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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