Specific Heat Capacities Of Kinetic Theory Of Gases

For an adiabatic process: $dQ = 0$ From first law: $dU + p\,dV = 0$ For ideal gas: $dU = nC_V\,dT$

Therefore: $nC_V\,dT + p\,dV = 0$

Using ideal gas law: $pV = nRT$ Differentiating: $p\,dV + V\,dp = nR\,dT$ So: $dT = \frac{p\,dV + V\,dp}{nR}$

Substituting: $C_V\left(\frac{p\,dV + V\,dp}{R}\right) + p\,dV = 0$ $(C_V + R)p\,dV + C_V V\,dp = 0$ $C_p p\,dV + C_V V\,dp = 0$

Rearranging: $\frac{C_p}{C_V}\frac{dV}{V} + \frac{dp}{p} = 0$ $\gamma\frac{dV}{V} + \frac{dp}{p} = 0$

Integrating: $\gamma\ln V + \ln p = \ln(\text{constant})$ $\ln(p V^\gamma) = \ln(\text{constant})$

Therefore: $pV^\gamma = \text{constant}$

For an ideal gas:

1. At constant volume: $(dQ)_V = dU$ (no work done)
2. At constant pressure: $(dQ)_p = dU + p\,dV$
3. For ideal gas: $p\,dV = nR\,dT$

From these equations: $(dQ)_p = (dQ)_V + nR\,dT$

Dividing by $n\,dT$: $\frac{1}{n}\frac{(dQ)_p}{dT} = \frac{1}{n}\frac{(dQ)_V}{dT} + R$

By definition: $C_p = C_V + R$

This shows the molar heat capacity at constant pressure exceeds the molar heat capacity at constant volume by the gas constant R.

When heat is supplied to a gas at constant volume:

• All heat goes into increasing internal energy (temperature rise)
• No work is done since volume remains fixed

When heat is supplied at constant pressure:

• Gas expands as temperature increases
• Part of the heat is used to do expansion work against external pressure
• Only the remaining heat increases internal energy

Therefore, more heat is required at constant pressure to achieve the same temperature increase as at constant volume, making $C_p > C_V$.

The quantitative difference for ideal gases is exactly the gas constant R: $C_p - C_V = R$

Starting with the adiabatic process equation: $pV^\gamma = \text{constant}$

Using the ideal gas law: $p = \frac{nRT}{V}$

Substituting: $\frac{nRT}{V}V^\gamma = \text{constant}$ $nRT \cdot V^{\gamma-1} = \text{constant}$

Since $n$ and $R$ are constants: $T \cdot V^{\gamma-1} = \text{constant}$

This relationship shows that:

• When a gas expands adiabatically ($V$ increases), temperature decreases
• When a gas is compressed adiabatically ($V$ decreases), temperature increases

The rate of temperature change depends on $\gamma$ (ratio of specific heats), which varies by gas type (monatomic: 1.67, diatomic: 1.4).

For a reversible adiabatic process: $TV^{\gamma-1} = \text{constant}$

This means: $T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}$

Solving for $T_2$: $T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}$

Alternatively, this can be written as: $T_2 = T_1\left(\frac{V_2}{V_1}\right)^{-(\gamma-1)}$

This equation reveals that:

• For expansion ($V_2 > V_1$), $T_2 < T_1$ (cooling)
• For compression ($V_2 < V_1$), $T_2 > T_1$ (heating)
• The temperature change is greater for gases with higher $\gamma$ values

Example: For a diatomic gas ($\gamma = 1.4$) expanding to twice its volume, $T_2 = T_1(0.5)^{0.4} = 0.76T_1$

Regnault's apparatus measures Cp through these key components and processes:

1. A large pressurized tank containing the experimental gas connects to two copper coils in series:

   • First coil in a hot oil bath (maintained at temperature θ₁)
   • Second coil in a calorimeter with water (initially at temperature θ₂)

2. Working principle:

   • Gas flows through the heated coil, then through the calorimeter coil
   • Gas transfers heat to the water, raising its temperature to θ₃
   • Flow rate is kept constant via a screw valve and monitored with manometers

3. Calculation method:

   • Cp = [(W + m)s(θ₃ - θ₂)]/[n(θ₁ - (θ₂+θ₃)/2)]
   • Where: W = water equivalent of calorimeter, m = mass of water, s = specific heat of water, n = moles of gas passed

4. The amount of gas (n) is determined by measuring the pressure difference in the tank before and after the experiment.

Regnault's apparatus fundamentally cannot measure Cv directly because:

1. Fundamental limitation: The apparatus relies on flowing gas, which requires changing volume and pressure, making constant volume conditions impossible.

2. Required modifications for Cv measurement:

   • Replace flow-based system with sealed chambers
   • Implement a differential calorimeter like Joly's device with two spheres
   • Add precise mechanisms to measure heat without volume change

3. Alternative approach:

   • First measure Cp using the existing setup
   • Measure the gas constant R independently
   • Calculate Cv using the relationship: Cp - Cv = R (for ideal gases)
   • Or determine γ = Cp/Cv through sound velocity measurements

4. Theoretical relationship:

   • For ideal gases: Cv = (1/n)(dU/dT), where internal energy changes without volume change
   • Cp = Cv + R because at constant pressure, additional energy goes into expansion work
   • Different experimental approaches are needed when volume must remain constant

Advantages of differential calorimeter method:

• Direct measurement of CV without conversion from CP
• Compensates for container heating by using identical spheres
• High precision due to differential measurement technique
• Works well for gases that are difficult to handle in flow systems

Sources of error:

• Slight expansion of metal spheres when heated (changes volume)
• Incomplete evacuation of reference sphere
• Uneven steam distribution around both spheres
• Water droplets forming on suspension wires (mitigated by heating coils)
• Imperfect thermal isolation from surroundings
• Non-ideal gas behavior at higher pressures

The method is particularly valuable for measuring CV of gases where P-V work would complicate other measurement approaches.

To determine γ = CP/CV from differential steam calorimeter data:

1. Calculate CV directly from the experiment using: CV = (M·m₂·L)/(m₁·(θ₂-θ₁)) where:

   • m₂ = extra mass of condensed steam
   • m₁ = mass of gas sample
   • M = molecular weight of gas
   • L = latent heat of vaporization
   • θ₁, θ₂ = initial and final temperatures

2. Calculate CP using the relation: CP - CV = R (gas constant)

   • CP = CV + R

3. Compute γ = CP/CV

4. For diatomic gases, expect γ ≈ 1.4 if only translational and rotational degrees of freedom are active

   • If vibrations are active, γ will be closer to 1.29
   • Deviation from expected values indicates additional energy storage modes

5. Use the result to verify the equipartition theorem prediction for the gas's molecular structure

Equipartition Theorem: Each degree of freedom in a molecular system in thermal equilibrium has an average energy of $\frac{1}{2}kT$.

Heat Capacity Predictions by Gas Type:

1. Monatomic gases (He, Ne, Ar):

   • 3 translational degrees of freedom → $E = \frac{3}{2}kT$ per molecule
   • $C_V = \frac{3}{2}R$, $C_p = \frac{5}{2}R$, $\gamma = \frac{5}{3} = 1.67$

2. Diatomic gases without vibration (H₂, O₂, N₂ at room temp):

   • 3 translational + 2 rotational degrees → $E = \frac{5}{2}kT$ per molecule
   • $C_V = \frac{5}{2}R$, $C_p = \frac{7}{2}R$, $\gamma = \frac{7}{5} = 1.40$

3. Diatomic gases with vibration (at high temperatures):

   • 3 translational + 2 rotational + 2 vibrational degrees → $E = \frac{7}{2}kT$ per molecule
   • $C_V = \frac{7}{2}R$, $C_p = \frac{9}{2}R$, $\gamma = \frac{9}{7} = 1.29$

Key Assumptions and Principles:

• Classical behavior (no quantum effects)
• System in thermal equilibrium
• Energy equally distributed among all degrees of freedom
• Molar internal energy = (degrees of freedom) × $\frac{1}{2}RT$
• $C_p = C_V + R$ (always true for ideal gases)

Limitations and Breakdowns:

1. Quantum effects at low temperatures "freeze out" certain modes (especially vibrations)
2. Temperature dependence contradicts prediction of constant heat capacities
3. Vibrational modes may not equally distribute energy between kinetic and potential components
4. Intermolecular forces store energy not accounted for by the theorem
5. Non-equilibrium conditions invalidate equal energy distribution assumption

Note: Experimental measurements match predictions well within appropriate temperature ranges but deviate significantly when quantum effects become dominant.