held in a 15.50-L vessel at \(15^{\circ} \mathrm{C}\) ?

Chlorine dioxide gas \(\left(\mathrm{ClO}_{2}\right)\) is used as a commercial bleaching agent. It bleaches materials by oxidizing them. In the course of these reactions, the \(\mathrm{ClO}_{2}\) is itself reduced. (a) What is the Lewis structure for \(\mathrm{ClO}_{2} ?\) (b) Why do you think that \(\mathrm{ClO}_{2}\) is reduced so readily? (c) When a \(\mathrm{ClO}_{2}\) molecule gains an electron, the chlorite ion, \(\mathrm{ClO}_{2}^{-}\), forms. Draw the Lewis structure for \(\mathrm{ClO}_{2}^{-}\). (d) Predict the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in the \(\mathrm{ClO}_{2}^{-}\) ion. (e) One method of preparing \(\mathrm{ClO}_{2}\) is by the reaction of chlorine and sodium chlorite: $$ \mathrm{Cl}_{2}(g)+2 \mathrm{NaClO}_{2}(s) \longrightarrow 2 \mathrm{ClO}_{2}(g)+2 \mathrm{NaCl}(s) $$ If you allow \(15.0 \mathrm{~g}\) of \(\mathrm{NaClO}_{2}\) to react with \(2.00 \mathrm{~L}\) of chlorine gas at a pressure of 1.50 atm at \(21^{\circ} \mathrm{C}\), how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

Complete the following table for an ideal gas: $$ \begin{array}{llll} \hline \boldsymbol{P} & \boldsymbol{v} & \boldsymbol{n} & \boldsymbol{T} \\ \hline 2.00 \mathrm{~atm} & 1.00 \mathrm{~L} & 0.500 \mathrm{~mol} & ? \mathrm{~K} \\ 0.300 \mathrm{~atm} & 0.250 \mathrm{~L} & ? \mathrm{~mol} & 27^{\circ} \mathrm{C} \\ 650 \text { torr } & ? \mathrm{~L} & 0.333 \mathrm{~mol} & 350 \mathrm{~K} \\ ? \mathrm{~atm} & 585 \mathrm{~mL} & 0.250 \mathrm{~mol} & 295 \mathrm{~K} \\ \hline \end{array} $$

What change or changes in the state of a gas bring about each of the following effects? (a) The number of impacts per unit time on a given container wall increases. (b) The average energy of impact of molecules with the wall of the container decreases. (c) The average distance between gas molecules increases. (d) The average speed of molecules in the gas mixture is increased.

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