First law of thermodynamics

01 / Heat, work and thermodynamic paths

Choose the system and name the energy transfers

Thermodynamics begins by drawing a boundary around a system. For a closed gas sample, matter does not cross the boundary, but energy can. Heat is energy transferred because of a temperature difference; work is energy transferred by other organized interactions such as a moving piston.

Internal energy U is a property of the system’s state. It includes microscopic kinetic and interaction energies, not the bulk kinetic energy of the whole container or its gravitational potential energy as a whole. A body contains internal energy; heat and work describe transfers across its boundary, not substances stored inside it.

In this chapter, Q is positive when heat enters the gas, and W is positive when the gas does work on its surroundings. With negligible changes in bulk kinetic and potential energy, the first law is ΔU = Q − W. If work is instead defined as work done on the gas, the equivalent form is ΔU = Q + W_on.

Write the convention before substituting numbers. Compression normally gives negative W in the by-the-gas convention. Cooling by heat loss gives negative Q. A negative result can be physically correct; its meaning depends on which transfer was defined positive.

\Delta U=Q-W,\qquad W_{\mathrm{on}}=-W

Closed system; changes of bulk kinetic and potential energy are neglected. Include other work modes explicitly if present.

Explain this equation

Read it in words

The change in stored internal energy equals heat entering the gas minus work the gas delivers.

Symbols and units

$\Delta U$: Final internal energy minus initial internal energyJ
$Q$: Heat transfer into the chosen gas systemJ
$W$: Work done by the chosen gas systemJ

Signs and operations

Positive Q enters the gas; positive W leaves it. Work done on the gas has negative W. Delta means final minus initial.

When it applies

Closed system; changes in bulk kinetic and gravitational energy negligible or treated separately. Identify any work modes beyond pressure–volume work.

Make it concrete

If 90 J enters as heat and the gas delivers 25 J of work, ΔU = 65 J. Compressing it with 25 J of work instead gives ΔU = 115 J.

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Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Equation (LaTeX): \Delta U=Q-W,\qquad W_{\mathrm{on}}=-W
Read it as: The change in stored internal energy equals heat entering the gas minus work the gas delivers.
Symbols:
\Delta U: Final internal energy minus initial internal energy; J
Q: Heat transfer into the chosen gas system; J
W: Work done by the chosen gas system; J
Signs and operations: Positive Q enters the gas; positive W leaves it. Work done on the gas has negative W. Delta means final minus initial.
Conditions: Closed system; changes in bulk kinetic and gravitational energy negligible or treated separately. Identify any work modes beyond pressure–volume work.

Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

WORKED EXAMPLE

Interpret a compression with heat loss

A gas is compressed while the surroundings do 90 J of work on it. During the process it transfers 35 J of heat to the surroundings. Find its internal-energy change.

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STOP & THINK

Which statement uses heat and internal energy correctly?

The gas stores 100 J of heat
The gas gains 100 J of internal energy through a specified energy-transfer process
Heat and temperature are the same quantity

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Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Heat, work and thermodynamic paths
Focus: Choose the system and name the energy transfers
Learning goals:
- Define heat, work and internal energy for a chosen closed system.
- Use a consistent sign convention through compression and expansion.
- Distinguish external-pressure work from an equilibrium p–V integral.
- Calculate ideal-gas internal-energy changes independently of path.
- Derive C_p − C_V = R and identify the limits of isothermal and adiabatic formulas.
- Explain why free expansion differs from reversible expansion.
- Construct a complete energy ledger for a cycle and verify closure.
Equations in this section (LaTeX):
\Delta U=Q-W,\qquad W_{\mathrm{on}}=-W
Meaning: Closed system; changes of bulk kinetic and potential energy are neglected. Include other work modes explicitly if present.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

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02 / Heat, work and thermodynamic paths

A moving boundary gives a pressure–volume work integral

For a piston of area A moving outward by dx against an external pressure p_ext, the displaced volume is dV = A dx. The work transferred against that external pressure is p_ext A dx = p_ext dV. Integrating over the actual volume change gives W = ∫p_ext dV for this boundary-work mode.

During a quasistatic mechanically reversible process, the external resisting pressure is arbitrarily close to the gas pressure, so W = ∫p dV along the equilibrium p–V path. This is the signed area under the curve. Expansion gives positive work, while compression reverses the integration direction and gives negative work.

During a rapid irreversible expansion, the gas may not have a single equilibrium pressure throughout its interior. Do not invent an equilibrium p–V path and integrate the initial gas pressure as though it remained constant. The actual external resisting pressure and the mechanical model determine the boundary work.

A rigid vessel has dV = 0, so its pressure–volume boundary work is zero. That does not forbid every other kind of work: a stirrer, electrical heater or rotating shaft can still transfer energy. The common statement constant volume means no work is valid only when pressure–volume work is the sole allowed work mode.

W=\int_{V_i}^{V_f}p_{\mathrm{ext}}\,dV\quad\longrightarrow\quad\int_{V_i}^{V_f}p\,dV\ \text{for a reversible quasistatic path}

Pressure must be in Pa and volume in m³ for work in joules. One kPa·L equals one joule.

WORKED EXAMPLE

Compute boundary work with consistent units

A gas expands from 2.0 L to 5.0 L against a constant external pressure of 80 kPa. Other work modes are absent. Find the work done by the gas.

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Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Heat, work and thermodynamic paths
Focus: A moving boundary gives a pressure–volume work integral
Learning goals:
- Define heat, work and internal energy for a chosen closed system.
- Use a consistent sign convention through compression and expansion.
- Distinguish external-pressure work from an equilibrium p–V integral.
- Calculate ideal-gas internal-energy changes independently of path.
- Derive C_p − C_V = R and identify the limits of isothermal and adiabatic formulas.
- Explain why free expansion differs from reversible expansion.
- Construct a complete energy ledger for a cycle and verify closure.
Equations in this section (LaTeX):
W=\int_{V_i}^{V_f}p_{\mathrm{ext}}\,dV\quad\longrightarrow\quad\int_{V_i}^{V_f}p\,dV\ \text{for a reversible quasistatic path}
Meaning: Pressure must be in Pa and volume in m³ for work in joules. One kPa·L equals one joule.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

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03 / Heat, work and thermodynamic paths

Internal energy depends on endpoints; heat and work depend on the route

Two different processes can connect the same equilibrium states. Because U is a state function, both have the same ΔU. Their work can differ because their pressure–volume paths differ. The first law then requires their heat transfers to differ by the same amount as their works.

For a classical ideal gas of fixed composition, U depends on temperature. If its molar heat capacity C_V is constant over the range, ΔU = nC_VΔT. The subscript V identifies how this heat capacity is defined; the internal-energy formula can be used between endpoints of other paths too.

For a monatomic ideal gas with only translational degrees of freedom active, C_V = 3R/2 and U = 3nRT/2 up to a constant. For a diatomic gas in a range where translation and rotation are active but vibration is negligible, C_V is approximately 5R/2. These are models tied to active energy modes, not universal constants for every gas and temperature.

For any fixed-composition ideal gas with a specified constant C_V, combining pV = nRT with the energy relation gives ΔU = (C_V/R)Δ(pV). This can be convenient when states are supplied as graph coordinates rather than temperatures.

\Delta U=nC_V(T_f-T_i)=\frac{C_V}{R}(p_fV_f-p_iV_i)

Use a constant molar C_V only when appropriate over the temperature range. For variable C_V, integrate nC_V(T)dT.

WORKED EXAMPLE

Compare two paths between the same gas states

A monatomic ideal gas goes from A: p = 100 kPa, V = 1.0 L to B: p = 200 kPa, V = 3.0 L. Path I expands at 100 kPa and then heats at fixed volume. Path II heats at fixed volume first and then expands at 200 kPa. Treat both as quasistatic and allow only pressure–volume work. Find ΔU, W and Q for each.

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Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Heat, work and thermodynamic paths
Focus: Internal energy depends on endpoints; heat and work depend on the route
Learning goals:
- Define heat, work and internal energy for a chosen closed system.
- Use a consistent sign convention through compression and expansion.
- Distinguish external-pressure work from an equilibrium p–V integral.
- Calculate ideal-gas internal-energy changes independently of path.
- Derive C_p − C_V = R and identify the limits of isothermal and adiabatic formulas.
- Explain why free expansion differs from reversible expansion.
- Construct a complete energy ledger for a cycle and verify closure.
Equations in this section (LaTeX):
\Delta U=nC_V(T_f-T_i)=\frac{C_V}{R}(p_fV_f-p_iV_i)
Meaning: Use a constant molar C_V only when appropriate over the temperature range. For variable C_V, integrate nC_V(T)dT.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

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04 / Heat, work and thermodynamic paths

Follow the energy while changing the route

A state is a destination; a process describes how the system got there. For an ideal gas with fixed composition, matching both pressure and volume at an endpoint also matches temperature. Start every comparison by checking that the amounts of gas and the endpoints really agree.

In the explorer, try the three routes between the same two states. At the end, the stored-energy changes agree. The route that expands against a larger pressure does more work; to reach the same final internal energy, it must also receive more heat. Move the cursor back to the corner of each route to explain each leg separately.

Then try the isotherm and adiabat. They begin at the same state, but generally finish at different pressures and temperatures for the same chosen final volume. Their different internal-energy changes no longer contradict the state-function rule. The final-pressure control is disabled because the selected process law supplies it.

INTERACTIVE LAB

Compare paths, not just formulas

Predict which of the first three routes transfers the most work for an expansion to a higher final pressure. Check at the endpoint, then choose a compression and explain the reversed work sign.

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WORKED EXAMPLE

Use a trapezium for a straight pressure–volume route

A monatomic ideal gas moves reversibly along a straight p–V segment from 100 kPa, 1 L to 200 kPa, 2 L. Find W, ΔU and Q. Only boundary work is allowed.

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STOP & THINK

For two routes with the same ideal-gas endpoints, route B does 80 J more work by the gas. How does its heat input compare?

80 J greater
80 J smaller
It must be identical

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Heat, work and thermodynamic paths
Focus: Follow the energy while changing the route
Learning goals:
- Define heat, work and internal energy for a chosen closed system.
- Use a consistent sign convention through compression and expansion.
- Distinguish external-pressure work from an equilibrium p–V integral.
- Calculate ideal-gas internal-energy changes independently of path.
- Derive C_p − C_V = R and identify the limits of isothermal and adiabatic formulas.
- Explain why free expansion differs from reversible expansion.
- Construct a complete energy ledger for a cycle and verify closure.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

05 / Heat, work and thermodynamic paths

Why constant-pressure heating requires more energy

Heating at constant volume with only pressure–volume work gives W = 0, so Q_V = ΔU = nC_VΔT. At constant pressure, the gas expands while warming, and part of the supplied heat also leaves as boundary work.

For fixed amount of ideal gas at constant pressure, pΔV = nRΔT. Therefore Q_p = ΔU + W = n(C_V + R)ΔT. Defining the molar constant-pressure heat capacity C_p gives C_p − C_V = R.

The heat capacities refer to specified process constraints. It is generally wrong to use Q = nC_VΔT for an arbitrary process merely because internal energy uses C_V. It is equally wrong to use C_p for every gas heating process. Determine work or the process condition first.

The ratio γ = C_p/C_V is greater than one in the ordinary ideal-gas model. For monatomic gas γ = 5/3; for the translation-plus-rotation diatomic model γ = 7/5. A temperature-dependent heat capacity may require more detailed integration and can make a constant-γ approximation inaccurate.

C_p-C_V=R,\qquad \gamma=\frac{C_p}{C_V}

These are molar heat capacities for an ideal gas of fixed composition. Specific heat capacities per kilogram obey the corresponding relation with the specific gas constant.

WORKED EXAMPLE

Separate energy stored from expansion work

One mole of monatomic ideal gas is heated quasistatically at constant pressure through 20 K. Use R = 8.31 J/(mol·K). Find ΔU, W and Q.

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Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Heat, work and thermodynamic paths
Focus: Why constant-pressure heating requires more energy
Learning goals:
- Define heat, work and internal energy for a chosen closed system.
- Use a consistent sign convention through compression and expansion.
- Distinguish external-pressure work from an equilibrium p–V integral.
- Calculate ideal-gas internal-energy changes independently of path.
- Derive C_p − C_V = R and identify the limits of isothermal and adiabatic formulas.
- Explain why free expansion differs from reversible expansion.
- Construct a complete energy ledger for a cycle and verify closure.
Equations in this section (LaTeX):
C_p-C_V=R,\qquad \gamma=\frac{C_p}{C_V}
Meaning: These are molar heat capacities for an ideal gas of fixed composition. Specific heat capacities per kilogram obey the corresponding relation with the specific gas constant.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

06 / Heat, work and thermodynamic paths

Constant temperature does not mean zero heat transfer

For a fixed amount of ideal gas, an isothermal process has ΔU = 0 because its internal energy depends only on T. The first law then gives Q = W. If the gas expands and does positive work while maintaining its temperature, an equal amount of heat must enter under this model.

For a reversible quasistatic isothermal path, substitute p = nRT/V into the work integral. Since T is constant, W = nRT ln(V_f/V_i). Expansion gives a positive logarithm and compression a negative one.

This logarithmic expression is not the work of every isothermal expansion. An irreversible expansion against a constant external pressure must use that external pressure for boundary work. The equation of state still describes the equilibrium endpoints, but it does not turn an irreversible path into a reversible one.

For a real gas whose internal energy also depends on volume, constant temperature does not necessarily imply ΔU = 0. The ideal-gas assumption is essential to that shortcut.

W_{\mathrm{rev,iso}}=nRT\ln\!\left(\frac{V_f}{V_i}\right),\qquad\Delta U=0,\qquad Q=W

The work expression assumes a fixed amount of ideal gas on a reversible isothermal path.

WORKED EXAMPLE

Check the logarithm with a doubling expansion

A fixed amount of ideal gas expands reversibly at constant temperature from V to 2V. Its constant nRT is 300 J. Find work, internal-energy change and heat input.

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Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Heat, work and thermodynamic paths
Focus: Constant temperature does not mean zero heat transfer
Learning goals:
- Define heat, work and internal energy for a chosen closed system.
- Use a consistent sign convention through compression and expansion.
- Distinguish external-pressure work from an equilibrium p–V integral.
- Calculate ideal-gas internal-energy changes independently of path.
- Derive C_p − C_V = R and identify the limits of isothermal and adiabatic formulas.
- Explain why free expansion differs from reversible expansion.
- Construct a complete energy ledger for a cycle and verify closure.
Equations in this section (LaTeX):
W_{\mathrm{rev,iso}}=nRT\ln\!\left(\frac{V_f}{V_i}\right),\qquad\Delta U=0,\qquad Q=W
Meaning: The work expression assumes a fixed amount of ideal gas on a reversible isothermal path.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

07 / Heat, work and thermodynamic paths

No heat transfer is a different condition from no temperature change

An adiabatic process has Q = 0. The first law gives ΔU = −W, so an ideal gas cools when it does positive expansion work and warms when it is compressed by work, assuming no other energy changes.

For a reversible quasistatic adiabatic process of an ideal gas with constant heat capacities, nC_V dT = −p dV. Substituting p = nRT/V and integrating gives TV^(γ−1) = constant, equivalently pV^γ = constant.

The extra conditions matter. Adiabatic only means no heat transfer; it does not automatically mean reversible or quasistatic. Free expansion into vacuum is adiabatic if the vessel is insulated, but it does not follow pV^γ = constant.

On a p–V diagram through the same initial state, the reversible adiabat is steeper than the isotherm for γ > 1. During expansion, the adiabatic gas loses internal energy, so its temperature and pressure fall more than for an isothermal expansion over the same volume change.

Q=0,\qquad pV^\gamma=\mathrm{constant},\qquad TV^{\gamma-1}=\mathrm{constant}

The power laws additionally require a reversible quasistatic ideal-gas process with constant γ.

Compare an isotherm and a reversible adiabat

Inspect a plotted point: V/V₁ = 1

Isothermal: 1
Adiabatic, γ = 5/3: 1

Both paths begin at the same state. For γ = 5/3, the adiabatic pressure falls faster as volume increases because expansion work comes from internal energy. Values are normalized to the initial state. Lines connect the plotted samples.

WORKED EXAMPLE

Use the appropriate volume power and confirm energy direction

A monatomic ideal gas expands reversibly and adiabatically from volume V₁ to 8V₁. Its initial temperature is 400 K. Find the final temperature under the constant-γ model.

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Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Heat, work and thermodynamic paths
Focus: No heat transfer is a different condition from no temperature change
Learning goals:
- Define heat, work and internal energy for a chosen closed system.
- Use a consistent sign convention through compression and expansion.
- Distinguish external-pressure work from an equilibrium p–V integral.
- Calculate ideal-gas internal-energy changes independently of path.
- Derive C_p − C_V = R and identify the limits of isothermal and adiabatic formulas.
- Explain why free expansion differs from reversible expansion.
- Construct a complete energy ledger for a cycle and verify closure.
Equations in this section (LaTeX):
Q=0,\qquad pV^\gamma=\mathrm{constant},\qquad TV^{\gamma-1}=\mathrm{constant}
Meaning: The power laws additionally require a reversible quasistatic ideal-gas process with constant γ.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

08 / Heat, work and thermodynamic paths

Use the process conditions before choosing a formula

Process names describe different constraints. Isothermal fixes temperature. Adiabatic rules out heat transfer. Isochoric fixes volume. Isobaric fixes pressure. These are not four mutually exclusive kinds of substance: a gas sample can follow one condition on one leg and another condition later.

Begin with the experimental sentence. A rigid container removes boundary work. Insulation removes heat flow across that boundary. A frictionless, slowly moving piston can justify an equilibrium pressure path, provided the driving imbalance is infinitesimal. Contact with a reservoir does not by itself guarantee the gas stays at that temperature during a rapid process.

Compute the endpoints, then internal-energy change, then the mechanically transferred work. Use the first law for the remaining quantity. This order prevents a common mistake: treating Cv as the coefficient of heat for every route. Cv controls ideal-gas internal energy, while heat also has to pay for work.

Translate a condition into a calculation
Condition and model	Work by gas	Heat into gas
Rigid; only p–V work	W = 0	Q = ΔU
Constant gas pressure; reversible	W = p(Vf − Vi)	Q = nCpΔT for ideal gas, constant Cp
Isothermal ideal gas; reversible	W = nRT ln(Vf/Vi)	Q = W
Adiabatic; only p–V work	W = −ΔU	Q = 0
Insulated free expansion into vacuum	W = 0	Q = 0; ΔT = 0 for ideal gas

WORKED EXAMPLE

Supply the same heat under different constraints

Two separate samples each contain 1 mol of the same ideal gas with Cv = 5R/2. Each receives 100 J. One is rigid; the other expands reversibly at constant pressure. Use R = 8 J/(mol K) for this model. Compare temperature rises.

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STOP & THINK

An insulated gas expands rapidly against a finite load. Which relation is justified without adding reversibility?

Q = 0
pV^γ = constant throughout
Temperature stays constant

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Heat, work and thermodynamic paths
Focus: Use the process conditions before choosing a formula
Learning goals:
- Define heat, work and internal energy for a chosen closed system.
- Use a consistent sign convention through compression and expansion.
- Distinguish external-pressure work from an equilibrium p–V integral.
- Calculate ideal-gas internal-energy changes independently of path.
- Derive C_p − C_V = R and identify the limits of isothermal and adiabatic formulas.
- Explain why free expansion differs from reversible expansion.
- Construct a complete energy ledger for a cycle and verify closure.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

09 / Heat, work and thermodynamic paths

An increasing occupied volume need not imply positive work

Imagine an insulated rigid container split into two chambers, one containing ideal gas and the other vacuum. Remove the partition and wait for a new equilibrium. No heat crosses the insulated outer boundary, and the gas expands against zero external pressure into the vacuum.

In the ideal free-expansion model, Q = 0 and W = 0, so ΔU = 0. For a fixed amount of ideal gas this gives the same final temperature as initially. Pressure falls because the gas occupies more volume at that temperature.

There is no sequence of uniform equilibrium gas states during the rapid spreading. A curve drawn by connecting the endpoints on an isotherm is not the actual process path, and integrating under that imagined curve would assign work that was not transferred.

The process is irreversible even though the ideal-gas temperature is unchanged. Energy conservation does not determine whether a process can spontaneously reverse; that question requires entropy and the second law.

STOP & THINK

An insulated ideal gas freely expands into vacuum within a rigid outer vessel. Which conclusion follows in the elementary model?

Q = 0, W = 0 and ΔU = 0
Q = 0 and W must be positive because volume grows
The process obeys pV^γ = constant

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Heat, work and thermodynamic paths
Focus: An increasing occupied volume need not imply positive work
Learning goals:
- Define heat, work and internal energy for a chosen closed system.
- Use a consistent sign convention through compression and expansion.
- Distinguish external-pressure work from an equilibrium p–V integral.
- Calculate ideal-gas internal-energy changes independently of path.
- Derive C_p − C_V = R and identify the limits of isothermal and adiabatic formulas.
- Explain why free expansion differs from reversible expansion.
- Construct a complete energy ledger for a cycle and verify closure.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

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10 / Heat, work and thermodynamic paths

The same endpoints can hide very different transfers

Take a gas from a smaller to a larger volume while keeping the same final temperature. A reversible isothermal expansion delivers work and absorbs matching heat. Removing a partition into a vacuum inside an insulated rigid enclosure delivers no work and absorbs no heat. Both can connect the same ideal-gas endpoints, but they are physically different experiments.

The reversible route is a sequence of equilibrium states, so drawing p = nRT/V throughout is meaningful. Free expansion has equilibrium endpoints but not a uniform equilibrium pressure at every instant in between. Draw the endpoints and describe the event; do not fill the space between them with a fictitious work-producing curve.

For the gas, entropy changes by nR ln(Vf/Vi) between these equal-temperature endpoints. On the reversible route, surroundings lose that entropy through reversible heat transfer. In insulated free expansion, surroundings supply none; the gas’s positive entropy change is produced internally. Computing entropy through a convenient reversible comparison path does not change the actual process into a reversible one.

WORKED EXAMPLE

Compare an isotherm with partition removal

An ideal gas starts with pV = 240 J and expands to three times its volume. Compare a reversible isothermal route with insulated free expansion. The final temperature equals the initial temperature for both routes.

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STOP & THINK

During insulated ideal-gas free expansion, Q = 0. Must the gas entropy change be zero?

Yes, because entropy always equals Q/T
No; entropy can increase through irreversibility
No; entropy must decrease

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Heat, work and thermodynamic paths
Focus: The same endpoints can hide very different transfers
Learning goals:
- Define heat, work and internal energy for a chosen closed system.
- Use a consistent sign convention through compression and expansion.
- Distinguish external-pressure work from an equilibrium p–V integral.
- Calculate ideal-gas internal-energy changes independently of path.
- Derive C_p − C_V = R and identify the limits of isothermal and adiabatic formulas.
- Explain why free expansion differs from reversible expansion.
- Construct a complete energy ledger for a cycle and verify closure.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

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11 / Heat, work and thermodynamic paths

A cycle returns the working substance to its initial state

A thermodynamic cycle returns all state properties of the working substance to their starting values. In particular ΔU_cycle = 0. Summing the first law over the cycle gives Q_net = W_net.

Individual legs can have nonzero internal-energy changes, heat inputs and heat outputs. Zero net ΔU does not mean each leg has zero ΔU, and zero net stored-energy change does not mean zero heat or work. The working substance can transfer energy from one part of its surroundings to another while returning to its initial state.

For a quasistatic p–V cycle with pressure–volume work only, the signed enclosed area gives net work by the gas. A clockwise loop gives positive work because expansion occurs on average at higher pressure than compression. Reversing the loop reverses the sign of work.

Heat input is the sum of positive heat transfers, while heat rejected is the magnitude of negative transfers. Neither is generally equal to net heat. This distinction becomes essential when computing an engine efficiency.

\Delta U_{\mathrm{cycle}}=0,\qquad Q_{\mathrm{net}}=W_{\mathrm{net}}=\oint p\,dV

The area interpretation uses a quasistatic p–V cycle and pressure–volume work. Other work modes require their own accounting.

WORKED EXAMPLE

Read a rectangular cycle from its directions

A gas follows a quasistatic cycle: expand from 1 L to 3 L at 200 kPa, cool at fixed 3 L to 100 kPa, compress to 1 L at 100 kPa, then heat at fixed 1 L back to 200 kPa. Find net work and net heat.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Heat, work and thermodynamic paths
Focus: A cycle returns the working substance to its initial state
Learning goals:
- Define heat, work and internal energy for a chosen closed system.
- Use a consistent sign convention through compression and expansion.
- Distinguish external-pressure work from an equilibrium p–V integral.
- Calculate ideal-gas internal-energy changes independently of path.
- Derive C_p − C_V = R and identify the limits of isothermal and adiabatic formulas.
- Explain why free expansion differs from reversible expansion.
- Construct a complete energy ledger for a cycle and verify closure.
Equations in this section (LaTeX):
\Delta U_{\mathrm{cycle}}=0,\qquad Q_{\mathrm{net}}=W_{\mathrm{net}}=\oint p\,dV
Meaning: The area interpretation uses a quasistatic p–V cycle and pressure–volume work. Other work modes require their own accounting.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

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12 / Heat, work and thermodynamic paths

Calculate each leg before adding heat inputs

A table of ΔU, W and Q for each leg prevents sign errors and distinguishes total heat input from net heat. First determine endpoint temperatures or pV products. Next compute work from the path. Finally use Q = ΔU + W for each leg.

For the rectangular cycle just described, a monatomic ideal gas has U changes proportional to changes in pV. The state products are 200 J, 600 J, 300 J and 100 J in order. Multiplying their differences by 3/2 gives the internal-energy changes.

The heat inputs occur during high-pressure expansion and the final constant-volume heating. Cooling and compression reject heat. Adding only the positive Q entries gives 1150 J of heat input, while the rejected magnitude is 950 J. Their difference is the 200 J net work.

A cycle calculation should close in three ways: the sum of ΔU is zero, the sum of Q equals the sum of W, and the p–V area agrees with net work. These checks catch errors that a single final number may hide.

Monatomic ideal-gas energy ledger for the preceding rectangular cycle
Leg	ΔU (J)	W by gas (J)	Q into gas (J)
A→B: high-pressure expansion	+600	+400	+1000
B→C: constant-volume cooling	−450	0	−450
C→D: low-pressure compression	−300	−200	−500
D→A: constant-volume heating	+150	0	+150
Cycle total	0	+200	+200

WORKED EXAMPLE

Find heat input without confusing it with net heat

For the monatomic rectangular cycle in the table, determine total heat absorbed and total heat rejected as positive magnitudes.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Heat, work and thermodynamic paths
Focus: Calculate each leg before adding heat inputs
Learning goals:
- Define heat, work and internal energy for a chosen closed system.
- Use a consistent sign convention through compression and expansion.
- Distinguish external-pressure work from an equilibrium p–V integral.
- Calculate ideal-gas internal-energy changes independently of path.
- Derive C_p − C_V = R and identify the limits of isothermal and adiabatic formulas.
- Explain why free expansion differs from reversible expansion.
- Construct a complete energy ledger for a cycle and verify closure.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

13 / Heat, work and thermodynamic paths

A smooth process can change the direction of heat transfer

Drawing one straight line does not guarantee that heat flows one way all along it. The energy stored and work delivered both change continuously. A positive total Q for a leg may be the difference between an earlier positive heat input and a later heat rejection.

For a fixed amount of ideal gas with constant Cv on a reversible pressure–volume path, δQ = (Cv/R)d(pV) + p dV. Differentiate the specified p(V) before deciding where heat enters. In an expansion, δQ/dV and δQ have the same sign; for compression the sign of dV must also be included.

For engine efficiency, integrate the positive heat portions over the entire cycle. Summing only positive net values of arbitrarily named legs can miss heat that entered and later left within one leg. The enclosed p–V area still gives net work, so it remains a useful independent check.

\frac{\delta Q}{dV}=\frac{C_V}{R}\left(p+V\frac{dp}{dV}\right)+p

This relation assumes reversible quasistatic p–V work, fixed ideal-gas composition and constant molar Cv. The derivative has pressure units when V is in m³.

Explain this equation

Read it in words

Heat added per small volume increase equals the change of ideal-gas internal energy per volume plus the pressure work per volume.

Symbols and units

$\delta Q$: Small heat transfer into the gasJ
$V$: Gas volumem³
$p$: Equilibrium gas pressure along the reversible pathPa
$C_V$: Molar heat capacity at constant volumeJ/(mol K)
$R$: Molar gas constantJ/(mol K)
$dp/dV$: Local slope of the specified pressure–volume pathPa/m³

Signs and operations

The product rule gives d(pV) = p dV + V dp. Positive δQ/dV means heat input on expansion; on compression dV is negative, so check the product. Heat is a transfer, not a stored state function.

When it applies

Fixed-composition ideal gas, constant Cv and reversible quasistatic boundary work only. Use SI pressure and volume, or explicitly keep the kPa·L = J conversion.

Make it concrete

For constant pressure, dp/dV = 0. A monatomic gas gives δQ/dV = (3/2 + 1)p = (5/2)p, agreeing with Cp = 5R/2.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Equation (LaTeX): \frac{\delta Q}{dV}=\frac{C_V}{R}\left(p+V\frac{dp}{dV}\right)+p
Read it as: Heat added per small volume increase equals the change of ideal-gas internal energy per volume plus the pressure work per volume.
Symbols:
\delta Q: Small heat transfer into the gas; J
V: Gas volume; m³
p: Equilibrium gas pressure along the reversible path; Pa
C_V: Molar heat capacity at constant volume; J/(mol K)
R: Molar gas constant; J/(mol K)
dp/dV: Local slope of the specified pressure–volume path; Pa/m³
Signs and operations: The product rule gives d(pV) = p dV + V dp. Positive δQ/dV means heat input on expansion; on compression dV is negative, so check the product. Heat is a transfer, not a stored state function.
Conditions: Fixed-composition ideal gas, constant Cv and reversible quasistatic boundary work only. Use SI pressure and volume, or explicitly keep the kPa·L = J conversion.

Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

WORKED EXAMPLE

Locate a hidden heat reversal

A monatomic ideal gas expands reversibly along p = 200 − 50V, with p in kPa and V in L, from V = 1 L to V = 3 L. Identify where heat changes direction and separate its positive input from its net heat.

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STOP & THINK

One process leg has net Q = +40 J. Can it have absorbed 70 J and rejected 30 J during that leg?

Yes
No

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Heat, work and thermodynamic paths
Focus: A smooth process can change the direction of heat transfer
Learning goals:
- Define heat, work and internal energy for a chosen closed system.
- Use a consistent sign convention through compression and expansion.
- Distinguish external-pressure work from an equilibrium p–V integral.
- Calculate ideal-gas internal-energy changes independently of path.
- Derive C_p − C_V = R and identify the limits of isothermal and adiabatic formulas.
- Explain why free expansion differs from reversible expansion.
- Construct a complete energy ledger for a cycle and verify closure.
Equations in this section (LaTeX):
\frac{\delta Q}{dV}=\frac{C_V}{R}\left(p+V\frac{dp}{dV}\right)+p
Meaning: This relation assumes reversible quasistatic p–V work, fixed ideal-gas composition and constant molar Cv. The derivative has pressure units when V is in m³.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

14 / Heat, work and thermodynamic paths

Use a fixed calculation order for an unfamiliar process

Start with the system boundary and sign convention. Determine whether the amount is fixed, whether the gas model is ideal and which energy modes or heat capacities are assumed. Identify the equilibrium endpoints without assuming a path that was not stated.

Compute ΔU from state information. Compute work from the actual mechanical process, using external pressure for an irreversible boundary movement and equilibrium gas pressure only for an appropriate quasistatic path. Then obtain the remaining transfer from the first law.

For multiple stages, create one row per stage and carry the final state of each into the next. For a cycle, check closure explicitly. Do not let a process label such as adiabatic, isothermal or constant volume silently erase other possible work modes or model assumptions.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Heat, work and thermodynamic paths
Focus: Use a fixed calculation order for an unfamiliar process
Learning goals:
- Define heat, work and internal energy for a chosen closed system.
- Use a consistent sign convention through compression and expansion.
- Distinguish external-pressure work from an equilibrium p–V integral.
- Calculate ideal-gas internal-energy changes independently of path.
- Derive C_p − C_V = R and identify the limits of isothermal and adiabatic formulas.
- Explain why free expansion differs from reversible expansion.
- Construct a complete energy ledger for a cycle and verify closure.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

BRING THE IDEAS TOGETHER

What should stay with you

With work by the gas positive, ΔU = Q − W.
Internal energy is a state function; heat and work depend on the process.
Boundary work depends on the resisting pressure and the actual path.
Constant volume eliminates p–V work but does not exclude electrical or shaft work.
An ideal-gas isotherm has ΔU = 0, while an adiabat has Q = 0.
The power law pV^γ = constant needs reversible quasistatic ideal-gas conditions and constant γ.
A cycle has zero net ΔU but can transfer heat and produce net work.
Add positive and negative heat transfers separately before interpreting an engine cycle.

RETRIEVE, THEN EXPLAIN

Try it without the examples

Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.

SELF CHECK 1

A gas absorbs 75 J of heat while 25 J of work is done on it. What is its internal-energy increase?

Answer in J.

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

SELF CHECK 2

A cycle absorbs 900 J and rejects 650 J as heat, with no net change of stored energy. What net work does it deliver?

Answer in J.

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

SELF CHECK 3

Two processes connect the same equilibrium states of the same closed system. Which quantity must agree?

Work
Heat
Internal-energy change

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

SELF CHECK 4

Which extra condition is needed before using pV^γ = constant for an ideal-gas adiabatic process with constant γ?

A reversible quasistatic path
A nonzero heat input
A fixed volume

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

SELF CHECK 5

A rigid sealed vessel contains a gas and an electrical heating element. Does fixed volume imply no work can enter the system?

Yes, all work requires a volume change
No; electrical work can enter despite zero p–V work
Only if pressure stays constant

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APPLY THE CHAPTER

Take the ideas into a problem

Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.

COME BACK TO THE IDEA

What can you explain later?

Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.

Next study session

For an ideal gas between the same two temperatures, what is fixed about internal-energy change, and what need not be fixed?

Compare your reasoning

For a given amount and composition, internal-energy change is fixed by the temperatures. Heat and work separately depend on the path. Different transfers can produce the same change in state.

Revisit the explanation →

A few days later: change the situation

A reversible isothermal expansion and insulated free expansion connect the same ideal-gas endpoints. Which of ΔU, Q and W must match?

Compare your reasoning

Only ΔU must match; it is zero here. The reversible route delivers positive work and absorbs matching heat. The free expansion delivers no work and absorbs no heat. Equal endpoints do not supply an actual reversible path.

Revisit the explanation →

Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.

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What you will learn to do

Bring back one idea

Choose the system and name the energy transfers

Read it in words

Symbols and units

Signs and operations

When it applies

Make it concrete

Interpret a compression with heat loss

Which statement uses heat and internal energy correctly?

A moving boundary gives a pressure–volume work integral

Compute boundary work with consistent units

Internal energy depends on endpoints; heat and work depend on the route

Compare two paths between the same gas states

Follow the energy while changing the route

Compare paths, not just formulas

Use a trapezium for a straight pressure–volume route

For two routes with the same ideal-gas endpoints, route B does 80 J more work by the gas. How does its heat input compare?

Why constant-pressure heating requires more energy

Separate energy stored from expansion work

Constant temperature does not mean zero heat transfer

Check the logarithm with a doubling expansion

No heat transfer is a different condition from no temperature change

Compare an isotherm and a reversible adiabat

Use the appropriate volume power and confirm energy direction

Use the process conditions before choosing a formula

Supply the same heat under different constraints

An insulated gas expands rapidly against a finite load. Which relation is justified without adding reversibility?

An increasing occupied volume need not imply positive work

An insulated ideal gas freely expands into vacuum within a rigid outer vessel. Which conclusion follows in the elementary model?

The same endpoints can hide very different transfers

Compare an isotherm with partition removal

During insulated ideal-gas free expansion, Q = 0. Must the gas entropy change be zero?

A cycle returns the working substance to its initial state

Read a rectangular cycle from its directions

Calculate each leg before adding heat inputs

Find heat input without confusing it with net heat

A smooth process can change the direction of heat transfer

Read it in words

Symbols and units

Signs and operations

When it applies

Make it concrete

Locate a hidden heat reversal

One process leg has net Q = +40 J. Can it have absorbed 70 J and rejected 30 J during that leg?

Use a fixed calculation order for an unfamiliar process

What should stay with you

Try it without the examples

A gas absorbs 75 J of heat while 25 J of work is done on it. What is its internal-energy increase?

A cycle absorbs 900 J and rejects 650 J as heat, with no net change of stored energy. What net work does it deliver?

Two processes connect the same equilibrium states of the same closed system. Which quantity must agree?

Which extra condition is needed before using pV^γ = constant for an ideal-gas adiabatic process with constant γ?

A rigid sealed vessel contains a gas and an electrical heating element. Does fixed volume imply no work can enter the system?

Take the ideas into a problem

What can you explain later?

Next study session

A few days later: change the situation