Kinetic Model and Perfect Gas Equation of State

Equation of State for a Perfect Gas

Goal

• To derive an equation of state for a collection of non-interacting gas particles, known as a perfect gas.

• An equation of state describes the relationship between the measurable properties of a system.

Measurable Properties of a Perfect Gas

• Volume ($V$)

  • Common units: Liters ($L$), cubic decimeters ($dm^3$), milliliters ($mL$), cubic centimeters ($cm^3$).

  • Examples: Volume of a box or a cylinder.

• Amount of Substance ($n$)

  • Common units: Moles ($mol$), millimoles ($mmol$).

• Temperature ($T$)

  • Common units: Kelvin ($K$), degrees Celsius ($^ ext{o}C$).

  • Definition: Temperature is a measure of the average molecular speeds of particles within the gas.

  • Thermal Equilibrium: This is achieved when the temperature of the system is equal to the temperature of the surroundings ($T<em>{ ext{system}} = T</em>{ ext{surroundings}}$). This implies no net energy transfer due to temperature differences.

• Pressure ($p$)

  • Dimensions: Force per unit Area ($rac{ ext{Force}}{ ext{Area}}$).

  • Common units: bar, atmospheres ($atm$), kilopascals ($kPa$).

  • Definition: Pressure is the force exerted by the collisions of gas particles with the container walls, divided by the area of those walls.

  • Mechanical Equilibrium: This is achieved when the pressure of the system is equal to the external pressure ($p<em>{ ext{system}} = p</em>{ ext{external}}$). This implies no net macroscopic force across the system boundary.

Variables of State for a Perfect Gas

• The four primary variables of state for a perfect gas are: pressure ($p$), volume ($V$), amount ($n$), and temperature ($T$).

• Equation of State: Expresses one variable as a function of the others, e.g., $p = f(V, n, T)$.

• If three variables of state are known, the fourth can be calculated using the equation of state.

Kinetic Model: Deriving the Equation of State

• The kinetic model aims to derive the equation of state for a perfect gas from fundamental mechanical laws and statistical principles.

Force Exerted by Gas Particles on the Container Wall

1. Newton's Second Law: Force ($F$) is the rate of change of momentum ($p<em>{ ext{momentum}}$). $F = rac{dp</em>{ ext{momentum}}}{dt}$ where $dp_{ ext{momentum}}$ represents the change in momentum over time $dt$.

2. Momentum Change per Collision: For a particle of mass $m$ and velocity $u<em>x$ colliding elastically with a wall perpendicular to the x-axis, its initial momentum is $mu</em>x$ and its final momentum is $-mu<em>x$. The change in momentum for one collision is $ext{Ap}</em>{ ext{collision}} = mu<em>x - (-mu</em>x) = 2mu_x$.

3. Number of Collisions: To find the total force, we need the total number of collisions with the wall in a given time interval $ext{Dt}$.

   • Consider particles within a small volume near the wall, defined by an area ($ext{Area}$) and a small distance $ext{Dx} = u<em>x ext{Dt}$ (where $u</em>x$ is the velocity component perpendicular to the wall).

   • The volume near the wall is $ext{Area} imes u_x ext{Dt}$.

   • The particle density is $rac{N}{V}$ (where $N$ is the total number of particles and $V$ is the total volume).

   • The number of particles in this volume is $rac{N}{V} imes ext{Area} imes u_x ext{Dt}$.

   • On average, half of these particles will be moving towards the wall and collide with it. So, the total number of collisions in time $ext{Dt}$ is:
     $ext{Number of collisions} = rac{1}{2} imes rac{N}{V} imes ext{Area} imes u_x ext{Dt}$.

4. Total Force on the Wall: The total force exerted is the total change in momentum divided by the time interval:
   $F = rac{ ext{Number of collisions} imes ext{Ap}<em>{ ext{collision}}}{ ext{Dt}} = rac{ rac{1}{2} imes rac{N}{V} imes ext{Area} imes u</em>x ext{Dt} imes 2mu<em>x}{ ext{Dt}} = rac{N m u</em>x^2 ext{Area}}{V}$.

Deriving Pressure and the Ideal Gas Law

1. Pressure: Pressure ($p$) is force per unit area:
   $p = rac{F}{ ext{Area}} = rac{N m u_x^2}{V}$.

2. Average Speeds: Gas particles do not all move with the same speed ($u<em>x$). Instead, there's a distribution of speeds (e.g., Maxwell-Boltzmann distribution). We must consider the average square speed, x^2>.

   • For gas particles moving randomly in three dimensions, the average square velocity components are equal: x^2> = y^2> = .

   • The total average square speed ($$) is = x^2> + y^2> + z^2> = 3x^2>.

   • Therefore, $= rac{1}{3}$.

3. Pressure with Average Square Speed: Substituting this into the pressure equation:
   $p = rac{N m }{V} = rac{N m }{3V}$.

4. Relating to Temperature: From the kinetic theory of gases, the average translational kinetic energy of a particle is directly proportional to temperature:
   $rac{1}{2}m = rac{3}{2}kT$
   where $k$ is the Boltzmann constant ($k = 1.3806 imes 10^{-23} ext{ J K}^{-1}$).
   Rearranging for $$: $= rac{3kT}{m}$.

5. Substituting into Pressure Equation:
   $p = rac{N m}{3V} imes rac{3kT}{m} = rac{NkT}{V}$.

6. Introducing Moles and the Gas Constant:

   • The total number of particles ($N$) can be expressed as the number of moles ($n$) multiplied by Avogadro's number ($NA$): $N = nNA$.

   • The universal gas constant ($R$) is defined as $R = k N_A$ ($R = 8.314 ext{ J mol}^{-1} ext{ K}^{-1}$).

   • Substituting these into the equation:
     $p = rac{nN_A kT}{V} = rac{nRT}{V}$.

7. Equation of State for a Perfect Gas (Ideal Gas Law):
   This rearranges to the familiar form:
   $pV = nRT$.

Boyle's Law

• When the amount of gas ($n$) and temperature ($T$) are kept constant, the product of pressure and volume is constant:
  $pV = ext{constant}$.

• This relationship can be expressed as $p1V1 = p2V2$ for two different states.

• This describes isotherms on a pressure-volume graph (hyperbolic curves at constant temperature).

Equipartition Theorem

• The Equipartition Theorem states that for each quadratic term that appears in the expression for the instantaneous energy ($E$) of a particle, there is an additional $rac{1}{2}kT$ of energy contributing to the internal energy ($U$) of the system.

Applications to Different Systems

• For 1 perfect gas particle in 1 Dimension (1D):

  • Instantaneous energy: $E = rac{1}{2}mu_x^2$ (one translational quadratic term).

  • Average energy: $= rac{1}{2}kT$.

• For 1 perfect gas particle in 3 Dimensions (3D):

  • Instantaneous energy: $E = rac{1}{2}mux^2 + rac{1}{2}muy^2 + rac{1}{2}mu_z^2$ (three translational quadratic terms).

  • Average energy: $= rac{3}{2}kT$.

• For $nN_A$ perfect gas particles in 3D (Total Internal Energy $U$):

  • Total instantaneous energy involves $3nN_A$ quadratic terms (for translational motion of all particles).

  • Total Average Energy: $U = nN_A imes rac{3}{2}kT = rac{3}{2}nRT$.

  • This $U$ is defined as the total internal energy of the system.

Internal Energy of Different Perfect Gases

• The internal energy ($U$) of a perfect gas depends on its molecular structure, specifically on the number of active quadratic terms (degrees of freedom).

• Perfect Gas of Atoms (Monatomic):

  • Energy only includes translational motion: $E = rac{1}{2}mux^2 + rac{1}{2}muy^2 + rac{1}{2}mu_z^2$ (3 quadratic terms).

  • Internal Energy: $U = nN_A imes rac{3}{2}kT = rac{3}{2}nRT$.

• Perfect Gas of Diatomics (e.g., $O_2$):

  • Energy includes translational and rotational motion (at typical temperatures; vibrational modes are often not active):
    $E = rac{1}{2}mux^2 + rac{1}{2}muy^2 + rac{1}{2}muz^2 + rac{1}{2}Ix ext{w}x^2 + rac{1}{2}Iy ext{w}_y^2$
    (3 translational + 2 rotational quadratic terms, totaling 5 quadratic terms).

  • Internal Energy: $U = nN_A imes rac{5}{2}kT = rac{5}{2}nRT$.

Molar Internal Energy ($U_m$)

• The molar internal energy of a perfect gas depends only on temperature ($T$).

  • For atomic (monatomic) perfect gases: $U_m = rac{U}{n} = rac{3}{2}RT$.

  • For diatomic perfect gases: $U_m = rac{U}{n} = rac{5}{2}RT$.

Example Calculation

• Problem: Calculate the change in internal energy ($ext{DU}$) when the temperature of $2.0 ext{ mol}$ of dilute $O_2$ gas is lowered from $25.0^ ext{o}C$ to $-40.0^ ext{o}C$.

• Given:

  • $n = 2.0 ext{ mol}$

  • $T_i = 25.0^ ext{o}C = 25.0 + 273.15 = 298.15 ext{ K}$

  • $T_f = -40.0^ ext{o}C = -40.0 + 273.15 = 233.15 ext{ K}$

  • $O_2$ is a diatomic gas, so $U = rac{5}{2}nRT$.

• Formula for Change in Internal Energy:
  $ext{DU} = Uf - Ui = rac{5}{2}nR(Tf - Ti)$

• Calculation:
  $ext{DU} = rac{5}{2} imes (2.0 ext{ mol}) imes (8.314 ext{ J mol}^{-1} ext{ K}^{-1}) imes (233.15 ext{ K} - 298.15 ext{ K})$
  $ext{DU} = rac{5}{2} imes (2.0 ext{ mol}) imes (8.314 ext{ J mol}^{-1} ext{ K}^{-1}) imes (-65.0 ext{ K})$
  $ext{DU} = -2702.05 ext{ J}$
  $ext{DU} ext{ ‰} -2.7 ext{ kJ}$