First Law and Thermodynamic Processes

01 / UNDERSTAND & APPLY

1. Draw a boundary before doing arithmetic

Thermodynamics describes a chosen system and its surroundings. A closed system exchanges energy but not matter; an open system can exchange both; an isolated system exchanges neither. A gas in a sealed piston can be closed while still exchanging heat and doing work. “Sealed” does not mean thermally insulated, and “insulated” does not necessarily mean rigid.

This lesson uses the chemistry convention: q is positive when heat enters the system, and w is positive when work is done on the system. The first law is ΔU=q+w. A physics convention often writes ΔU=Q−W, with W positive for work done by the system. Both express the same conservation statement. Translate the definition of work before comparing formulas from different sources.

Heat is energy transferred because of a temperature difference; work is energy transfer through another organized interaction, such as a moving boundary or electrical circuit. Neither is a substance stored inside a body. Internal energy U is a state property associated with the system’s microscopic energy. A system has internal energy; it does not contain a uniquely defined amount of “heat.”

Temperature is not total energy. Two samples at the same temperature can have different internal energies because their amounts, phases or compositions differ. Heating during a phase change can transfer energy without changing temperature. Conversely, compressing a gas adiabatically can raise its temperature without heat entering it.

02 / UNDERSTAND & APPLY

2. Internal energy is a state function; heat and work are paths

For a closed system with negligible changes in macroscopic kinetic and potential energy, the first law states ΔU=q+w. The change depends only on the initial and final thermodynamic states, while q and w depend on how those states are connected. Two paths can deliver different heat and work yet give the same sum.

If a system absorbs 700 J of heat and does 250 J of work on the surroundings, q=+700 J and w=−250 J, so ΔU=+450 J. The word “does” identifies the direction of work; do not assign the sign from whether the process sounds energetic. Similarly, heat released by the system has negative q even if the surroundings become warmer.

In a cycle the system returns to its initial state, so ΔU_cycle=0 and q_cycle=−w_cycle. The individual heat and work totals need not vanish. A heat engine can deliver net work over a cycle while receiving net heat, without storing a permanent increase in internal energy after every cycle.

The first law conserves energy but does not decide whether a process is spontaneous. A proposed conversion can satisfy energy conservation and still violate the second law. It also does not specify how quickly a process occurs. Thermodynamic direction and kinetics require additional principles beyond this energy balance.

03 / UNDERSTAND & APPLY

3. Boundary work uses the opposing pressure

For expansion against an external pressure, the chemistry-sign boundary work is w=−∫P_ext dV. Expansion has dV>0 and therefore negative work: energy leaves the system as it pushes the surroundings. Compression has dV<0 and positive work. The pressure in this expression is the opposing external pressure along the path.

If P_ext is constant, w=−P_ext(V₂−V₁). A gas pressure quoted only at the initial state is not automatically the external pressure throughout an irreversible expansion. In a reversible mechanical limit, the system and external pressures differ only infinitesimally and the equilibrium gas pressure can be used in the integral.

A pressure–volume product is energy, but the numerical conversion depends on units. One litre-bar equals 100 J, while one litre-atmosphere is approximately 101.325 J. Using pascals and cubic metres gives joules directly. The ideal-gas constant must use the same pressure–volume units as the calculation.

Free expansion into a vacuum has P_ext=0 and hence w=0 for this boundary-work model, even though volume increases. “Expansion” alone does not prove nonzero work. If the vessel is also insulated, q=0 and ΔU=0. For an ideal gas this implies no temperature change; for a real gas, internal energy can depend on volume as well as temperature.

04 / UNDERSTAND & APPLY

4. Why constant-pressure heat differs from constant-volume heat

Enthalpy is defined as H=U+PV. It is a state function, not simply “the heat content.” In a closed system with only pressure–volume work, q_V=ΔU for a rigid constant-volume process because boundary work vanishes. At constant pressure with the usual mechanical conditions, q_P=ΔH. Electrical or other non-PV work requires additional terms and can invalidate these simple heat identifications.

A heat capacity relates an incremental temperature change to energy transfer under specified conditions. Molar C_V and C_P have units J mol⁻¹ K⁻¹; a whole-sample heat capacity has units J K⁻¹. For a fixed amount of ideal gas of unchanged composition, dU=nC_V dT and dH=nC_P dT. Integrate if heat capacities vary with temperature; use nCΔT only when taking C approximately constant.

For an ideal gas, PV=nRT, so H=U+nRT at fixed amount. Differentiating gives C_P−C_V=R for molar heat capacities. Constant-pressure heating requires extra energy because the gas expands and does work while its internal energy rises. The difference is R per mole, not nR unless the heat capacities refer to the entire n-mole sample.

For a monatomic ideal gas in the usual temperature regime, C_V=3R/2 and C_P=5R/2. For a rigid diatomic ideal gas with rotational modes active and vibrational modes neglected, C_V≈5R/2 and C_P≈7R/2. These are conditional molecular models; at sufficiently different temperatures additional modes or other effects change the values.

05 / UNDERSTAND & APPLY

5. Isothermal ideal-gas work depends on the path

For an ideal gas of fixed composition, internal energy depends only on temperature. An isothermal process therefore has ΔU=0 and ΔH=0, while q and w can each be nonzero. The first law then gives q=−w. Do not infer q=0 merely because temperature stays constant: heat can enter exactly as energy leaves through expansion work.

For reversible isothermal expansion, substitute P=nRT/V into w=−∫P dV to obtain w_rev=−nRT ln(V₂/V₁). Because T is fixed, it comes outside the integral. Expansion to larger volume gives a positive logarithm and negative w. Compression gives a negative logarithm and positive w.

For an irreversible expansion against a constant P_ext, use −P_extΔV instead. If both paths connect the same ideal-gas temperatures and volumes, ΔU is the same but the work differs, so heat differs too. In the reversible isothermal limit, the gas delivers the maximum expansion-work magnitude between those volumes at that temperature among the usual comparable pressure-opposed paths.

The logarithm requires a dimensionless volume ratio, so any common volume unit cancels. Temperature must be kelvin when used in nRT. A pressure ratio can replace the inverse volume ratio for an isothermal ideal gas, but not indiscriminately for an adiabatic or real-gas process.

06 / UNDERSTAND & APPLY

6. Adiabatic does not mean isothermal

Adiabatic means q=0, so ΔU=w in the chemistry convention. If an ideal gas expands while doing nonzero work without heat input, its internal energy and temperature fall. Adiabatic compression raises its temperature. Insulation can make heat transfer negligible, but it does not prevent energy transfer as work.

For a reversible adiabatic ideal-gas process with constant heat capacities, nC_V dT=−P dV=−nRT dV/V. Divide by nT and integrate to obtain ln(T₂/T₁)=−(R/C_V)ln(V₂/V₁). Writing γ=C_P/C_V and using C_P−C_V=R gives TV^(γ−1)=constant and PV^γ=constant.

The word reversible is essential. An irreversible adiabatic expansion does not generally obey PV^γ=constant. Instead, combine q=0 with the actual work expression and the ideal-gas state relation as needed. Free expansion into vacuum is an extreme example: q=w=0 gives no ideal-gas temperature change, even though a reversible adiabatic expansion between the same volumes would cool the gas.

For the reversible constant-heat-capacity case, w=nC_V(T₂−T₁)=(P₂V₂−P₁V₁)/(γ−1). Expansion gives T₂<T₁ and negative work. Checking that sign is quicker than memorizing two superficially different formulas that may use opposite work conventions.

07 / UNDERSTAND & APPLY

7. Compare common paths using one state-function calculation

An isochoric path has ΔV=0 and zero PV work, while an isobaric path has constant pressure and w=−PΔV. For fixed-composition ideal gas at constant pressure, PΔV=nRΔT. Thus q_P=nC_PΔT, w=−nRΔT and ΔU=nC_VΔT. The three expressions satisfy the first law because C_P−R=C_V.

For the same initial and final ideal-gas temperatures, ΔU is the same even if pressure and volume histories differ. This does not mean any chosen isobaric and isochoric path reach the same final state: their final pressures and volumes generally differ. State-function comparisons must specify which state variables actually match.

On a reversible P–V graph, the area under the curve gives work done by the gas; the chemistry-sign work on the gas is its negative. An isochoric segment is vertical and contributes zero area. An isobaric segment is horizontal and contributes a rectangular area. A cycle’s enclosed signed area gives the net boundary work, with direction determining its sign.

For irreversible paths, plotting only equilibrium endpoint pressures and joining them with a guessed curve does not establish the external-pressure work. Use the actual specified opposing pressure history. Graphical area is meaningful only when the plotted pressure corresponds to the work integral being used.

08 / UNDERSTAND & APPLY

8. A calorimeter measures an energy balance

For a sample heated without phase change and with approximately constant specific heat c, its heat change is q=mcΔT. A calorimeter with total heat capacity C_cal has q_cal=C_calΔT. In an effectively insulated assembly, the sum of heat changes of all included parts is zero. The reaction’s heat has the opposite sign to the heat gained by the solution and calorimeter.

A rigid bomb calorimeter measures a constant-volume energy change under its experimental model, so the reaction heat relates to ΔU. A constant-pressure solution calorimeter commonly relates heat to ΔH when PV work is the only relevant work. Do not interchange these quantities merely because both instruments report a temperature rise.

If a phase transition occurs, include its latent energy separately rather than apply one mcΔT expression across the transition. If reactants have different starting temperatures, include their sensible heat changes. Heat lost to the room, evaporation or ignored apparatus heat capacity can bias the inferred reaction energy.

The thermodynamic system must stay consistent throughout the calculation. Calling the reacting chemicals the system gives one heat sign; calling the warmed water the system gives the opposite sign. Both are valid perspectives if clearly declared. A sign should be traced to energy flow, not inferred from whether the final numerical temperature is positive.

09 / UNDERSTAND & APPLY

9. Original process and sign practice

These original problems use the chemistry work convention and closed systems unless stated otherwise. Ideal-gas equations require fixed composition and the assumptions attached to each formula. A concise answer should state why a term vanishes before substituting numbers.

Use one final audit: the heat and work signs should match their physical directions; q+w must equal the state-function change; no reversible-only formula should have been applied to an unspecified irreversible path. Distinguish a model assumption from a conclusion proved by the first law.