Thermodynamics · Lesson 7 of 7
Chapter Summary and Practice
“Connect equilibrium, energy accounting, gas processes and second-law limits, then apply the full chapter to mixed problems.”
• Connect the zeroth, first and second laws in a single thermodynamic description. • Select equations from the process conditions rather than from the desired answer. • Compare state functions, transfers, heat capacities and reversible limits. • Use formula conditions, units and graph areas to check solutions. • Solve mixed energy and engine problems while explaining physical assumptions.
From temperature to the limits of energy conversion
Start by defining the system and its boundary. Decide whether matter is fixed, heat can cross and the boundary can move. Establish whether the states are equilibrium states so that P, V and T have well-defined bulk meanings. The zeroth law gives a consistent temperature comparison; the equation of state links state variables. The first law then balances energy transfers. Finally, the second law checks whether the proposed overall direction and performance are possible. Passing the energy balance is necessary but not sufficient.
Equilibrium of bulk variables does not freeze molecular motion. Internal energy includes microscopic motion and interactions but excludes the centre-of-mass kinetic energy of the whole sample. For fixed ideal-gas amount and composition U depends only on T; this statement does not extend unchanged to a phase transition or arbitrary real material. Temperature is intensive, whereas total U depends on how much material is present.
Distinctions that prevent most mistakes
| Pair | How to distinguish them |
|---|---|
| Temperature / internal energy | T indicates thermal equilibrium; U is an extensive stored-energy state function. |
| Internal energy / heat and work | U belongs to a state; Q and W describe transfers along a process. |
| Isothermal / adiabatic | Isothermal fixes T; adiabatic fixes Q = 0. They do not generally describe the same path. |
| Quasi-static / reversible | Slow near-equilibrium change is not sufficient if friction or other dissipation persists. |
| Zero ΔU / zero heat | An ideal-gas isotherm can have ΔU = 0 with nonzero Q and W. |
| Return to initial volume / complete cycle | A cycle must restore every state variable, including P, T and U. |
| Engine efficiency / refrigerator α | η = useful work/heat absorbed; α = cooling/work supplied. α can exceed 1. |
| Adiabatic free expansion / reversible adiabat | Both can have Q = 0; only the reversible ideal-gas path obeys PV^γ = constant. |
Formula reference with conditions
| Quantity | Unit / dimension check | Reminder |
|---|---|---|
| P, V, T, n | Pa; m³; K; mol | Gas pressure and temperature must be absolute. |
| Q, W, U | J; [M L² time⁻²] | Pa·m³ = J; Q and W are not state functions. |
| s | J kg⁻¹ K⁻¹; [L² time⁻² temperature⁻¹] | Capacity per mass; distinct from total S. |
| C and R | J mol⁻¹ K⁻¹ | Use molar quantities consistently. |
| Power / heat-transfer rate | J s⁻¹ = watt | A rate is not an energy. |
| γ, η, α and logarithm arguments | Dimensionless | A log requires a ratio of like units. |
Two earlier thermal-properties quantities sometimes included in a chapter recap remain distinct from this chapter’s energy balance. The volume-expansion coefficient αV has unit K⁻¹ and, for a small isotropic expansion, αV ≈ 3αL where αL is the linear coefficient. Thermal conductivity k has unit W m⁻¹ K⁻¹; one-dimensional conduction is expressed by heat-transfer rate Qdot = −kA dT/dx, with the minus sign indicating transfer toward lower temperature. These are background reminders, not additional gas-process laws or new derivations.
Choose a method from the physical conditions
First list the constraints, then choose an equation. “Insulated” supplies Q = 0, not automatically PV^γ = constant. “Rigid” supplies zero boundary work, not necessarily zero total work if electrical or stirring work is present. “Ideal gas at fixed temperature” supplies ΔU = 0, not automatically the quasi-static work formula. When given a valid P–V path, calculate signed work from its area. When given only endpoints, do not assume work is path-independent.
Check three levels separately. Units: energy terms must all be joules. Signs and limits: expansion normally gives positive boundary work; a reversible adiabatic expansion cools; an engine between nearly equal temperatures has a small maximum efficiency. Scope: using Celsius in temperature ratios, applying a reversible formula to free expansion, or treating a two-leg return to the same volume as a cycle can produce a numerically neat but physically wrong answer.
Worked synthesis: a rectangular gas cycle
Problem
One mole of ideal gas with Cᵥ = 3R/2 follows A→B→C→D→A. The states are A: (V = 1.0 L, P = 1.0×10⁵ Pa), B: (3.0 L, 1.0×10⁵ Pa), C: (3.0 L, 3.0×10⁵ Pa), D: (1.0 L, 3.0×10⁵ Pa). All legs are quasi-static and pressure–volume work is the only work. Find work and ΔU on each leg, and net heat.
- 1.Use 1 L = 10⁻³ m³. For this gas U changes by (3/2)Δ(PV), since nRT = PV and Cᵥ = 3R/2. The endpoint PV values are 100, 300, 900 and 300 J.
- 2.A→B is an isobaric expansion: WAB = (1.0×10⁵)(0.002) = +200 J; ΔUAB = (3/2)(300−100) = +300 J; QAB = 500 J.
- 3.B→C is isochoric heating: WBC = 0; ΔUBC = (3/2)(900−300) = +900 J; QBC = +900 J.
- 4.C→D is an isobaric compression: WCD = (3.0×10⁵)(−0.002) = −600 J; ΔUCD = (3/2)(300−900) = −900 J; QCD = −1500 J.
- 5.D→A is isochoric cooling: WDA = 0; ΔUDA = (3/2)(100−300) = −300 J; QDA = −300 J.
- 6.Total ΔU = 0, Wnet = −400 J, and Qnet = −400 J. The path is anticlockwise: it requires 400 J net work input and rejects 400 J net heat.
- 7.This cycle does not operate only on two isotherms. Do not assign it the two-reservoir Carnot formula merely because it is a closed loop.
Worked synthesis: an engine-limit check
Problem
A device between 800 K and 320 K absorbs 2500 J per cycle and claims 1400 J net work output. Determine its rejected heat, claimed efficiency and status relative to the reversible ceiling.
- 1.First-law balance gives QC = QH − Wout = 2500−1400 = 1100 J.
- 2.The claimed efficiency is η = 1400/2500 = 0.56 = 56%.
- 3.The Carnot maximum is ηC = 1−320/800 = 0.60 = 60%. The claimed efficiency does not exceed the second-law bound.
- 4.The maximum possible work for that input is 0.60×2500 = 1500 J, corresponding to QC,min = 1000 J in the reversible limit.
- 5.Passing these two tests does not by itself establish that a practical machine with the claimed design exists. It shows that its quoted energy transfers do not violate the first law or this two-reservoir second-law bound.
Connect the different question types
A flowing-water heater uses mass flow × specific heat × temperature rise. Heating a gas at constant pressure uses nCₚΔT rather than nCᵥΔT because expansion also performs work. Insulated compression changes temperature through work, whereas insulated free expansion into vacuum has no work and, for an ideal gas, no temperature change. A graphical problem asks for signed area; an engine question additionally asks whether the implied efficiency respects its reservoir temperatures. Identifying the physical constraints is the common step behind these apparently different problems.
Quick Quiz
Quiz
A fixed ideal gas changes temperature by ΔT along two different paths. For constant Cᵥ, which must be the same?
U depends only on T for a fixed ideal-gas sample, while Q and W can differ between paths.
A gas receives 200 J heat while expanding against constant external pressure 1.0×10⁵ Pa through 1.0 L. ΔU is
Work is (10⁵)(10⁻³) = 100 J, so ΔU = 200−100 = 100 J.
Which conclusion about an insulated ideal-gas free expansion into vacuum is correct?
With Q = W = 0, ΔU = 0; ideal-gas U(T) then gives ΔT = 0. Intermediate states are not an equilibrium curve.
A cyclic engine runs only between 800 K and 320 K. Its reversible maximum efficiency is
ηC = 1−320/800 = 0.60.
A clockwise rectangular quasi-static P–V loop spans ΔP = 2.0×10⁵ Pa and ΔV = 2.0 L. Its net work is
The signed enclosed area is positive for clockwise traversal: ΔPΔV = (2×10⁵)(2×10⁻³) = 400 J.
Practice Problems
- In your own words connect the zeroth, first and second laws. For each law give one question it can answer and one that it cannot answer on its own.
- A 0.50 kg water sample warms by 20 K, with s = 4200 J kg⁻¹ K⁻¹ and negligible expansion work. Compare its heat input with that for 1.0 mol ideal gas warmed by 20 K at constant pressure with Cₚ = 7R/2 and R = 8.314 J mol⁻¹ K⁻¹. For the gas also find W and ΔU.
- Two moles of ideal gas with γ = 1.4 start at 300 K and expand to twice their volume by (a) a quasi-static isotherm and (b) a reversible adiabat. Calculate final temperatures, work, heat and ΔU. Use R = 8.314 J mol⁻¹ K⁻¹ and explain why the two endpoint states differ.
- A system changes from A to B adiabatically when 60 J of work is done on it. A second path from A to B absorbs 150 J heat. Find the work done by the system on the second path and explain the path-independent quantity.
- A quasi-static gas follows a straight-line expansion from (2.0 L,4.0×10⁵ Pa) to (5.0 L,1.0×10⁵ Pa), then returns at 1.0×10⁵ Pa to 2.0 L, then at fixed volume to the initial pressure. Find each leg’s work, cycle work and net heat.
- A cyclic engine between 700 K and 350 K absorbs 1800 J and claims 1000 J work per cycle. Find rejected heat and claimed efficiency, then test the two-reservoir Carnot limit. Determine the maximum permitted work.
- Explain why two insulated equal chambers, one containing ideal gas and one initially evacuated, give different expansion work from a quasi-static gas expansion between the same initial and final volumes. Identify the first-law consequences.
- A refrigerator removes 900 J per cycle with 300 J work input. Find heat rejected and α. Explain why this does not imply an engine efficiency of 300%.
- An ideal reversible engine has hot temperature 600 K and efficiency 40%. Determine its cold temperature. If run backward to remove 1200 J from that cold reservoir, calculate its work input and heat delivered to the hot reservoir.
Key Takeaways
• Define the system, constraints and equilibrium status before choosing equations. • The zeroth law organises temperature; the first law balances energy; the second law limits overall direction and performance. • Heat and work are path quantities; ΔU is fixed by endpoints, and for an ideal gas by the temperature change. • Process formulas have conditions: reversible adiabats, quasi-static isotherms and free expansion must be distinguished. • Use SI units, signed P–V area, kelvin ratios and the Carnot bound as independent checks.
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The Carnot Engine and Maximum Efficiency
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