8.2 First Law of Thermodynamics: Heat, Work, Internal Energy, and Enthalpy

First Law of Thermodynamics: Heat, Work, Internal Energy, and Enthalpy

Thermodynamics governs the macroscopic transformations of energy, work, and heat in chemical and physical systems. On the PRC Chemist Licensure Examination, mastery of the First Law requires rigorous facility with system boundary definitions, mathematical criteria for state functions, differential calculus of thermodynamic properties, reversible versus irreversible $PV$ work, and thermochemical enthalpy manipulations.

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1. Thermodynamic Systems and State Functions

A thermodynamic system is the specific portion of the universe under experimental observation, separated from the surroundings by defined boundaries.

Boundary Characteristics

• Diathermal (Diathermic) Wall: Permits heat conduction to establish thermal equilibrium with surroundings ($q \ne 0$).
• Adiabatic Wall: Thermally insulating boundary preventing heat transfer ($q = 0$).
• Rigid Wall: Inflexible boundary preventing boundary displacement work ($w_{PV} = 0, \Delta V = 0$).
• Permeable Wall: Allows chemical transport of specific chemical species.

State Functions vs. Path Functions

• State Functions ($U, H, S, G, A, T, P, V$): Properties whose numerical values depend strictly on the current equilibrium state of the system, independent of the historical preparation pathway. The differential of a state function $X$ is an exact differential ($dX$), satisfying: $\oint dX = 0 \quad \text{and} \quad \Delta X = \int_{1}^{2} dX = X_2 - X_1$
• Path Functions ($q, w$): Quantities whose integrated values depend explicitly on the mechanical or thermal pathway traversed between initial and final states. Their differentials are inexact differentials (denoted $\delta q, \delta w$ or $dq, dw$ with a stroke), satisfying: $\oint \delta w \ne 0, \quad \oint \delta q \ne 0$

Mathematical Criterion of Exact Differentials (Euler's Reciprocity)

For a continuous function $z = f(x, y)$, its total differential is: $dz = M(x, y)\,dx + N(x, y)\,dy \quad \text{where } M = \left(\frac{\partial z}{\partial x}\right)_y, \; N = \left(\frac{\partial z}{\partial y}\right)_x$ The differential $dz$ is exact if and only if Clairaut-Schwarz's theorem holds for mixed second partial derivatives: $\left(\frac{\partial M}{\partial y}\right)_x = \left(\frac{\partial N}{\partial x}\right)_y \iff \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y}$ This condition underpins the derivation of all Maxwell relations in thermodynamic potentials.

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2. The First Law of Thermodynamics and Sign Conventions

The First Law formalizes the principle of conservation of energy: internal energy ($U$) of an isolated system is constant. For a closed system undergoing an infinitesimal process:

$dU = \delta q + \delta w$ $\Delta U = q + w \quad \text{[Integrated Form]}$

The Modern IUPAC Sign Convention

The IUPAC convention adopts the system-centric perspective:

• Heat ($q$): $q > 0$ when heat is absorbed by the system (endothermic); $q < 0$ when heat is evolved from the system to surroundings (exothermic).
• Work ($w$): $w > 0$ when work is done on the system by the surroundings (compression); $w < 0$ when work is done by the system on the surroundings (expansion).

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3. Pressure-Volume ($PV$) Expansion Work

Expansion work occurs when a system displaces its movable boundaries against an external opposing pressure ($P_{ext}$):

$\delta w = -P_{ext}\,dV \implies w = -\int_{V_1}^{V_2} P_{ext}\,dV$

Case 1: Free Expansion (Gay-Lussac Expansion)

Expansion into an evacuated vacuum where $P_{ext} = 0$: $w = -\int_{V_1}^{V_2} (0)\,dV = 0$ For an ideal gas, because $(\frac{\partial U}{\partial V})_T = 0$, free isothermal expansion into a vacuum has $w = 0, \Delta U = 0$, and therefore $q = 0$.

Case 2: Irreversible Expansion Against Constant External Pressure

When a gas expands against constant atmospheric pressure or a fixed mechanical weight ($P_{ext} = \text{constant}$): $w_{irr} = -P_{ext} \int_{V_1}^{V_2} dV = -P_{ext}(V_2 - V_1) = -P_{ext}\,\Delta V$

Case 3: Reversible Isothermal Expansion of an Ideal Gas

A reversible process is carried out quasi-statically through a continuous sequence of equilibrium states where the external opposing pressure differs from internal system pressure by an infinitesimal differential ($P_{ext} = P_{int} \pm dP = \frac{nRT}{V}$):

$w_{rev} = -\int_{V_1}^{V_2} \frac{nRT}{V}\,dV = -nRT \ln\left(\frac{V_2}{V_1}\right) = -nRT \ln\left(\frac{P_1}{P_2}\right)$

Because the expansion curve follows the highest possible opposing pressure at every increment, a reversible isothermal expansion delivers the maximum possible expansion work to the surroundings: $|w_{rev}| > |w_{irr}|$

Worked Step-by-Step Calculation: Reversible vs. Irreversible Work

Scenario: A $1.000\text{ mol}$ sample of an ideal gas at $298.15\text{ K}$ expands from $V_1 = 10.00\text{ L}$ ($0.01000\text{ m}^3$) to $V_2 = 30.00\text{ L}$ ($0.03000\text{ m}^3$). Calculate $w$, $q$, and $\Delta U$ for:

• (a) A reversible isothermal expansion.
• (b) A single-stage irreversible expansion against a constant external pressure equal to the final gas pressure $P_2$.

Constants: $R = 8.31446\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.

Part (a) Reversible Path: $w_{rev} = -nRT \ln\left(\frac{V_2}{V_1}\right) = -(1.000)(8.31446)(298.15) \ln\left(\frac{30.00}{10.00}\right)$ $w_{rev} = -(2478.96\text{ J}) \ln(3.000) = -(2478.96)(1.09861) = -2723.4\text{ J} \approx -2.723\text{ kJ}$ Because internal energy of an ideal gas depends solely on temperature, $\Delta U = 0$ for an isothermal process: $q_{rev} = \Delta U - w_{rev} = 0 - (-2723.4\text{ J}) = +2723.4\text{ J} \approx +2.723\text{ kJ}$

Part (b) Irreversible Path: First calculate the final pressure $P_2 = P_{ext}$: $P_2 = \frac{nRT}{V_2} = \frac{(1.000\text{ mol})(8.31446\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1})(298.15\text{ K})}{0.03000\text{ m}^3} = 82632\text{ Pa} = 0.8263\text{ bar}$ Now compute irreversible expansion work: $w_{irr} = -P_{ext}(V_2 - V_1) = -(82632\text{ Pa})(0.03000 - 0.01000\text{ m}^3) = -(82632)(0.02000) = -1652.6\text{ J} \approx -1.653\text{ kJ}$ $q_{irr} = \Delta U - w_{irr} = 0 - (-1652.6\text{ J}) = +1652.6\text{ J} \approx +1.653\text{ kJ}$ Thermodynamic Deduction: The reversible expansion extracted $2.723\text{ kJ}$ of work from the surroundings' heat bath, whereas the single-stage irreversible drop delivered only $1.653\text{ kJ}$—wasting $1.071\text{ kJ}$ through dissipative non-equilibrium expansion.

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4. Heat Capacities and Molecular Degrees of Freedom

The heat capacity ($C$) represents heat absorbed per unit temperature change:

• Constant-Volume Heat Capacity ($C_v$): Because $\delta w_{PV} = 0$ at constant volume ($dV = 0$), $dU = \delta q_v$: $C_v = \left(\frac{\partial U}{\partial T}\right)_v \implies dU = C_v\,dT$
• Constant-Pressure Heat Capacity ($C_p$): At constant pressure, heat exchanged equals enthalpy change ($dH = \delta q_p$): $C_p = \left(\frac{\partial H}{\partial T}\right)_p \implies dH = C_p\,dT$

General Thermodynamic Relation Between $C_p$ and $C_v$

For any homogeneous substance: $C_p - C_v = \left[P + \left(\frac{\partial U}{\partial V}\right)_T\right] \left(\frac{\partial V}{\partial T}\right)_p = \frac{T V \alpha^2}{\kappa_T}$ where $\alpha = \frac{1}{V}(\frac{\partial V}{\partial T})_p$ is isobaric thermal expansivity, and $\kappa_T = -\frac{1}{V}(\frac{\partial V}{\partial P})_T$ is isothermal compressibility.

For an ideal gas, $(\frac{\partial U}{\partial V})_T = 0$ and $(\frac{\partial V}{\partial T})_p = \frac{nR}{P}$, reducing exactly to Mayer's Relation: $C_p - C_v = nR \iff C_{p,m} - C_{v,m} = R$

Heat Capacity Ratio ($\gamma$) and Equipartition Theorem

The heat capacity ratio is defined as $\gamma = \frac{C_p}{C_v} = \frac{C_{p,m}}{C_{v,m}}$. Based on the classical equipartition of energy (contributing $\frac{1}{2}R$ to $C_{v,m}$ per quadratic degree of freedom):

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5. Reversible Adiabatic Expansion of an Ideal Gas

In an adiabatic process, no heat enters or leaves the system ($q = 0$). From the First Law: $dU = \delta w \implies n C_{v,m}\,dT = -P\,dV = -\frac{nRT}{V}\,dV$ Separating variables and integrating between $(T_1, V_1)$ and $(T_2, V_2)$: $\int_{T_1}^{T_2} \frac{C_{v,m}}{T}\,dT = -\int_{V_1}^{V_2} \frac{R}{V}\,dV \implies C_{v,m} \ln\left(\frac{T_2}{T_1}\right) = -R \ln\left(\frac{V_2}{V_1}\right)$ Recalling that $R = C_{p,m} - C_{v,m}$ and dividing by $C_{v,m}$: $\ln\left(\frac{T_2}{T_1}\right) = -(\gamma - 1) \ln\left(\frac{V_2}{V_1}\right) = \ln\left(\frac{V_1}{V_2}\right)^{\gamma - 1}$ $\implies T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$

Applying the ideal gas equation of state ($P = \frac{nRT}{V}$) yields the complete set of Poisson adiabatic equations:

$P_1 V_1^\gamma = P_2 V_2^\gamma$ $T_1^\gamma P_1^{1-\gamma} = T_2^\gamma P_2^{1-\gamma} \iff T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{\frac{\gamma - 1}{\gamma}}$

Worked Step-by-Step Adiabatic Calculation

Problem: A $1.000\text{ mol}$ sample of argon gas ($\gamma = 5/3 = 1.667$) at $T_1 = 400.0\text{ K}$ and $P_1 = 5.00\text{ bar}$ undergoes a reversible adiabatic expansion until the pressure falls to $P_2 = 1.00\text{ bar}$. Determine the final temperature $T_2$ and the boundary work $w$.

Step 1: Calculate $T_2$: $\frac{\gamma - 1}{\gamma} = \frac{5/3 - 1}{5/3} = \frac{2/3}{5/3} = \frac{2}{5} = 0.400$ $T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{0.400} = (400.0\text{ K}) \left(\frac{1.00\text{ bar}}{5.00\text{ bar}}\right)^{0.400} = (400.0) (0.200)^{0.400}$ $(0.200)^{0.400} = 0.52530 \implies T_2 = (400.0)(0.52530) = 210.12\text{ K}$

Step 2: Calculate Work $w$: Because $q = 0$, $w = \Delta U = n C_{v,m}(T_2 - T_1)$: $C_{v,m} = \frac{3}{2}R = 1.5 \times 8.31446 = 12.4717\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$ $w = (1.000\text{ mol})(12.4717\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1})(210.12 - 400.0\text{ K}) = (12.4717)(-189.88\text{ K}) = -2368.1\text{ J} \approx -2.368\text{ kJ}$ Physical Rationale: During adiabatic expansion, the gas expends its own internal energy to do work against surroundings, driving substantial temperature drop ($400.0\text{ K} \rightarrow 210.1\text{ K}$).

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6. Enthalpy and the Joule-Thomson Effect

Enthalpy ($H$) is defined mathematically as: $H \equiv U + PV$ For an infinitesimal process in a closed system with $PV$ work only: $dH = dU + P\,dV + V\,dP = (\delta q - P\,dV) + P\,dV + V\,dP = \delta q + V\,dP$ At constant pressure ($dP = 0$), $dH = \delta q_p$, confirming that enthalpy represents constant-pressure heat transfer.

The Joule-Thomson Throttling Process

In a Joule-Thomson expansion, gas flows continuously through an insulated porous plug or throttling valve from high pressure $P_1$ to low pressure $P_2$. The process is strictly isenthalpic ($\Delta H = 0$).

The Joule-Thomson Coefficient ($\mu_{JT}$) is defined as: $\mu_{JT} \equiv \left(\frac{\partial T}{\partial P}\right)_H = -\frac{1}{C_p} \left(\frac{\partial H}{\partial P}\right)_T = \frac{1}{C_p}\left[T\left(\frac{\partial V}{\partial T}\right)_p - V\right]$

• Ideal Gas: Because $V = \frac{nRT}{P}$, $(\frac{\partial V}{\partial T})_p = \frac{nR}{P} = \frac{V}{T}$. Substituting into $\mu_{JT}$ yields: $\mu_{JT} = \frac{1}{C_p}\left[T\left(\frac{V}{T}\right) - V\right] = 0$ An ideal gas undergoes zero temperature change during Joule-Thomson throttling.
• Real Gas Inversion Temperature ($T_i$):
  • If $\mu_{JT} > 0$: Because expansion entails $dP < 0$, $dT = \mu_{JT}\,dP < 0$. The gas cools on expansion (basis of the Linde liquefaction cycle for $\text{N}_2, \text{O}_2$).
  • If $\mu_{JT} < 0$: The gas heats on throttling ($dT > 0$).
  • The boundary where $\mu_{JT} = 0$ is the inversion curve. For a van der Waals gas, the maximum inversion temperature is approximately: $T_i \approx \frac{2a}{Rb}$ For hydrogen ($T_i \approx 202\text{ K}$) and helium ($T_i \approx 40\text{ K}$), inversion temperatures lie far below room temperature. Consequently, throttling $\text{H}_2$ or $\text{He}$ at $298\text{ K}$ results in warming, presenting severe combustion hazards during industrial depressurization.

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7. Thermochemistry and Kirchhoff's Law

Standard States and Hess's Law

• Standard State ($^\circ$): The pure substance in its standard physical state at a pressure of exactly $P^\circ = 1\text{ bar}$ ($10^5\text{ Pa}$) and specified temperature (usually $298.15\text{ K}$). For solutes, standard state is a hypothetical ideal solution of unit molality ($m^\circ = 1\text{ mol/kg}$).
• Standard Enthalpy of Formation ($\Delta H_f^\circ$): Enthalpy change when 1 mole of substance is synthesized from its constituent elements in their standard reference states. By convention: $\Delta H_f^\circ(\text{element in reference state, } 298.15\text{ K}) = 0\text{ kJ/mol}$ Examples: $\Delta H_f^\circ[\text{C(graphite)}] = 0$, but $\Delta H_f^\circ[\text{C(diamond)}] = +1.90\text{ kJ/mol}$; $\Delta H_f^\circ[\text{O}_2(g)] = 0$, but $\Delta H_f^\circ[\text{O}_3(g)] = +142.7\text{ kJ/mol}$.
• Hess's Law of Constant Heat Summation: Because enthalpy is a state function, standard reaction enthalpy is the algebraic sum of step enthalpies: $\Delta H_{rxn}^\circ = \sum \nu_p \Delta H_f^\circ(\text{products}) - \sum \nu_r \Delta H_f^\circ(\text{reactants})$

Kirchhoff's Law of Temperature Dependence

Differentiating standard reaction enthalpy with respect to temperature at constant pressure: $\left(\frac{\partial \Delta H_{rxn}^\circ}{\partial T}\right)_p = \Delta C_p^\circ = \sum \nu_p C_{p,m}^\circ(\text{products}) - \sum \nu_r C_{p,m}^\circ(\text{reactants})$ Integrating from $T_1$ to $T_2$ yields Kirchhoff's Law: $\Delta H_{rxn}^\circ(T_2) = \Delta H_{rxn}^\circ(T_1) + \int_{T_1}^{T_2} \Delta C_p^\circ\,dT$ If $\Delta C_p^\circ$ is temperature-independent over the interval: $\Delta H_{rxn}^\circ(T_2) = \Delta H_{rxn}^\circ(T_1) + \Delta C_p^\circ(T_2 - T_1)$

Worked Step-by-Step Kirchhoff's Law Calculation

Scenario: Calculate standard reaction enthalpy at $373.15\text{ K}$ for the formation of liquid water: $\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l)$ Given Data at $298.15\text{ K}$:

• $\Delta H_f^\circ[\text{H}_2\text{O}(l)] = -285.83\text{ kJ/mol}$
• $C_{p,m}^\circ[\text{H}_2(g)] = 28.82\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
• $C_{p,m}^\circ[\text{O}_2(g)] = 29.36\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
• $C_{p,m}^\circ[\text{H}_2\text{O}(l)] = 75.29\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$

Step 1: Compute $\Delta C_p^\circ$: $\Delta C_p^\circ = C_{p,m}^\circ[\text{H}_2\text{O}(l)] - \left(C_{p,m}^\circ[\text{H}_2(g)] + \frac{1}{2} C_{p,m}^\circ[\text{O}_2(g)]\right)$ $\Delta C_p^\circ = 75.29 - \left(28.82 + 0.5 \times 29.36\right) = 75.29 - (28.82 + 14.68) = 75.29 - 43.50 = +31.79\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$

Step 2: Apply Kirchhoff's formula for $\Delta T = 373.15 - 298.15 = 75.0\text{ K}$: $\Delta H_{373.15}^\circ = \Delta H_{298.15}^\circ + \Delta C_p^\circ\,\Delta T$ $\Delta H_{373.15}^\circ = -285.83\text{ kJ/mol} + \frac{(+31.79\text{ J/mol}\cdot\text{K})(75.0\text{ K})}{1000\text{ J/kJ}}$ $\Delta H_{373.15}^\circ = -285.83 + 2.384 = -283.45\text{ kJ/mol}$ Note: Because liquid water has a significantly larger heat capacity than the gaseous reactants, $\Delta C_p^\circ > 0$, making the reaction enthalpy slightly less exothermic at elevated temperature.