a) Write the assumptions of kinetic theory of gases. Derive the following expression of the pressure exerted by an ideal gas:
p=(1)/(3)mn bar(v)^(2)p=\frac{1}{3} m n \bar{v}^2
Also, using this expression deduce Avogadro’s law. What is the kinetic interpretation temperature?
Answer:
The kinetic theory of gases is a fundamental theory that explains the behavior of gases and the properties of gases at the molecular level. It is based on several key assumptions that simplify the complex interactions of gas particles, allowing for the derivation of important gas laws and principles. Here are the assumptions of the kinetic theory of gases:
Gas consists of a large number of small particles (atoms or molecules), which are in constant, random motion.
The volume of the individual particles is negligible compared to the total volume of the gas. This means that the actual volume of the gas molecules is so small relative to the space between them that it can be assumed to be zero.
There are no forces of attraction or repulsion between the particles of the gas. This assumption implies that the gas particles do not interact with each other except during collisions.
The collisions between gas particles and between particles and the walls of the container are perfectly elastic. This means that there is no net loss of kinetic energy in the system of particles after the collisions.
The average kinetic energy of the gas particles is directly proportional to the absolute temperature of the gas. This implies that as the temperature increases, the average kinetic energy of the gas particles increases, and vice versa.
The gas particles obey Newton’s laws of motion, and their motion can be described by the laws of mechanics (classical mechanics).
The time of collision between the particles is much shorter than the time between successive collisions, meaning that the actual duration of a collision is negligible compared to the time particles spend flying freely.
These assumptions allow for the derivation of the ideal gas law and the explanation of gas properties such as pressure, temperature, and volume. It’s important to note that these assumptions are idealizations; real gases may deviate from these assumptions under certain conditions, such as high pressure or low temperature.
To derive the expression for the pressure exerted by an ideal gas, p=(1)/(3)mn bar(v)^(2)p = \frac{1}{3} m n \bar{v}^2, we’ll follow the kinetic theory of gases and its assumptions. Here, pp represents the pressure of the gas, mm is the mass of a single molecule, nn is the number density of the gas molecules (number of molecules per unit volume), and bar(v)^(2)\bar{v}^2 is the mean square velocity of the gas molecules.
Step 1: Consider a Gas in a Cubic Container
Imagine a cubic container of volume VV with sides of length LL. Gas molecules are moving randomly in all directions. We’ll focus on the molecules moving perpendicular to one of the walls of the cube.
Step 2: Calculate the Change in Momentum for a Single Molecule
When a molecule with velocity v_(x)v_x in the x-direction collides elastically with a wall, it rebounds with the same speed but in the opposite direction. The change in momentum (Delta p\Delta p) for this molecule during a collision with the wall is:
Delta p=2mv_(x)\Delta p = 2mv_x
Step 3: Determine the Rate of Collisions Against a Wall
The time (tt) it takes for a molecule to hit the wall and return to the same spot (after hitting the opposite wall and coming back) is (2L)/(v_(x))\frac{2L}{v_x}. The rate of collisions of one molecule against a wall is the inverse of this time:
“Rate of collisions”=(1)/(t)=(v_(x))/(2L)\text{Rate of collisions} = \frac{1}{t} = \frac{v_x}{2L}
Step 4: Calculate the Force Exerted by a Single Molecule
The force (FF) exerted by a single molecule on the wall is the rate of change of momentum:
Considering all molecules and averaging over all directions (since the container is cubic and the gas molecules’ movements are isotropic), the average force due to all molecules is proportional to the mean square velocity (bar(v)^(2)\bar{v}^2) and the number of molecules (NN). The total force (F_(“total”)F_{\text{total}}) exerted by NN molecules is:
F_(“total”)=N*(m bar(v)^(2))/(3L)F_{\text{total}} = N \cdot \frac{m\bar{v}^2}{3L}
Here, we divide by 3 to account for the three-dimensional motion of the molecules (x, y, and z directions).
Step 6: Relate Force to Pressure
Pressure (pp) is defined as force per unit area. For one face of the cube, the area (AA) is L^(2)L^2, so:
Since L^(3)L^3 is the volume (VV) of the container and n=(N)/(V)n = \frac{N}{V} is the number density of the molecules, we can rewrite the pressure as:
p=(1)/(3)mn bar(v)^(2)p = \frac{1}{3} m n \bar{v}^2
This derivation shows how the pressure exerted by an ideal gas can be related to the mass of the molecules, their mean square velocity, and their number density, under the assumptions of the kinetic theory of gases.
Deduction of Avogadro’s Law
Avogadro’s law states that equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules. From the derived pressure formula, we see that the pressure is directly proportional to the number of molecules (nn) and their average kinetic energy ((1)/(2)m bar(v)^(2)\frac{1}{2}m\bar{v}^2), which is a measure of temperature.
For a given pp and TT (temperature), if two gases have the same volume VV, the number of molecules nn must be the same, according to the formula, regardless of the type of gas. This is because the kinetic energy component ((1)/(2)m bar(v)^(2)\frac{1}{2}m\bar{v}^2) reflects the temperature, which is the same for both gases. Thus, Avogadro’s law is a direct consequence of the kinetic theory of gases.
Kinetic Interpretation of Temperature
The kinetic interpretation of temperature is that it is a measure of the average kinetic energy of the molecules in a substance. From the kinetic theory, the average kinetic energy of a molecule is given by:
And since the average kinetic energy is directly proportional to the absolute temperature (TT) of the gas, we can write:
“Average kinetic energy”prop T\text{Average kinetic energy} \propto T
This relationship indicates that the temperature of a gas is a measure of the average kinetic energy of its particles. Higher temperatures correspond to higher average kinetic energies and faster-moving particles.
b) The expression for the number of molecules in a Maxwellian gas having speeds in the range vv to v+dvv+d v is given by
dN_(v)=4pi N((m)/(2pik_(B)T))^(3//2)v^(2)exp[-((mv^(2))/(2k_(B)T))]dvd N_v=4 \pi N\left(\frac{m}{2 \pi k_{\mathrm{B}} T}\right)^{3 / 2} v^2 \exp \left[-\left(\frac{m v^2}{2 k_{\mathrm{B}} T}\right)\right] d v
Obtain an expression of average speed and root mean square speed of a molecule.
Answer:
To find the average speed and root mean square (rms) speed of a molecule in a Maxwellian gas, we’ll use the distribution function given:
dN_(v)=4pi N((m)/(2pik_(B)T))^(3//2)v^(2)exp[-((mv^(2))/(2k_(B)T))]dvd N_v = 4 \pi N \left(\frac{m}{2 \pi k_B T}\right)^{3/2} v^2 \exp\left[-\left(\frac{m v^2}{2 k_B T}\right)\right] dv
where:
NN is the total number of molecules,
mm is the mass of a molecule,
k_(B)k_B is the Boltzmann constant,
TT is the temperature in Kelvin,
vv is the speed of a molecule.
Average Speed (bar(v)\bar{v})
The average speed is given by the first moment of the speed distribution:
bar(v)=(1)/(N)int_(0)^(oo)vdN_(v)\bar{v} = \frac{1}{N} \int_{0}^{\infty} v \, d N_v
Substituting the expression for dN_(v)d N_v:
bar(v)=(1)/(N)int_(0)^(oo)4pi N((m)/(2pik_(B)T))^(3//2)v^(3)exp[-((mv^(2))/(2k_(B)T))]dv\bar{v} = \frac{1}{N} \int_{0}^{\infty} 4 \pi N \left(\frac{m}{2 \pi k_B T}\right)^{3/2} v^3 \exp\left[-\left(\frac{m v^2}{2 k_B T}\right)\right] dv
So, the average speed and root mean square speed of a molecule in a Maxwellian gas are given by sqrt((8k_(B)T)/(pi m))\sqrt{\frac{8 k_B T}{\pi m}} and sqrt((3k_(B)T)/(m))\sqrt{\frac{3 k_B T}{m}}, respectively.