Mixed Thermodynamics Problems

Introduction

Mixed problems are hard because several familiar formulas may appear relevant. The solution begins by asking what is measured, which boundary is chosen and which constraints are stated. A rigid bomb, pressure-held cup, reversible ideal-gas piston and equilibrium mixture each call for a different first equation.

Core explanation

For a closed system, always begin with ΔU = q + w when heat and work are involved. If a rigid vessel has no other work, w = 0 and q V = ΔU. If the process is at constant pressure with only P–V work, q p = ΔH. If an ideal gas has fixed composition and temperature, ΔU = 0 even though q and w may be nonzero. These are conditional simplifications of one energy balance, not formulas to choose by keyword alone.

For work, identify the mechanical path. Constant external pressure gives w = −P extΔV. A reversible isothermal ideal-gas path gives w = −nRT ln(V f/V i). Free expansion into a vacuum has P–V work zero. Using the reversible formula for a sudden free expansion merely because both have the same endpoints confuses a path function with a state function.

For reaction enthalpy, use balanced thermochemical equations, formation enthalpies or a Hess cycle. First state product phases and the scale of the reaction. If a calorimeter gives heat for an actual sample, divide by reaction extent to obtain a molar value. If comparing ΔH and ΔU for an ideal-gas reaction at common temperature, count gaseous coefficients to use ΔH ≈ ΔU + Δn gRT under the relevant approximation.

For direction at constant temperature and pressure, use actual ΔG, not ΔH alone. If ΔH and ΔS are supplied for a specified process, compute ΔG = ΔH − TΔS with compatible units. If a standard ΔG° and current mixture are supplied, use Δ rG = Δ rG° + RT ln Q. At equilibrium actual Δ rG = 0 and Δ rG° = −RT ln K. None of these Gibbs results alone gives reaction speed.

Before arithmetic, track units. A heat capacity in J K⁻¹ times ΔT gives J. A specific heat in J g⁻¹ K⁻¹ needs grams. An entropy in J mol⁻¹ K⁻¹ multiplied by T gives J mol⁻¹, requiring conversion if ΔH is in kJ mol⁻¹. A pressure-volume product in kPa L gives joules, while atm L requires a conversion factor. Dimensional analysis often catches the wrong formula immediately.

Check signs from the system viewpoint. Heat entering is positive q; expansion work done by the gas is negative w. An exothermic reaction warms surroundings but has negative q for the reaction. Products-minus-reactants sets state-function signs; reversing a reaction reverses ΔH, ΔS and ΔG for matched states.

A complete solution ends with a physical check. Does the reaction calorimeter warm for a negative reaction heat? Does a compression have positive work on the gas? Is a negative Gibbs value interpreted only at the stated T, P and composition? This last pass is often more valuable than carrying extra decimal places.

Step-by-step reasoning

1. Define the system, initial and final states and asked quantity. 2. Identify rigid, pressure-held, isothermal, reversible or equilibrium constraints. 3. Choose a state-function or path-transfer equation accordingly. 4. Convert units and reaction extent before substitution. 5. Interpret the sign and check it against the physical process.

Visual explanation

Draw a decision flow: “heat/work?” leads to first law; “calorimeter?” branches to rigid ΔU or constant-pressure ΔH; “reaction data?” leads to Hess or formation values; “direction?” leads to Gibbs and Q. Put units and system boundary as checks at the top of every branch.

Real-world analogy

A toolbox contains different instruments for length, mass and temperature. Choosing a familiar tool without identifying the quantity gives a confident but wrong answer. Thermodynamic equations likewise have specific measurements and constraints.

Real-world example

A fuel burns in a bomb calorimeter, warming a calibrated bath. The bath heat gives reaction q V and hence an estimate of ΔU; a gas-mole correction relates it to ΔH. If the question then asks whether a different reaction is favorable at a stated temperature, Gibbs data are required. One experiment can connect several quantities without making them interchangeable.

Why?

Why is “exothermic” insufficient to answer a spontaneity question? It gives an enthalpy or heat sign under conditions but omits entropy and composition, both of which influence actual Gibbs energy.

Common misconception

“The longest formula in the unit is the most advanced and therefore best.” The correct relation is the one whose assumptions match the defined system and path; a simple heat balance can be more appropriate than a Gibbs equation.

Worked example

One mole ideal gas expands reversibly and isothermally at 300 K from 1.0 L to 2.0 L. First identify the path, giving w = −nRT ln 2 ≈ −1.73 kJ. The ideal gas is isothermal, so ΔU = 0. First law gives q = +1.73 kJ. Do not use q p = ΔH without a constant-pressure claim; pressure changes along this reversible isotherm. The signs show heat entering to replace energy exported as work.

Quick check

1. Which relation directly interprets a rigid bomb's measured reaction heat under P–V-only assumptions? Answer: q V ≈ ΔU for the reacting system, after reversing calorimeter heat sign.

Exam focus

Write assumptions beside each equation. Show units and reaction scale, then explain the sign in words. Distinguish standard from actual Gibbs values and path functions from state functions.

Advanced insight

Many apparent contradictions disappear when one distinguishes state endpoints, process path and chosen boundary. A rigorous mixed-problem solution can be organized as a diagram of states and arrows before any equation is selected.

Summary

Method selection follows constraints: first law for energy balance, P–V path for work, calorimetry for heat, Hess cycles for reaction enthalpy and Gibbs energy for direction at fixed T and P. Units, phases, extent and signs are part of the chemistry, not optional formatting.

Practice questions

1. A rigid sealed reaction releases heat. Is its measured q more directly ΔU or ΔH? Answer: ΔU under negligible non-P–V work, because volume does not change. 2. What formula fits reversible isothermal ideal-gas work? Answer: w = −nRT ln(V f/V i), with constant n and T. 3. A reaction has ΔH < 0 but ΔS < 0. Can temperature matter to forward feasibility? Answer: Yes. At high T the positive −TΔS term can outweigh favorable negative ΔH.