First law for closed and open systems

Closed- and open-system energy balances with consistent sign conventions, steady-flow devices and tank filling.

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Why it matters

Every heat exchanger duty, compressor power, reactor cooling load and flash-drum heat requirement in a chemical plant starts from an energy balance. The first law is that balance. If you can set up the closed-system and the open (control-volume) forms correctly, with consistent signs, most of process thermodynamics becomes bookkeeping.

Key ideas

Energy is conserved. The total energy of a system plus its surroundings is constant. Energy crosses a system boundary only as heat (driven by a temperature difference), as work (any other boundary interaction: shaft, piston, electrical), or carried with flowing mass.

Sign convention used here. Heat Q is positive when added to the system; work W is positive when done by the system. Some chemical engineering textbooks (Smith, Van Ness and Abbott) take W positive when done on the system and write ΔU = Q + W. Both are correct; mixing them is the error. Always state your convention.

Internal energy U is a state function: it depends only on the state, not on how the system got there. Q and W are path functions: they depend on the process. That is why a cycle has ΔU = 0 while the net heat and net work need not be zero individually; they are equal to each other.

Closed system (control mass). No mass crosses the boundary. For a quasi-static (reversible) expansion or compression the only boundary work is W = ∫P dV. At constant volume W = 0 so Q = ΔU; at constant pressure Q = ΔH, which is why enthalpy is the natural property for constant-pressure heating and for heats of reaction measured at constant pressure.

Open system (control volume). Mass flows in and out. Each kilogram pushed into the control volume brings its internal energy u plus the flow work Pv needed to push it in; together these are the enthalpy h = u + Pv. So enthalpy, not internal energy, appears in flow balances. Work in an open system is split into flow work (already inside h) and shaft work Ws (turbines, pumps, compressors).

Steady flow. Nothing inside the control volume changes with time, and ṁ_in = ṁ_out. Useful special cases:

  • Adiabatic turbine or compressor: Ẇs = −ṁΔh (ΔKE, ΔPE negligible).
  • Throttling valve: no work, negligible heat and KE change, so h_out = h_in (isenthalpic). For an ideal gas this means no temperature change.
  • Nozzle: no work, adiabatic, so the drop in enthalpy appears as kinetic energy.
  • Heat exchanger: no shaft work; heat lost by the hot stream equals heat gained by the cold stream if the shell is insulated.

Unsteady flow (filling and emptying). The energy stored in the control volume changes with time. A classic result: filling an initially evacuated, rigid, insulated tank from a supply line at constant T_line gives u_final = h_line. For an ideal gas with constant heat capacities, T_final = γ·T_line, so the tank gas is hotter than the line gas because flow work is converted into internal energy.

Ideal-gas results. For an ideal gas U and H depend on T only: dU = Cv dT and dH = Cp dT, with Cp − Cv = R (molar). These hold for any process of an ideal gas, not just constant-volume or constant-pressure ones.

Formulas

ΔU = Q − W (closed system, W done by the system) W = ∫P dV (reversible boundary work) Q = ΔU (closed, constant volume); Q = ΔH (closed, constant pressure, PV work only) H = U + PV dU = Cv dT, dH = Cp dT, Cp − Cv = R (ideal gas) Q̇ − Ẇs = Σṁ_out(h + c²/2 + g·z)_out − Σṁ_in(h + c²/2 + g·z)_in (steady flow) Q̇ − Ẇs = ṁ·Δh (single stream, steady, KE and PE negligible) d(mu)_cv/dt = ṁ_in·h_in − ṁ_out·h_out + Q̇ − Ẇs (unsteady, KE and PE negligible) T_final = γ·T_line (evacuated insulated rigid tank filled with an ideal gas, constant heat capacities)

Symbols: U internal energy (J), H enthalpy (J), lower case u, h per unit mass (J/kg); Q heat (J), Q̇ heat rate (W); W work (J), Ẇs shaft power (W); P pressure (Pa); V volume (m³); ṁ mass flow (kg/s); c velocity (m/s); g = 9.81 m/s²; z elevation (m); Cv, Cp heat capacities (J/kg·K or J/mol·K); R gas constant (8.314 J/mol·K, or 287 J/kg·K for air); γ = Cp/Cv.

Worked examples

Example 1 (standard): constant-pressure heating in a piston-cylinder. Given: 2 kg of air (ideal gas) heated at a constant 200 kPa from 300 K to 400 K. Take R = 0.287 kJ/kg·K, Cv = 0.718 kJ/kg·K, Cp = 1.005 kJ/kg·K. Find W, ΔU and Q.

  1. Boundary work at constant P: W = PΔV = m·R·ΔT = 2 × 0.287 × 100 = 57.4 kJ (done by the gas).
  2. ΔU = m·Cv·ΔT = 2 × 0.718 × 100 = 143.6 kJ.
  3. Q = ΔU + W = 143.6 + 57.4 = 201.0 kJ.
  4. Check: Q = ΔH = m·Cp·ΔT = 2 × 1.005 × 100 = 201.0 kJ. Consistent. Answer: W = 57.4 kJ, ΔU = 143.6 kJ, Q = 201.0 kJ

Example 2 (GATE level): filling an evacuated tank. Given: a rigid, insulated, initially evacuated tank of 0.5 m³ is connected to a line carrying air at 300 K. The valve is opened until the tank pressure reaches 500 kPa. Treat air as an ideal gas with γ = 1.4, R = 0.287 kJ/kg·K. Find the final temperature and mass in the tank.

  1. Unsteady balance with no outflow, Q = 0, Ws = 0, initial mass zero: m_f·u_f = m_f·h_line, so u_f = h_line.
  2. With u = Cv·T and h = Cp·T (same reference): Cv·T_f = Cp·T_line, so T_f = γ·T_line = 1.4 × 300 = 420 K.
  3. Mass from the ideal-gas law: m = P·V / (R·T) = 500 × 0.5 / (0.287 × 420) = 2.074 kg. Answer: T_f = 420 K (147 °C), m = 2.07 kg

Example 3: compressor with heat loss. Given: 0.5 kg/s of air is compressed steadily from 300 K to 450 K and loses 5 kW of heat to the surroundings. Cp = 1.005 kJ/kg·K; neglect KE and PE. Find the power input.

  1. Q̇ − Ẇs = ṁ·Cp·ΔT, with Q̇ = −5 kW.
  2. ṁ·Cp·ΔT = 0.5 × 1.005 × 150 = 75.4 kW.
  3. Ẇs = Q̇ − ṁ·Cp·ΔT = −5 − 75.4 = −80.4 kW (negative: work is done on the air). Answer: power input = 80.4 kW

Common mistakes

  • Mixing sign conventions: using ΔU = Q − W with W taken as work done on the system (or the textbook form ΔU = Q + W with work done by the system).
  • Using ΔU in a flow balance instead of Δh, which drops the flow work Pv.
  • Writing Q = m·Cv·ΔT for a constant-pressure process; Cv gives ΔU for any ideal-gas process, but Q equals m·Cv·ΔT only at constant volume.
  • Assuming a throttled real fluid keeps its temperature; only h is constant. Real gases and flashing liquids change temperature.
  • Forgetting that the filling-tank result uses h_line, not u_line, giving T_final = T_line instead of γ·T_line.
  • Mixing kJ and J when adding kinetic energy (c²/2 is in J/kg when c is in m/s).

For GATE CH

  • NAT problems on compressor, turbine or pump power from an enthalpy or ideal-gas temperature change, often with a stated heat loss.
  • Tank filling or discharge problems (unsteady first law) and adiabatic mixing of two streams.
  • Conceptual MCQs on state versus path functions, throttling (isenthalpic) and the meaning of Q = ΔH at constant pressure.
  • Practise: write the general balance first, strike out terms with reasons, then substitute. Check which sign convention the question uses.

Quick check

  1. What is ΔU for a closed system over one complete cycle?
  2. For steady adiabatic throttling, which property stays constant?
  3. 1 kg of an ideal gas with Cv = 0.718 kJ/kg·K is heated at constant volume by 50 K. How much heat is added?
  4. Why does enthalpy, rather than internal energy, appear in the open-system energy balance?
  5. An evacuated insulated tank is filled with helium (γ = 5/3) from a line at 300 K. What is the final temperature?

Answers: 1. Zero. 2. Enthalpy. 3. 35.9 kJ. 4. Because flowing mass carries flow work Pv as well as u, and u + Pv = h. 5. 500 K.

Interview questions

All Chemical Engineering Thermodynamics interview questions

Try answering each one aloud before you open it.

  1. 1.What is the first law of thermodynamics for a closed system?Concept

    The first law of thermodynamics for a closed system states that the change in internal energy of the system is equal to the heat added to the system minus the work done by the system. Mathematically, it is expressed as ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added, and W is the work done by the system.

  2. 2.Explain the first law of thermodynamics for an open system.Concept

    For a control volume, the rate of change of stored energy equals heat in minus shaft work out plus the energy carried in by flowing mass minus that carried out. Each stream carries h + c²/2 + g·z per kilogram, where h = u + Pv includes the flow work needed to push the mass across the boundary. At steady state the storage term vanishes and, for a single stream with negligible kinetic and potential energy changes, Q̇ − Ẇs = ṁ·Δh.

  3. 3.How does the first law of thermodynamics apply to a cyclic process?Concept

    In a cyclic process, the system returns to its initial state at the end of the cycle. According to the first law of thermodynamics, the net change in internal energy over one complete cycle is zero. Therefore, the net heat added to the system is equal to the net work done by the system, i.e., Q_net = W_net.

  4. 4.Why is enthalpy used instead of internal energy in the first law for open systems?Application

    Enthalpy is used in open systems because it accounts for the flow of energy due to mass entering or leaving the system. In open systems, mass flow can carry energy in the form of enthalpy, which includes both internal energy and the energy associated with pressure-volume work. This makes enthalpy a more convenient property for analyzing energy changes in open systems.

  5. 5.What happens if heat is added to a closed system at constant volume?Application

    If heat is added to a closed system at constant volume, the work done by the system is zero because there is no volume change. According to the first law of thermodynamics, all the heat added goes into increasing the internal energy of the system. Mathematically, ΔU = Q, where ΔU is the change in internal energy and Q is the heat added.

  6. 6.Describe a real-world example where the first law of thermodynamics is applied to an open system.Application

    A common real-world example is a steam turbine in a power plant. In this system, steam enters the turbine, does work on the blades, and exits at a lower enthalpy. The first law of thermodynamics helps in calculating the work output of the turbine by considering the change in enthalpy of the steam as it passes through the turbine.

  7. 7.What is the significance of the first law of thermodynamics in chemical reactions?Application

    The first law lets us compute the heat that must be added or removed to run a reaction at given conditions. At constant pressure with only PV work, the heat released or absorbed equals the enthalpy change, so standard heats of reaction (from heats of formation) are used together with sensible-heat terms to size reactor heating or cooling duty. It also gives the adiabatic reaction temperature when Q = 0.

  8. 8.Calculate the change in internal energy for a closed system where 500 J of heat is added and 200 J of work is done by the system.Numerical

    Using the first law of thermodynamics for a closed system, ΔU = Q - W. Here, Q = 500 J and W = 200 J. Therefore, ΔU = 500 J - 200 J = 300 J. The change in internal energy is 300 J.

  9. 9.What assumptions are typically made when applying the first law of thermodynamics to a closed system?Concept

    No mass crosses the boundary, and the initial and final states are equilibrium states so that U is defined. To evaluate work as ∫P dV the process is taken as quasi-static (reversible); otherwise work must be known some other way. Kinetic and potential energy changes of the system are usually neglected, so ΔU = Q − W with W done by the system.

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