Physics Thermal Physics 100% Free Open Access
Chapter 1 • Theory & Derivations

Zeroth and First Law of Thermodynamics

Foundations of macroscopic thermodynamics: thermodynamic equilibrium, empirical temperature, state functions, work and heat energy, conservation of energy, Mayer's relation, and atmospheric adiabatic lapse rates.

1.1Thermodynamic Equilibrium & State Variables

1. The Nature of Thermodynamic Systems

A thermodynamic system is defined as any macroscopic quantity of matter or radiation separated from the rest of the universe (the surroundings) by an identifiable real or hypothetical boundary. Thermodynamic systems are classified according to the permeability of their boundaries:

1. Isolated Systems: Exchange neither matter nor energy (heat or work) with surroundings (e.g., matter enclosed within an idealized rigid, non-radiating adiabatic vacuum chamber).

2. Closed Systems: Exchange energy with surroundings via mechanical work or heat conduction, but have boundaries impermeable to mass ($dN = 0$).

3. Open Systems: Exchange both energy and matter with their environment across permeable control surfaces.

2. Multi-Criteria Thermodynamic Equilibrium

For a macroscopic system to be in genuine thermodynamic equilibrium, three independent physical conditions must be satisfied simultaneously:

  • Thermal Equilibrium: There exists no net heat exchange between any sub-regions within the system or across its boundary, implying uniform empirical temperature throughout ($T_A = T_B = \dots = T$).
  • Mechanical Equilibrium: No unbalanced macroscopic forces exist within the system or at the boundary, ensuring uniform hydrostatic pressure ($P_A = P_B = \dots = P$) in the absence of external fields, or hydrostatic balance $

abla P = ho \vec{g}$ under gravitational fields.

  • Chemical Equilibrium: No spontaneous chemical reactions or net diffusive species transport occur between phases, ensuring uniform chemical potentials ($\mu_i^{(A)} = \mu_i^{(B)} = \dots = \mu_i$) for every species $i$.

When all three conditions hold, the system's state remains invariant over time in the absence of external perturbations.

3. Intensive vs. Extensive State Coordinates

Thermodynamic state variables are categorized by their scaling behavior under spatial partitioning or scaling of total mass by a dimensionless scalar $\lambda$:

  • Extensive Variables: Scale linearly with system size such that $X(\lambda N, \lambda V) = \lambda X(N, V)$. Examples include volume $V$, internal energy $U$, enthalpy $H$, entropy $S$, Helmholtz free energy $F$, and Gibbs free energy $G$.
  • Intensive Variables: Invariant under system scaling such that $Y(\lambda N, \lambda V) = Y(N, V)$. Examples include temperature $T$, hydrostatic pressure $P$, chemical potential $\mu$, density $ ho = M/V$, and molar volumes $V_m = V/n$.

By Euler's theorem for homogeneous functions of degree 1, any fundamental extensive thermodynamic function, say $U(S, V, N_i)$, satisfies the exact relation:

$$U = S \left(\frac{\partial U}{\partial S}\right)_{V, N} + V \left(\frac{\partial U}{\partial V}\right)_{S, N} + \sum_i N_i \left(\frac{\partial U}{\partial N_i}\right)_{S, V} = T S - P V + \sum_i \mu_i N_i$$

1.2Zeroth Law of Thermodynamics & Empirical Temperature

1. The Postulate of Thermal Transitivity

The Zeroth Law of Thermodynamics formalizes the operational and mathematical basis for the existence of temperature. Historically recognized after the First and Second Laws, Ralph H. Fowler in 1935 noted its logical priority:

$$\text{If system } A \text{ is in thermal equilibrium with system } C, \text{ and system } B \text{ is independently in thermal equilibrium with } C, \text{ then system } A \text{ is in thermal equilibrium with } B.$$

Mathematically, let the thermodynamic state of system $A$ be specified by coordinates $(P_A, V_A)$, system $B$ by $(P_B, V_B)$, and thermometer system $C$ by $(P_C, V_C)$. Thermal equilibrium between $A$ and $C$ establishes an implicit functional constraint:

$$f_{AC}(P_A, V_A; P_C, V_C) = 0$$

Solving for $P_C$:

$$P_C = \phi_A(P_A, V_A; V_C)$$

Similarly, thermal equilibrium between $B$ and $C$ requires:

$$P_C = \phi_B(P_B, V_B; V_C)$$

Equating both expressions yields:

$$\phi_A(P_A, V_A; V_C) = \phi_B(P_B, V_B; V_C)$$

The Zeroth Law asserts that thermal equilibrium between $A$ and $B$ depends strictly on coordinates $(P_A, V_A)$ and $(P_B, V_B)$ without reference to the arbitrary coordinates $V_C$ of the intermediary. Hence, the parameter $V_C$ must factor out algebraically:

$$\theta_A(P_A, V_A) = \theta_B(P_B, V_B)$$

2. Definition of Empirical Temperature

The scalar function $\theta(P, V)$ is called the empirical temperature. Thermal equilibrium is an equivalence relation possessing:

1. Reflexivity: $A \sim A$.

2. Symmetry: $A \sim B \implies B \sim A$.

3. Transitivity: $A \sim C \land B \sim C \implies A \sim B$.

This partitions all thermodynamic equilibrium states into disjoint equivalence classes (isotherms). A thermometer is any physical system possessing a measurable thermometric property $X$ (e.g., mercury column height, electrical resistance of platinum, thermoelectric EMF, gas pressure at constant volume) that monotonically tracks empirical temperature $\theta = a X$.

1.3Work, Heat Energy & Exact vs Inexact Differentials

1. Macroscopic Work Interactions

Thermodynamic work represents energy transfer driven by a generalized force operating through a conjugate generalized displacement. Unlike mechanical work on a point mass, thermodynamic work alters the external configuration or boundary of a macroscopic ensemble:

  • Hydrostatic Boundary Work: When a fluid exerts normal pressure $P$ on a movable piston of area $A$, an infinitesimal displacement $dx$ produces volume change $dV = A dx$. The work performed by the system is:

$$\delta W = P dV$$

  • Surface Film Work: For a 2D interface with surface tension $\gamma$, expanding interfacial area by $dA$ requires work:

$$\delta W = -\gamma dA$$

  • Magnetic Work: In a paramagnetic medium subjected to external magnetic field $\vec{H}$, altering total magnetization $\vec{M}$ performs work:

$$\delta W = -\mu_0 \vec{H} \cdot d\vec{M}$$

2. Inexact Differentials and Path Dependence

Work $\delta W$ and heat $\delta Q$ are path functions, not state functions. They describe energy in transit across boundaries during a process, rather than properties stored within a state.

Mathematically, let a differential form in two independent state variables $(x, y)$ be written as:

$$\delta Z = M(x, y) dx + N(x, y) dy$$

By Euler's criterion, $\delta Z$ is an exact differential ($dZ$) if and only if:

$$\left(\frac{\partial M}{\partial y}\right)_x = \left(\frac{\partial N}{\partial x}\right)_y$$

When exact, the line integral between state 1 and state 2 is independent of path:

$$\int_{\text{Path } A}^{(1) \to (2)} dZ = \int_{\text{Path } B}^{(1) \to (2)} dZ = Z_2 - Z_1, \quad \oint dZ = 0$$

For boundary work $\delta W = P dV + 0 dP$, we have $M(V, P) = P$ and $N(V, P) = 0$. The cross-derivatives give:

$$\left(\frac{\partial M}{\partial P}\right)_V = \frac{\partial P}{\partial P} = 1, \quad \left(\frac{\partial N}{\partial V}\right)_P = 0 \implies 1 \neq 0$$

Hence $\delta W$ is an inexact differential, denoted by $\delta W$ or $\mathrm{d}\\!\bar{\;\,}W$. Consequently:

$$W_{1 \to 2} = \int_{\text{Path}} P dV \neq W_2 - W_1$$

The integral equals the area under the process curve on a $P$-$V$ indicator diagram, which explicitly depends on the transformation trajectory.

1.4First Law of Thermodynamics & Internal Energy

1. Conservation of Energy in Macroscopic Systems

The First Law of Thermodynamics is the universal law of conservation of energy extended to incorporate thermal phenomena. In any thermodynamic transformation between equilibrium states (1) and (2), the individual quantities of heat absorbed $Q$ and work performed $W$ depend heavily on the specific path. However, experimental investigations by Joule (1843–1850) established that their algebraic difference $(Q - W)$ is strictly identical for every conceivable path connecting the two states:

$$\Delta U \equiv U_2 - U_1 = Q - W$$

In infinitesimal differential form:

$$dU = \delta Q - \delta W = \delta Q - P dV$$

Because $dU$ is an exact differential, the Internal Energy $U$ is a true thermodynamic state function.

2. Microscopic Nature of Internal Energy

Microscopically, internal energy $U$ represents the total microscopic kinetic and potential energies of all constituent particles within the rest frame of the center of mass:

$$U = \sum_{i=1}^N \frac{\vec{p}_i^2}{2m} + \sum_{i < j} V(\vec{r}_i - \vec{r}_j) + \sum_{i=1}^N \left( E_{\text{rot}, i} + E_{\text{vib}, i} + E_{\text{electronic}, i} \right)$$

For a monoatomic ideal gas, intermolecular potential energies are zero ($V(r) \equiv 0$), and translational kinetic energy dominates:

$$U = \frac{3}{2} N k_B T = \frac{3}{2} n R T$$

3. Joule's Free Expansion Experiment

In 1845, James Prescott Joule tested whether internal energy depends on volume at constant temperature. Two copper vessels—one containing gas at high pressure ($P_1$) and the other evacuated—were immersed in a thermally insulated water calorimeter and connected by a stopcock. When the stopcock was opened, gas expanded freely into the vacuum without moving any external boundary:

  • $W = 0$ (no external resistance against expansion into vacuum).
  • $Q = 0$ (no net heat transfer observed from calorimeter water).
  • By the First Law: $\Delta U = Q - W = 0$.

Joule detected no measurable change in calorimeter water temperature ($\Delta T = 0$). Hence, for an ideal gas:

$$\left(\frac{\partial U}{\partial V}\right)_T = 0 \implies U = U(T)$$

Internal energy of an ideal gas depends solely on absolute temperature, fundamentally independent of volume or pressure.

1.5Thermodynamic Processes & Work Calculations

1. Reversible Isochoric Process ($V = \text{constant}$)

In a rigid container, volume remains fixed ($dV = 0$):

  • Work performed: $W = \int P dV = 0$.
  • Heat added: $Q = \Delta U = \int_{T_1}^{T_2} C_v dT$.
  • Specific heat at constant volume: $C_v \equiv \left(\frac{\partial U}{\partial T}\right)_V$.

2. Reversible Isobaric Process ($P = \text{constant}$)

In a cylinder with freely moving weighted piston:

  • Work performed: $W = \int_{V_1}^{V_2} P dV = P (V_2 - V_1) = n R (T_2 - T_1)$.
  • Heat added: $Q = \Delta U + W = (U_2 - U_1) + P(V_2 - V_1) = (U_2 + P V_2) - (U_1 + P V_1) = H_2 - H_1 = \Delta H$.
  • The Enthalpy $H \equiv U + P V$ serves as the heat content state function under constant pressure conditions.
  • Specific heat at constant pressure: $C_p \equiv \left(\frac{\partial H}{\partial T}\right)_P$.

3. Reversible Isothermal Process ($T = \text{constant}$)

For an ideal gas at constant temperature $T$:

  • Internal energy change: $\Delta U = 0$ (since $U = U(T)$).
  • Heat and work equality: $Q = W$.
  • Work integration from ideal gas equation $P = nRT / V$:

$$W = \int_{V_1}^{V_2} \frac{n R T}{V} dV = n R T \ln\left(\frac{V_2}{V_1}\right) = n R T \ln\left(\frac{P_1}{P_2}\right)$$

4. Reversible Adiabatic Process ($Q = 0$)

When thermally insulated from surroundings ($\delta Q = 0$):

  • First law: $dU = -\delta W \implies C_v dT = -P dV$.
  • Substituting $P = nRT / V$:

$$C_v dT = -\frac{n R T}{V} dV \implies \frac{dT}{T} + \frac{R}{C_v} \frac{dV}{V} = 0$$

Recalling Mayer's relation $R = C_p - C_v$ and the adiabatic index $\gamma = C_p / C_v$, the ratio $R / C_v = \gamma - 1$:

$$\ln T + (\gamma - 1) \ln V = \text{const} \implies T V^{\gamma - 1} = \text{constant}$$

Using $T = P V / nR$:

$$P V^\gamma = \text{constant}, \quad T^\gamma P^{1 - \gamma} = \text{constant}$$

  • Adiabatic Work Integration:

$$W = -\Delta U = -C_v (T_2 - T_1) = \frac{n R (T_1 - T_2)}{\gamma - 1} = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$$

1.6Specific Heats, Compressibility & Thermal Expansion

1. General Thermodynamic Relation Between $C_p$ and $C_v$

Let internal energy be expressed as a function of temperature and volume, $U = U(T, V)$:

$$dU = \left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV = C_v dT + \left(\frac{\partial U}{\partial V}\right)_T dV$$

Substituting into the First Law $\delta Q = dU + P dV$:

$$\delta Q = C_v dT + \left[ P + \left(\frac{\partial U}{\partial V}\right)_T \right] dV$$

Dividing by $dT$ at constant pressure $P$:

$$C_p = \left(\frac{\delta Q}{dT}\right)_P = C_v + \left[ P + \left(\frac{\partial U}{\partial V}\right)_T \right] \left(\frac{\partial V}{\partial T}\right)_P$$

Therefore, the universal relation between heat capacities is:

$$C_p - C_v = \left[ P + \left(\frac{\partial U}{\partial V}\right)_T \right] \left(\frac{\partial V}{\partial T}\right)_P$$

For an ideal gas, Joule's experiment proves $(\partial U/\partial V)_T = 0$, and from $P V = n R T$, $(\partial V/\partial T)_P = n R / P$:

$$C_p - C_v = P \cdot \left(\frac{n R}{P}\right) = n R \quad \implies \quad C_{p, m} - C_{v, m} = R$$

This is Mayer's classical relation.

2. Response Coefficients: Compressibility and Expansion

To express thermodynamic relations in terms of directly measurable material properties, we define three response coefficients:

1. Isobaric Thermal Expansion Coefficient $\alpha$:

$$\alpha \equiv \frac{1}{V} \left(\frac{\partial V}{\partial T}\right)_P$$

2. Isothermal Compressibility $\kappa_T$:

$$\kappa_T \equiv -\frac{1}{V} \left(\frac{\partial V}{\partial P}\right)_T$$

3. Isochoric Pressure Coefficient $\beta$:

$$\beta \equiv \frac{1}{P} \left(\frac{\partial P}{\partial T}\right)_V$$

By the cyclic triple product identity for $(P, V, T)$:

$$\left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial T}{\partial V}\right)_P \left(\frac{\partial V}{\partial P}\right)_T = -1 \implies \left(\frac{\partial P}{\partial T}\right)_V = -\frac{(\partial V/\partial T)_P}{(\partial V/\partial P)_T} = \frac{\alpha V}{\kappa_T V} = \frac{\alpha}{\kappa_T}$$

Using Maxwell's relation $(\partial U/\partial V)_T = T (\partial P/\partial T)_V - P$, the bracketed term in $C_p - C_v$ becomes:

$$P + \left(\frac{\partial U}{\partial V}\right)_T = T \left(\frac{\partial P}{\partial T}\right)_V = T \frac{\alpha}{\kappa_T}$$

Multiplying by $(\partial V/\partial T)_P = V \alpha$:

$$C_p - C_v = T V \frac{\alpha^2}{\kappa_T}$$

This rigorous formula holds for any substance in any phase (solids, liquids, and real gases). Because absolute temperature $T > 0$, volume $V > 0$, and mechanical stability requires $\kappa_T > 0$, it follows that $C_p \ge C_v$ always, with $C_p = C_v$ occurring only at $T = 0\text{ K}$ or where $\alpha = 0$ (such as liquid water at $3.98^{\circ}\text{C}$).

1.7Atmospheric Thermodynamics & Adiabatic Lapse Rate

1. Hydrostatic Equation of the Atmosphere

Consider a column of dry atmospheric air under gravity. For a horizontal air parcel of cross-sectional area $A$ and thickness $dz$, mechanical balance between upward pressure force, downward pressure force, and parcel weight gives:

$$P A - (P + dP) A = \rho A g dz \implies dP = -\rho g dz$$

where $\rho$ is local air density and $g$ is gravitational acceleration. Using the ideal gas equation of state $\rho = \frac{P M}{R T}$ where $M$ is the effective molar mass of dry air ($M \approx 28.97\text{ g/mol}$):

$$\frac{dP}{P} = -\frac{M g}{R T(z)} dz$$

2. Derivation of the Dry Adiabatic Lapse Rate

When an air parcel rises rapidly through the troposphere, thermal conduction across its boundary is negligible compared to convective transit times; the parcel expands adiabatically ($Q = 0$). For an adiabatic parcel:

$$T^\gamma P^{1-\gamma} = \text{const} \implies \gamma \frac{dT}{T} + (1-\gamma) \frac{dP}{P} = 0$$

Solving for $dP/P$:

$$\frac{dP}{P} = \frac{\gamma}{\gamma - 1} \frac{dT}{T}$$

Equating this with the hydrostatic relation $\frac{dP}{P} = -\frac{M g}{R T} dz$:

$$\frac{\gamma}{\gamma - 1} \frac{dT}{T} = -\frac{M g}{R T} dz \implies \frac{dT}{dz} = -\frac{\gamma - 1}{\gamma} \frac{M g}{R}$$

Recalling that $C_{p, m} = \frac{\gamma R}{\gamma - 1}$, the specific heat capacity per unit mass is $c_p = \frac{C_{p, m}}{M} = \frac{\gamma R}{M (\gamma - 1)}$. Thus:

$$\Gamma_{\text{dry}} \equiv -\frac{dT}{dz} = \frac{g}{c_p}$$

For Earth's atmosphere, $g \approx 9.807\text{ m/s}^2$ and dry air specific heat $c_p \approx 1005\text{ J/(kg}\cdot\text{K)}$:

$$\Gamma_{\text{dry}} = \frac{9.807}{1005} \approx 0.00976\text{ K/m} \approx 9.76\text{ K/km} \approx 9.8^{\circ}\text{C/km}$$

3. Atmospheric Stability Criteria

The environmental lapse rate $\Gamma_{\text{env}} = -\frac{dT_{\text{env}}}{dz}$ determines atmospheric convective stability:

  • Unstable Atmosphere ($\Gamma_{\text{env}} > \Gamma_{\text{dry}}$): A displaced air parcel becomes warmer and less dense than surrounding ambient air, experiencing positive buoyant acceleration, generating strong convective storms and cumulonimbus clouds.
  • Neutral Atmosphere ($\Gamma_{\text{env}} = \Gamma_{\text{dry}}$): Displaced parcel remains at ambient density.
  • Stable Atmosphere ($\Gamma_{\text{env}} < \Gamma_{\text{dry}}$): Parcel becomes colder and denser than surrounding air, experiencing restoring buoyant forces that suppress vertical air motion (temperature inversions and smog trapping).

Standard University Exam Examination Problems

Rigorous Analytical & Numerical Solved Problems

Comprehensive step-by-step mathematical proofs, dimensional evaluations, and calculations matching B.Sc. Honors university examinations.

StandardExample 1.1: Polytropic Process Gas Work, Internal Energy & Heat Capacity

Two moles of an ideal diatomic gas ($\\gamma = 1.40$) undergo a reversible polytropic expansion according to the relation $P V^{1.25} = \\text{constant}$, expanding from an initial pressure of $P_1 = 4.00\\times 10^5\\text{ Pa}$ and volume $V_1 = 0.020\\text{ m}^3$ to a final volume of $V_2 = 0.060\\text{ m}^3$. Calculate: (a) the final pressure $P_2$, (b) the total work done $W$, (c) the change in internal energy $\\Delta U$, and (d) the molar heat capacity $C$ for this specific polytropic trajectory.

Step 1: Determine Final Pressure $P_2$

For a polytropic process $P V^n = \\text{const}$ with $n = 1.25$:

$$P_2 = P_1 \\left(\\frac{V_1}{V_2}\\right)^n = (4.00 \\times 10^5\\text{ Pa}) \\left(\\frac{0.020}{0.060}\\right)^{1.25} = \\frac{4.00 \\times 10^5}{3^{1.25}} = \\frac{4.00 \\times 10^5}{3.9482} = 1.0131 \\times 10^5\\text{ Pa}$$

This confirms the gas expands to nearly standard atmospheric pressure.

Step 2: Calculate Boundary Work Done $W$

The boundary work performed during a polytropic path is:

$$W = \\int_{V_1}^{V_2} P dV = \\frac{P_1 V_1 - P_2 V_2}{n - 1}$$

Evaluating the pressure-volume products:

$$P_1 V_1 = (4.00 \\times 10^5)(0.020) = 8000\\text{ J}$$
$$P_2 V_2 = (1.0131 \\times 10^5)(0.060) = 6078.7\\text{ J}$$
$$W = \\frac{8000 - 6078.7}{1.25 - 1} = \\frac{1921.3}{0.25} = +7685.2\\text{ J}$$

Step 3: Calculate Change in Internal Energy $\\Delta U$

Using the ideal gas relation $\\Delta U = n C_v \\Delta T = \\frac{P_2 V_2 - P_1 V_1}{\\gamma - 1}$:

$$\\Delta U = \\frac{6078.7 - 8000}{1.40 - 1} = \\frac{-1921.3}{0.40} = -4803.3\\text{ J}$$

The gas internal energy decreases by $4.80\\text{ kJ}$ due to thermal cooling during expansion.

Step 4: Determine Net Heat Absorbed $Q$ and Polytropic Heat Capacity $C_m$

By the First Law of Thermodynamics:

$$Q = \\Delta U + W = -4803.3 + 7685.2 = +2881.9\\text{ J}$$

The molar polytropic heat capacity is derived via $C_m = C_{v, m} + \\frac{R}{1 - n}$:

$$C_{v, m} = \\frac{R}{\\gamma - 1} = \\frac{8.314}{0.40} = 20.785\\text{ J/(mol}\\cdot\\text{K)}$$
$$\\frac{R}{1 - n} = \\frac{8.314}{1 - 1.25} = \\frac{8.314}{-0.25} = -33.256\\text{ J/(mol}\\cdot\\text{K)}$$
$$C_m = 20.785 - 33.256 = -12.47\\text{ J/(mol}\\cdot\\text{K)}$$

Notice that the polytropic molar heat capacity is negative: as the gas absorbs heat, its temperature decreases because the work done on the surroundings exceeds the heat supplied.

Final Answer & Physical Insight

Final pressure $P_2 = 1.01\\times 10^5\\text{ Pa}$, work done $W = +7.69\\text{ kJ}$, internal energy change $\\Delta U = -4.80\\text{ kJ}$, heat absorbed $Q = +2.88\\text{ kJ}$, and molar heat capacity $C_m = -12.47\\text{ J/(mol}\\cdot\\text{K)}$.

StandardExample 1.2: Exact Calculation of $C_p - C_v$ for a Van der Waals Gas

Starting from the fundamental thermodynamic relation $C_p - C_v = \\left[ P + \\left(\\frac{\\partial U}{\\partial V}\\right)_T \\right] \\left(\\frac{\\partial V}{\\partial T}\\right)_P$, derive an explicit formula for $C_p - C_v$ for one mole of a real gas obeying the Van der Waals equation of state $\\left(P + \\frac{a}{V_m^2}\\right)(V_m - b) = R T$, and evaluate the percentage deviation from Mayer's ideal relation ($R$) for carbon dioxide at $T = 300\\text{ K}$ and $V_m = 1.00\\times 10^{-3}\\text{ m}^3\\text{/mol}$, given $a = 0.364\\text{ J}\\cdot\\text{m}^3\\text{/mol}^2$ and $b = 4.27\\times 10^{-5}\\text{ m}^3\\text{/mol}$.

Step 1: Evaluate the Internal Pressure Term $\\left(\\frac{\\partial U}{\\partial V}\\right)_T$

From thermodynamic energy equations (derived from Maxwell's relations):

$$\\left(\\frac{\\partial U}{\\partial V_m}\\right)_T = T \\left(\\frac{\\partial P}{\\partial T}\\right)_{V_m} - P$$

For a Van der Waals gas, $P = \\frac{R T}{V_m - b} - \\frac{a}{V_m^2}$. Taking the derivative with respect to $T$ at constant $V_m$:

$$\\left(\\frac{\\partial P}{\\partial T}\\right)_{V_m} = \\frac{R}{V_m - b}$$

Substituting back:

$$\\left(\\frac{\\partial U}{\\partial V_m}\\right)_T = T \\left(\\frac{R}{V_m - b}\\right) - \\left(\\frac{R T}{V_m - b} - \\frac{a}{V_m^2}\\right) = \\frac{a}{V_m^2}$$

Therefore, the bracketed term simplifies beautifully to:

$$P + \\left(\\frac{\\partial U}{\\partial V_m}\\right)_T = \\frac{R T}{V_m - b}$$

Step 2: Evaluate the Thermal Expansion Derivative $\\left(\\frac{\\partial V_m}{\\partial T}\\right)_P$

Differentiating the Van der Waals equation $\\left(P + \\frac{a}{V_m^2}\\right)(V_m - b) = R T$ implicitly with respect to $T$ at constant $P$:

$$\\left(-\\frac{2a}{V_m^3} \\frac{\\partial V_m}{\\partial T}\\right)(V_m - b) + \\left(P + \\frac{a}{V_m^2}\\right)\\frac{\\partial V_m}{\\partial T} = R$$

Factoring $\\frac{\\partial V_m}{\\partial T}$:

$$\\frac{\\partial V_m}{\\partial T} \\left[ \\frac{R T}{V_m - b} - \\frac{2a(V_m - b)}{V_m^3} \\right] = R$$
$$\\left(\\frac{\\partial V_m}{\\partial T}\\right)_P = \\frac{R}{\\frac{R T}{V_m - b} - \\frac{2a(V_m - b)}{V_m^3}} = \\frac{R (V_m - b)}{R T - \\frac{2a(V_m - b)^2}{V_m^3}}$$

Step 3: Combine Expressions for $C_p - C_v$

Multiplying the two terms:

$$C_p - C_v = \\left(\\frac{R T}{V_m - b}\\right) \\left[ \\frac{R (V_m - b)}{R T - \\frac{2a(V_m - b)^2}{V_m^3}} \\right] = \\frac{R}{1 - \\frac{2a(V_m - b)^2}{R T V_m^3}}$$

Notice that when $a \\to 0$, $C_p - C_v \\to R$ (ideal gas).

Step 4: Numerical Evaluation for Carbon Dioxide

Given parameters: $T = 300\\text{ K}$, $V_m = 1.00 \\times 10^{-3}\\text{ m}^3\\text{/mol}$, $a = 0.364$, $b = 4.27 \\times 10^{-5}$.

$$V_m - b = 1.00 \\times 10^{-3} - 0.0427 \\times 10^{-3} = 0.9573 \\times 10^{-3}\\text{ m}^3$$
$$(V_m - b)^2 = 9.164 \\times 10^{-7}\\text{ m}^6$$
$$2a(V_m - b)^2 = 2(0.364)(9.164 \\times 10^{-7}) = 6.671 \\times 10^{-7}$$
$$R T V_m^3 = (8.314)(300)(1.00 \\times 10^{-3})^3 = 2494.2 \\times 10^{-9} = 2.4942 \\times 10^{-6}$$

The denominator dimensionless correction term is:

$$\\delta = \\frac{2a(V_m - b)^2}{R T V_m^3} = \\frac{6.671 \\times 10^{-7}}{2.4942 \\times 10^{-6}} = 0.2675$$

Thus:

$$C_p - C_v = \\frac{R}{1 - 0.2675} = \\frac{8.314}{0.7325} = 11.35\\text{ J/(mol}\\cdot\\text{K)}$$

Percentage deviation from ideal gas Mayer value:

$$\\Delta\\% = \\frac{11.35 - 8.314}{8.314} \\times 100\\% = +36.5\\%$$

Final Answer & Physical Insight

Exact relation is $C_p - C_v = \\frac{R}{1 - \\frac{2a(V_m - b)^2}{R T V_m^3}}$. For $\\text{CO}_2$ at $300\\text{ K}$, $C_p - C_v = 11.35\\text{ J/(mol}\\cdot\\text{K)}$, representing a $+36.5\\%$ enhancement above the ideal gas constant $R$ due to intermolecular attractive forces.

StandardExample 1.3: Atmospheric Parcel Ascent & Dry Adiabatic Lapse Rate

An air parcel with an initial temperature of $T_0 = 25.0^{\circ}\\text{C}$ ($298.15\\text{ K}$) at sea level ($z = 0\\text{ m}$, $P_0 = 101.3\\text{ kPa}$) is forced to rise over a mountain ridge of height $h = 3200\\text{ m}$. Assuming dry adiabatic ascent ($c_p = 1005\\text{ J/(kg}\\cdot\\text{K)}$, $g = 9.81\\text{ m/s}^2$, $M = 28.97\\text{ g/mol}$): (a) calculate the parcel temperature at the mountain summit, (b) calculate the barometric pressure at the summit, and (c) determine the buoyant force per unit mass if the ambient environment exhibits an environmental lapse rate of $\\Gamma_{\\text{env}} = 6.50^{\circ}\\text{C/km}$.

Step 1: Calculate Parcel Temperature at Summit

The dry adiabatic lapse rate is:

$$\\Gamma_{\\text{dry}} = \\frac{g}{c_p} = \\frac{9.81\\text{ m/s}^2}{1005\\text{ J/(kg}\\cdot\\text{K)}} = 0.009761\\text{ K/m} = 9.761\\text{ K/km}$$

At altitude $h = 3200\\text{ m} = 3.20\\text{ km}$:

$$T_{\\text{parcel}}(h) = T_0 - \\Gamma_{\\text{dry}} h = 298.15 - (9.761)(3.20) = 298.15 - 31.24 = 266.91\\text{ K} = -6.24^{\circ}\\text{C}$$

Step 2: Calculate Summit Atmospheric Pressure

For an adiabatic ascent, pressure and temperature are related via:

$$P(h) = P_0 \\left(\\frac{T_{\\text{parcel}}(h)}{T_0}\\right)^{\\frac{\\gamma}{\\gamma - 1}}$$

For dry air ($\\gamma = 1.40$), $\\frac{\\gamma}{\\gamma - 1} = \\frac{1.40}{0.40} = 3.50$.

$$P(h) = (101.3\\text{ kPa}) \\left(\\frac{266.91}{298.15}\\right)^{3.50} = (101.3) (0.89522)^{3.50} = (101.3)(0.6781) = 68.69\\text{ kPa}$$

Step 3: Calculate Ambient Temperature and Buoyancy Acceleration

The surrounding ambient environmental temperature at $3200\\text{ m}$ with $\\Gamma_{\\text{env}} = 6.50^{\circ}\\text{C/km}$ is:

$$T_{\\text{env}}(h) = 298.15 - (6.50)(3.20) = 298.15 - 20.80 = 277.35\\text{ K} = +4.20^{\circ}\\text{C}$$

Notice that the adiabatic parcel ($-6.24^{\circ}\\text{C}$) is colder than the environment ($+4.20^{\circ}\\text{C}$). The buoyant acceleration (Archimedes force per unit mass) is:

$$a_b = g \\left(\\frac{\\rho_{\\text{env}} - \\rho_{\\text{parcel}}}{\\rho_{\\text{parcel}}}\\right) = g \\left(\\frac{T_{\\text{parcel}} - T_{\\text{env}}}{T_{\\text{env}}}\\right)$$
$$a_b = 9.81 \\left(\\frac{266.91 - 277.35}{277.35}\\right) = 9.81 \\left(\\frac{-10.44}{277.35}\\right) = -0.369\\text{ m/s}^2$$

The negative sign signifies a restoring downward buoyant force, proving that the atmosphere is stable against dry convection.

Final Answer & Physical Insight

At $3200\\text{ m}$, parcel temperature $T_{\\text{parcel}} = -6.24^{\circ}\\text{C}$ ($266.9\\text{ K}$), atmospheric pressure $P = 68.7\\text{ kPa}$, and restoring buoyant acceleration $a_b = -0.369\\text{ m/s}^2$ indicating a statically stable atmosphere.