- Thermodynamics studies heat, temperature, work and their interconversion. It is macroscopic: a system is described by measurable bulk variables rather than tracking each molecule.
- Mechanical motion of a body as a whole is distinct from its internal energy. When a moving bullet stops in wood, organized kinetic energy becomes disordered internal energy of the bullet and nearby wood.
- The old caloric-fluid model treated heat as a material fluid. Rumford’s cannon-boring observation showed heat production depended on work done, supporting heat as energy transfer and conversion.
- Heat flows spontaneously from higher to lower temperature. Rubbing hands converts work into internal energy; a steam engine converts some transferred heat into useful work.
CHAPTER 11
Learn the chapter
Start with heat and internal energy, then follow the laws through gas processes and the Carnot cycle.
11.1
Thermodynamics and heat as energy
Thermodynamics describes bulk energy transfers and transformations.
11.2–11.3
Thermal equilibrium and the Zeroth Law
Thermal equilibrium is the basis for defining temperature.
- A system is in thermodynamic equilibrium when its macroscopic variables remain constant in time. Mechanical equilibrium (zero net force/torque) is a different idea.
- An adiabatic wall prevents heat transfer; a diathermic wall permits it. Systems across a diathermic wall exchange energy until no net heat flows; they are then in thermal equilibrium.
- Zeroth Law: if A and B are each in thermal equilibrium with C, then A and B are in thermal equilibrium with each other. It makes temperature a well-defined common property.
- At thermal equilibrium, temperatures are equal. A thermometer is a third system that comes to equilibrium with the body; reproducible thermometric properties enable a temperature scale.
11.4
Heat, internal energy and work
Q and W describe energy transfer; U describes the system’s state.
- Internal energy U is the sum of molecular kinetic and potential energies in the centre-of-mass frame. It includes random translational, rotational and vibrational energy, not kinetic energy of the system moving as a whole.
- U is a state variable: its change depends only on initial and final states, not the path. Pressure, volume and temperature are also state variables for a gas.
- Heat Q is energy transferred because of a temperature difference; work W is energy transferred by other macroscopic means (such as a piston). Heat and work are not properties “contained” in a state.
- Sign convention throughout: Q>0 when heat enters the system; W>0 when the system does work on surroundings. For constant external pressure, W=PΔV; in a quasi-static process, W=∫P dV.
11.5
First Law of Thermodynamics
Energy supplied becomes internal energy and/or work.
- For the stated sign convention, the First Law is Q=ΔU+W, or ΔU=Q−W. It is conservation of energy applied to a thermodynamic system.
- ΔU is path-independent, while Q and W individually depend on the path. Their difference Q−W is fixed by the change of state.
- If ΔU=0 (for an ideal gas in an isothermal process), then Q=W: supplied heat is converted into work. During a complete cycle, ΔU=0, so net heat equals net work.
- For constant pressure, W=PΔV. Example from the chapter: vaporising 1 g water at 1 atm absorbs 2256 J; ΔV=(1671−1) cm³=1.670×10⁻³ m³, so W≈1.013×10⁵×1.670×10⁻³=169.2 J and ΔU≈2256−169.2=2086.8 J.
WORKED
Chapter calculation
1 g of water becomes steam at atmospheric pressure. Q=2256 J, Vᵢ=1 cm³, V𝒇=1671 cm³. Therefore W=PΔV≈169.2 J and ΔU=Q−W≈2086.8 J; most supplied energy raises internal energy.
11.6
Specific heat capacity
Heat capacity depends on amount, substance, temperature and process.
- Heat capacity S=ΔQ/ΔT; specific heat s=S/m=ΔQ/(mΔT), in J kg⁻¹ K⁻¹. Molar heat capacity C=ΔQ/(μΔT), in J mol⁻¹ K⁻¹.
- Specific and molar heat capacities depend on the material and temperature. For gases they also depend on the constraint: Cᵥ at constant volume and Cₚ at constant pressure.
- For a monatomic solid modelled as atoms vibrating in three dimensions, mean energy per atom is 3kᵦT. Per mole U=3RT. Since solid expansion work PΔV is small, C≈ΔU/ΔT=3R (Dulong–Petit law); agreement fails at low T and for some materials such as carbon.
- A calorie was precisely defined using water from 14.5°C to 15.5°C; SI uses joules. Water’s specific heat near room temperature is about 4186 J kg⁻¹ K⁻¹.
- For an ideal gas, Cₚ−Cᵥ=R. At constant volume Cᵥ=(ΔU/ΔT)₍ᵥ₎. At constant pressure Cₚ=ΔU/ΔT+PΔV/ΔT; for one mole PV=RT gives PΔV/ΔT=R. Subtracting gives Mayer’s relation.
PROOF
Proof
For one mole at fixed volume, Qᵥ=ΔU, hence Cᵥ=ΔU/ΔT. At fixed pressure, Qₚ=ΔU+PΔV, hence Cₚ=ΔU/ΔT+PΔV/ΔT. Since PV=RT, at fixed P, PΔV/ΔT=R. Therefore Cₚ=Cᵥ+R.
11.7
Thermodynamic state variables and equation of state
State variables describe equilibrium states; an equation of state links them.
- An equilibrium gas state is specified by macroscopic variables such as P,V,T and amount/composition. During rapid free expansion or an explosive reaction, pressure or temperature may be non-uniform, so no single equilibrium state describes the system.
- An equation of state relates state variables. For an ideal gas PV=μRT; for fixed μ, only two of P,V,T are independent. A fixed-temperature P–V curve is an isotherm.
- Thermodynamic state variables also include entropy S and enthalpy H. Entropy is a state function related to energy dispersal and accessible microscopic arrangements; enthalpy is defined H=U+PV and is useful for constant-pressure changes.
- Extensive quantities scale with system size and halve when an equilibrium system is divided equally: U,V,total mass and entropy. Intensive quantities remain unchanged: P,T and density.
- Dimensional and extensive/intensive checks help test equations. In Q=ΔU+PΔV each term is extensive because intensive P multiplies extensive ΔV.
11.8
Quasi-static and special thermodynamic processes
The path determines work and heat, even between the same states.
- A quasi-static process is an ideal limit carried out infinitely slowly so the system remains arbitrarily close to equilibrium at every stage. The system and surroundings differ only infinitesimally in pressure or temperature. Real slow processes can approximate it.
- Isothermal: T constant. For an ideal gas, PV=constant and P=μRT/V; integrating W=∫P dV from V₁ to V₂ gives W=μRT ln(V₂/V₁). Since ideal-gas U depends only on T, ΔU=0 and Q=W. Expansion absorbs heat; compression releases it.
- Adiabatic: thermally insulated, Q=0. For an ideal gas, PVᵞ=constant (γ=Cₚ/Cᵥ). With P=C/Vᵞ, integrate W=∫P dV to obtain W=(P₂V₂−P₁V₁)/(1−γ)=μR(T₁−T₂)/(γ−1). Expansion work lowers U and T; compression raises them.
- Isochoric: V constant, W=0, so Q=ΔU. Isobaric: P constant, W=P(V₂−V₁)=μR(T₂−T₁); heat changes both U and volume. Cyclic: returns to initial state, ΔU=0 and net Q=W.
- On a P–V graph, quasi-static work is signed area under the path. Isotherms are less steep than adiabats through the same state; adiabatic compression heats more than isothermal compression.
WORKED
Worked comparison
For an isothermal ideal-gas expansion from V₁ to 2V₁, W=μRT ln 2 and Q=W. In an adiabatic expansion between the same initial state and a larger volume, Q=0 and the gas cools; its pressure falls more steeply than on the isotherm.
11.9
Second Law of Thermodynamics
The First Law allows energy bookkeeping; the Second Law constrains direction.
- Energy conservation alone permits imagined processes never observed, such as a book spontaneously rising while its internal energy converts entirely into mechanical energy. The Second Law rules out such directions.
- Kelvin–Planck statement: no cyclic process can have as its sole result absorption of heat from one reservoir and complete conversion of that heat into work. Thus a heat engine cannot be 100% efficient.
- Clausius statement: no process can have as its sole result transfer of heat from a colder body to a hotter body. A refrigerator therefore requires work; its coefficient of performance cannot be infinite.
- The statements are equivalent. If free cold→hot transfer were possible, it could return a real engine’s rejected heat to the hot source at no work cost, leaving the sole net effect to convert heat from one reservoir wholly into work (Kelvin–Planck violation). If a perfect engine existed, its work could drive a refrigerator; the combined device would move heat cold→hot with no other net effect (Clausius violation).
11.10
Reversible and irreversible processes
Reversibility requires restoring both system and surroundings without other effects.
- A process is reversible only if it can be reversed so that system and surroundings both return to their original states and nothing else changes. This is an ideal limit.
- A reversible process must be quasi-static and free of dissipative effects. A very slow isothermal expansion with a frictionless piston is an ideal example.
- Natural processes are generally irreversible: heat flows down a temperature gradient, gases freely expand/diffuse, combustion occurs, and friction or viscosity dissipates mechanical energy.
- Irreversibility arises when a system passes through nonequilibrium states and from dissipation (friction, viscosity, turbulence). These effects can be reduced but not fully eliminated in practice.
- Reversible engines set the maximum efficiency between two reservoirs; any irreversible engine operating between them is less efficient.
11.11
Carnot engine and Carnot’s theorem
A reversible cycle sets the greatest possible engine efficiency between two temperatures.
- A Carnot engine is reversible and operates between hot source T₁ and cold sink T₂ (T in kelvin). Its cycle consists of: isothermal expansion at T₁ absorbing Q₁; adiabatic expansion cooling to T₂; isothermal compression at T₂ rejecting Q₂; adiabatic compression returning to T₁.
- For an ideal gas, isothermal expansion work and heat absorbed are Q₁=μRT₁ ln(V₂/V₁). Isothermal compression rejects Q₂=μRT₂ ln(V₃/V₄) in magnitude. The adiabatic legs obey TVᵞ⁻¹=constant, which gives V₂/V₁=V₃/V₄.
- Efficiency η=W/Q₁=(Q₁−Q₂)/Q₁. Using the volume-ratio equality gives η=1−T₂/T₁. The efficiency depends only on reservoir temperatures, not the working substance.
- Carnot’s theorem: no engine between the same two reservoirs can be more efficient than a reversible engine; all reversible engines between them have the same efficiency. Otherwise coupling a better engine to a reversible refrigerator would violate Kelvin–Planck.
- Proof: suppose engine I were more efficient than reversible engine R. Let I run forward and R backward, adjusted to return to the hot source exactly the heat I removed. Since I produces more work than R consumes, the coupled device’s only net effect is to draw heat from the cold sink and convert it fully to work, contradicting Kelvin–Planck. Interchanging the working substances of two reversible engines gives the same contradiction if their efficiencies differ; hence all reversible engines between fixed reservoirs share the Carnot efficiency.
- Reversing a Carnot cycle makes a refrigerator: work is supplied to remove heat from the cold reservoir and deliver it to the hot one. In a Carnot cycle Q₂/Q₁=T₂/T₁, a universal temperature relation.
PROOF
Proof of Carnot efficiency
Isothermal legs give Q₁=μRT₁ ln(V₂/V₁) and Q₂=μRT₂ ln(V₃/V₄). Adiabatic legs obey TVᵞ⁻¹=constant, so V₂/V₁=V₃/V₄. Thus Q₂/Q₁=T₂/T₁ and η=(Q₁−Q₂)/Q₁=1−T₂/T₁.
MODEL
Thermodynamics in motion
See how pressure changes along isothermal and adiabatic paths, and how reservoir temperatures set Carnot efficiency.
1 · P–V process and work
Change final volume and compare isothermal and adiabatic pressure.
P–V area represents work; the adiabatic path is steeper.
2 · Carnot engine efficiency
Adjust source and sink temperatures; temperatures are kelvin.
η = 1 − T₂/T₁; no real engine reaches this reversible limit.
3 · Energy accounting
Adjust heat supplied and work by the gas to see the First Law balance.
Q = ΔU + W. Positive W means expansion work by the gas.
Use the sliders to change the conditions, or play each model to watch how the system responds.
QUIZ
Concept quiz
Try the questions and check each explanation after you answer.
0 / 30 AnsweredCorrect: 0 (0%)
11.1Thermodynamics and heat as energy
1. Thermodynamics is chiefly a study of:
It uses measurable bulk variables and energy transfers.
Remember: Thermodynamics studies bulk states and heat–work conversion.
2. Internal energy excludes:
Internal energy is defined in the centre-of-mass frame.
Remember: Thermodynamics studies bulk states and heat–work conversion.
3. Rumford’s cannon boring observation supported the view that:
Continued mechanical work produced continued heating.
Remember: Thermodynamics studies bulk states and heat–work conversion.
11.2–11.3Thermal equilibrium and the Zeroth Law
1. A diathermic wall:
Diathermic walls conduct heat.
Remember: Equilibrium variables do not change with time.
2. If A is in equilibrium with C and B is in equilibrium with C, then:
This is the Zeroth Law.
Remember: Equilibrium variables do not change with time.
3. Thermal equilibrium of two systems means:
The common equilibrium property is temperature.
Remember: Equilibrium variables do not change with time.
11.4Heat, internal energy and work
1. Which is a state variable?
U depends on the state, not its history.
Remember: U is a state property; Q and W are path-dependent energy transfers.
2. With W positive for work done by the system, expansion against constant P gives:
Expansion has positive ΔV and positive work by the gas.
Remember: U is a state property; Q and W are path-dependent energy transfers.
3. Heat is best described as:
Heat is energy in transit.
Remember: U is a state property; Q and W are path-dependent energy transfers.
11.5First Law of Thermodynamics
1. Using Q=ΔU+W, if Q=500 J and W=180 J, ΔU is:
ΔU=Q−W=320 J.
Remember: First Law: Q=ΔU+W (W is work by system).
2. For a complete thermodynamic cycle:
The system returns to its initial state.
Remember: First Law: Q=ΔU+W (W is work by system).
3. Vaporisation example: when 2256 J heat is supplied and 169 J is work, ΔU is about:
ΔU=Q−W≈2087 J.
Remember: First Law: Q=ΔU+W (W is work by system).
11.6Specific heat capacity
1. Specific heat capacity has SI units:
It is heat per mass per temperature rise.
Remember: S=ΔQ/ΔT; s=S/m; C=S/μ.
2. For a classical monatomic solid model, molar heat capacity is approximately:
Three-dimensional oscillator energy gives U=3RT.
Remember: S=ΔQ/ΔT; s=S/m; C=S/μ.
3. Mayer’s relation for an ideal gas is:
Constant-pressure work contributes R per kelvin per mole.
Remember: S=ΔQ/ΔT; s=S/m; C=S/μ.
11.7Thermodynamic state variables and equation of state
1. An ideal gas with fixed amount has how many independent variables among P,V,T?
PV=μRT constrains the three variables.
Remember: State variables describe equilibrium; nonequilibrium states may lack uniform P,T.
2. Which set contains only intensive quantities?
Intensive quantities do not scale with system size.
Remember: State variables describe equilibrium; nonequilibrium states may lack uniform P,T.
3. During rapid free expansion, a gas may lack a single well-defined:
The transient state is not uniform equilibrium.
Remember: State variables describe equilibrium; nonequilibrium states may lack uniform P,T.
11.8Quasi-static and special thermodynamic processes
1. A quasi-static process proceeds:
The system remains infinitesimally close to equilibrium.
Remember: Quasi-static paths pass through equilibrium states.
2. For an ideal-gas isothermal change:
Ideal-gas internal energy depends only on T.
Remember: Quasi-static paths pass through equilibrium states.
3. In an adiabatic process:
Adiabatic means no heat transfer.
Remember: Quasi-static paths pass through equilibrium states.
4. For an isochoric process, work by the gas is:
ΔV=0.
Remember: Quasi-static paths pass through equilibrium states.
11.9Second Law of Thermodynamics
1. Kelvin–Planck statement rules out:
No heat engine can be perfectly efficient.
Remember: First Law does not set process direction; Second Law does.
2. Clausius statement forbids as the sole effect:
A refrigerator needs external work.
Remember: First Law does not set process direction; Second Law does.
3. The two standard formulations of the Second Law are:
A violation of either permits a violation of the other.
Remember: First Law does not set process direction; Second Law does.
11.10Reversible and irreversible processes
1. A reversible process must be:
Both requirements are needed for reversibility.
Remember: Reversible: system and surroundings return, with no other net change.
2. Which process is irreversible?
Free expansion cannot be undone without other changes.
Remember: Reversible: system and surroundings return, with no other net change.
3. Viscosity makes a process irreversible because it:
Dissipation converts mechanical energy into internal energy.
Remember: Reversible: system and surroundings return, with no other net change.
11.11Carnot engine and Carnot’s theorem
1. A Carnot cycle contains:
The reversible heat exchange occurs isothermally; temperature changes adiabatically.
Remember: Carnot cycle: hot isotherm → adiabatic expansion → cold isotherm → adiabatic compression.
2. A Carnot engine runs between 600 K and 300 K. Its efficiency is:
η=1−300/600=0.5.
Remember: Carnot cycle: hot isotherm → adiabatic expansion → cold isotherm → adiabatic compression.
EXERCISES
Chapter exercises
Try each question before opening its solution. Question 11.3 has four parts.
Ex 11.1NCERT
11.1A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 10⁴ J/g?
Related concept: Review the matching notes
- Step 1. Assume water density 1 kg/L: mass flow =3.0/60=0.050 kg/s.
- Step 2. Heat rate =ṁsΔT=0.050×4186×50=1.0465×10⁴ J/s.
- Step 3. Fuel rate = heat rate/(4.0×10⁴ J/g)=0.262 g/s.
Answer: 0.26 g/s (about 15.7 g/min).
Ex 11.2NCERT
11.2What amount of heat must be supplied to 2.0 × 10⁻² kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N₂ = 28; R = 8.3 J mol⁻¹ K⁻¹.)
Related concept: Review the matching notes
- Step 1. Amount n=0.020/0.028=0.714 mol. At room temperature N₂ is diatomic, so Cₚ≈7R/2.
- Step 2. Q=nCₚΔT=0.714×(7/2×8.3)×45.
Answer: ≈9.33×10² J (about 0.93 kJ).
Ex 11.3NCERT
11.3Explain why (a) two bodies at different temperatures T₁ and T₂ if brought in thermal contact do not necessarily settle at the mean temperature (T₁ + T₂)/2; (b) the coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat; (c) air pressure in a car tyre increases during driving; (d) the climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
Related concept: Review the matching notes
- Step 1. (a) Final equilibrium follows energy balance, not a simple temperature average; it depends on masses and specific heats (and possible losses/phase changes).
- Step 2. (b) High specific heat lets coolant absorb much energy with a smaller temperature rise.
- Step 3. (c) Driving heats the tyre and enclosed air; at nearly fixed volume, ideal-gas pressure rises with absolute temperature.
- Step 4. (d) Water’s high specific heat moderates coastal temperature changes by absorbing/releasing substantial heat with small temperature changes.
Answer: Each explanation follows energy conservation and the relevant heat-capacity/ideal-gas relation.
Ex 11.4NCERT
11.4A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?
Related concept: Review the matching notes
- Step 1. Insulation makes the compression adiabatic. For diatomic hydrogen at room temperature γ=Cₚ/Cᵥ≈7/5.
- Step 2. P₁V₁ᵞ=P₂V₂ᵞ, with V₂=V₁/2; therefore P₂/P₁=(V₁/V₂)ᵞ=2^(7/5).
Answer: Pressure increases by about 2.64 times.
Ex 11.5NCERT
11.5In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)
Related concept: Review the matching notes
- Step 1. Adiabatic path: Q=0 and work by system W=−22.3 J, so ΔU=Q−W=+22.3 J. This endpoint change is path-independent.
- Step 2. Other path absorbs Q=9.35×4.19=39.1765 J. From ΔU=Q−W, W=Q−ΔU=39.1765−22.3.
Answer: 16.9 J of work done by the system.
Ex 11.6NCERT
11.6Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following: (a) what is the final pressure of the gas in A and B? (b) what is the change in internal energy of the gas? (c) what is the change in the temperature of the gas? (d) do the intermediate states of the system (before settling to the final equilibrium state) lie on its P–V–T surface?
Related concept: Review the matching notes
- Step 1. This is free expansion into vacuum in an insulated enclosure: Q=0 and external work W=0. First Law gives ΔU=0.
- Step 2. For an ideal gas U depends only on T, so ΔT=0. The gas fills twice the original volume; at unchanged T, PV=nRT gives final equilibrium pressure Pᶠ=Pⁱ/2.
- Step 3. During rapid expansion the state is non-uniform and not an equilibrium state; intermediate states cannot be represented as points on an equilibrium P–V–T surface.
Answer: (a) Half the initial pressure (about 0.5 atm if initial is 1 atm); (b) ΔU=0; (c) ΔT=0 for ideal gas; (d) No.
Ex 11.7NCERT
11.7An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second, at what rate is the internal energy increasing?
Related concept: Review the matching notes
- Step 1. Use the rate form of the First Law: Q̇=Ů+Ẇ.
- Step 2. Ů=100−75=25 J/s.
Answer: 25 W.
Ex 11.8NCERT
11.8A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (11.13). Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.
Related concept: Review the matching notes
- Step 1. Work along D→E is the signed area under the straight P–V segment: Wᴅᴇ=(Pᴅ+Pᴇ)(Vᴇ−Vᴅ)/2.
- Step 2. For the isobaric leg E→F, Wᴇғ=Pᴇ(Vғ−Vᴇ), with Vғ=Vᴅ.
- Step 3. Add the two: Wtotal=(Pᴅ+Pᴇ)(Vᴇ−Vᴅ)/2−Pᴇ(Vᴇ−Vᴅ)=(Pᴅ−Pᴇ)(Vᴇ−Vᴅ)/2. Substitute graph values with consistent units.
Answer: Find the pressures and volumes at D and E from the P–V graph, then substitute them in Wtotal=(Pᴅ−Pᴇ)(Vᴇ−Vᴅ)/2. The figure is needed for the numerical answer.
KEY IDEAS
Revision
Review the main ideas and equations here, then return to the notes for details.
11.1 · Thermodynamics and heat as energy
Review this concept- Thermodynamics studies bulk states and heat–work conversion.
- Internal energy is disordered microscopic energy; whole-body motion is mechanical energy.
- Rumford’s boring experiment supported conversion of work into heat.
11.2–11.3 · Thermal equilibrium and the Zeroth Law
Review this concept- Equilibrium variables do not change with time.
- Adiabatic walls insulate; diathermic walls conduct heat.
- Zeroth Law: Tₐ=T꜀ and Tᵦ=T꜀ imply Tₐ=Tᵦ.
11.4 · Heat, internal energy and work
Review this concept- U is a state property; Q and W are path-dependent energy transfers.
- Heat requires temperature difference; work is transfer by other means.
- Q>0 into system; W>0 by system; quasi-static W=∫P dV.
11.5 · First Law of Thermodynamics
Review this concept- First Law: Q=ΔU+W (W is work by system).
- ΔU depends only on endpoints; Q and W depend on path.
- At constant P, W=PΔV; vaporising 1 g water gives W≈169 J and ΔU≈2087 J.
11.6 · Specific heat capacity
Review this concept- S=ΔQ/ΔT; s=S/m; C=S/μ.
- Solid oscillator model gives C≈3R; low-temperature departures occur.
- Ideal gas Mayer relation: Cₚ−Cᵥ=R.
11.7 · Thermodynamic state variables and equation of state
Review this concept- State variables describe equilibrium; nonequilibrium states may lack uniform P,T.
- Ideal gas equation: PV=μRT; H=U+PV.
- U,V,S,mass are extensive; P,T,ρ are intensive.
11.8 · Quasi-static and special thermodynamic processes
Review this concept- Quasi-static paths pass through equilibrium states.
- Isothermal ideal gas: PV const, W=μRT ln(V₂/V₁), Q=W.
- Adiabatic: Q=0, PVᵞ const and W=μR(T₁−T₂)/(γ−1); isochoric W=0; cyclic ΔU=0.
11.9 · Second Law of Thermodynamics
Review this concept- First Law does not set process direction; Second Law does.
- Kelvin–Planck forbids a perfect single-reservoir heat engine.
- Clausius forbids heat transfer cold→hot as the only effect; the statements are equivalent.
11.10 · Reversible and irreversible processes
Review this concept- Reversible: system and surroundings return, with no other net change.
- Requires quasi-static evolution and no dissipation.
- Friction, viscosity, free expansion, diffusion and finite gradients make processes irreversible.
11.11 · Carnot engine and Carnot’s theorem
Review this concept- Carnot cycle: hot isotherm → adiabatic expansion → cold isotherm → adiabatic compression.
- ηCarnot=1−T₂/T₁ (kelvin); no engine between these reservoirs can exceed it.
- Carnot theorem proves substance independence; reversed cycle is a refrigerator.
TEST YOURSELF
Test yourself
Choose a level, answer the questions, then check your score.
1. Internal energy excludes:
2. If A is in equilibrium with C and B is in equilibrium with C, then:
3. With W positive for work done by the system, expansion against constant P gives:
4. Using Q=ΔU+W, if Q=500 J and W=180 J, ΔU is:
5. For a complete thermodynamic cycle: