core24 min studyLesson 313 in Physics

The first law and heat capacities

Use heat-in and work-on energy accounting, distinguish constant-volume and constant-pressure heating, and derive the ideal-gas capacity difference.

On this page

What you’ll understand

  • Apply the first law with a consistent sign convention.
  • Interpret heat capacities as derivatives under specified constraints.
  • Derive C pressure minus C volume for a fixed ideal-gas sample.

Heating a gas at fixed volume raises its internal energy without moving the walls. Heating it at fixed pressure generally expands it, so some energy leaves through mechanical work. These two protocols explain why a heat-capacity number needs a constraint label. The first law organizes the energy accounting even when the actual process is not reversible, while equilibrium derivatives define reproducible material properties. We will keep the distinction between a process heat and a state-function derivative visible in every calculation.

Energy changes are state changes

For a closed system at rest, neglecting changes in bulk kinetic and gravitational energy, write ΔU=Q+Won\Delta U=Q+W_{\mathrm{on}}. Heat Q is positive into the system, and work on it is positive into the system. Boundary expansion work is therefore negative. The law applies between equilibrated endpoints even when the intermediate state cannot be assigned a single pressure or temperature. Internal energy excludes the separately tracked motion or elevation of the apparatus as a whole.

A cyclic device returns to its original internal state, so its net internal-energy change is zero. Net heat then balances minus net work on the device. An adiabatic process has zero heat exchange, not necessarily zero temperature change: compression can increase internal energy by positive work input. Conversely, ideal-gas free expansion has both heat and work zero and leaves temperature unchanged because that model’s internal energy depends only on temperature. These are different mechanisms, despite both being adiabatic.

dU=δQ+δWon,ΔU=Q+Won,δQrev=dU+p dVdU=\delta Q+\delta W_{\mathrm{on}},\qquad \Delta U=Q+W_{\mathrm{on}},\qquad \delta Q_{\mathrm{rev}}=dU+p\,dV
The first-law balance uses heat-in and work-on signs. The final form additionally assumes reversible quasistatic pressure-volume work with no other work channels; all energy transfers are in joules.↔ Scroll wide equations horizontally.

Capacity labels specify the experiment

At fixed volume and amount, pressure-volume work vanishes. Reversible heating then gives CV=(∂U/∂T)V,νC_V=(\partial U/\partial T)_{V,\nu}. At fixed pressure define enthalpy H=U+pVH=U+pV; its differential is dH=dU+p dV+V dpdH=dU+p\,dV+V\,dp. Along a constant-pressure path with only pressure-volume work, the first law gives heat equal to the enthalpy change. Thus Cp=(∂H/∂T)p,νC_p=(\partial H/\partial T)_{p,\nu}. Both are total sample capacities in joules per kelvin.

Molar capacities divide these total quantities by the amount in moles; mass-specific capacities divide by mass. A numerical comparison is meaningless until the normalization agrees. Capacities can depend on temperature and on the equilibrium state, so multiplying one constant value by a large temperature interval is an approximation unless constancy is part of the model. Phase changes also require latent heat and cannot generally be represented by a finite smooth single-phase capacity across the transition.

CV=(∂U∂T)V,ν,Cp=(∂H∂T)p,ν,Cp−CV=νR(ideal gas)C_V=\left(\frac{\partial U}{\partial T}\right)_{V,\nu},\qquad C_p=\left(\frac{\partial H}{\partial T}\right)_{p,\nu},\qquad C_p-C_V=\nu R\quad\text{(ideal gas)}
The ideal-gas difference assumes fixed amount and an equilibrium ideal gas with U depending only on temperature. It concerns total capacities; molar capacities instead differ by R.↔ Scroll wide equations horizontally.

Derive the difference rather than memorize a ratio

For an ideal gas, H(T)=U(T)+νRTH(T)=U(T)+\nu RT. Differentiating with respect to temperature gives Cp=CV+νRC_p=C_V+\nu R. The extra term is the energy that accompanies expansion during constant-pressure heating. It does not require that the gas be monatomic. A monatomic classical ideal gas further has U=3νRT/2U=3\nu RT/2 and therefore a specific capacity ratio, whereas molecular rotations or vibrations change U and the capacities without changing the ideal-gas difference.

For a nonideal fluid, internal energy can depend on volume as well as temperature, and the difference is determined by expansion and compressibility responses rather than simply the gas constant. Later Maxwell relations provide that general identity. Even here the first law diagnoses common errors: isothermal does not mean zero heat, and adiabatic does not mean zero work. Decide which coordinate or transfer is held fixed before substituting a capacity formula into the energy balance.

Two ways to warm an ideal gas

A fixed ideal-gas sample has νR=10 J K−1\nu R=10\,\mathrm{J\,K^{-1}} and constant CV=15 J K−1C_V=15\,\mathrm{J\,K^{-1}}. Warm it by 20 K, first at fixed volume and then in a separate reversible constant-pressure run. Find heat, work on the gas, and internal-energy change for each.

  1. For either run the ideal-gas internal-energy change is 15(20)=300 J15(20)=300\,\mathrm J. The identical temperature change fixes this state-function difference.

  2. At fixed volume, work is zero, so the heat input is 300 J. The first law gives 300=300+0300=300+0.

  3. At fixed pressure, Cp=15+10=25 J K−1C_p=15+10=25\,\mathrm{J\,K^{-1}}, so heat input is 500 J. Expansion work on the gas is −νRΔT=−200 J-\nu R\Delta T=-200\,\mathrm J, giving 300=500−200300=500-200.

The result

Both protocols raise internal energy by 300 J. Constant-pressure heating requires 200 J more heat because the gas exports that energy as expansion work.

Explore the model

Interactive lab

Interactive lab · function

A local slope is a response

Use the parabola only as an analogy for a state-function response derivative. Reset, set Coefficient a = 0.5 and Position x = 1, and predict slope 1. For a fresh comparison Reset, set a = 1 and x = 1, and predict slope 2.

A local slope is a responseThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe lab differentiates y = a x². It does not model an ideal gas, calculate heat capacities, or define a heating path; the physical constraint labels come from the lesson equations.

Check your understanding

A closed system receives 120 J of heat and does 50 J of work on its surroundings. What is its internal-energy change?

Try it yourself

An insulated closed gas sample has constant CV=20 J K−1C_V=20\,\mathrm{J\,K^{-1}} and receives 160 J of compression work. Find heat exchange, internal-energy change, and temperature rise, assuming its ideal-gas caloric model applies.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Adiabatic fixes heat, while positive work on the gas can still change temperature.

Compare with the solution

Insulation gives Q = 0. The first law gives ΔU=0+160=160 J\Delta U=0+160=160\,\mathrm J. With constant capacity, ΔT=160/20=8 K\Delta T=160/20=8\,\mathrm K. A temperature increase is fully compatible with adiabatic compression: the added internal energy enters mechanically rather than through a temperature-driven heat transfer.

Take it with you

  • The first law fixes the combined heat and work accounting, while each transfer depends on the process.
  • Constant-pressure heat tracks enthalpy when pressure-volume work is the only work channel.
  • The ideal-gas capacity difference is universal within that model, but the individual capacities need more information.

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