Questions: First Law for Control Mass Systems

Question 1

A rigid sealed tank contains steam. Heat is added until the steam's internal energy increases by 500 kJ. How much boundary work was done, and how much heat was added?

• A. W = 500 kJ, Q = 1000 kJ — boundary work equals the heat required to maintain pressure
• B. W = 0, Q = 500 kJ — a rigid tank has no volume change, so boundary work is zero and all heat input becomes internal energy change
• C. W = 500 kJ, Q = 0 — the process is adiabatic so work must supply the internal energy
• D. W = −500 kJ, Q = 0 — internal energy increases by work done on the system

Answer: (B)

For a rigid tank, the volume cannot change (ΔV = 0), so boundary work W = ∫P dV = 0. The first law for a closed system reduces to ΔU = Q − W = Q − 0 = Q. Therefore Q = ΔU = 500 kJ: all the heat added goes directly into increasing the internal energy of the steam. This is the isochoric (constant-volume) process — the simplest application of the first law because the work term vanishes completely.

Question 2

A piston-cylinder device compresses a gas at constant pressure. An engineer wants to calculate the heat transferred during this compression. Which approach is most direct?

• A. Use ΔU = Q − W, then separately calculate W = PΔV and subtract from ΔU to find Q
• B. Use ΔH = Q_P directly — at constant pressure, the enthalpy change equals the heat transferred, automatically accounting for the boundary work
• C. Set Q = 0 since compression processes are adiabatic by assumption
• D. Use dE/dt = Q̇ − Ẇ with the rate form since piston problems require differential analysis

Answer: (B)

At constant pressure, the first law ΔU = Q − PΔV can be rearranged to Q = ΔU + PΔV = ΔH (since H = U + PV, so ΔH = ΔU + PΔV at constant pressure). This means the heat transferred in any constant-pressure process equals the change in enthalpy — no separate calculation of boundary work is needed. This is exactly why enthalpy is tabulated in property tables: it bundles internal energy and the PV work of expansion together into one state property, making constant-pressure energy balances a single lookup rather than a two-step calculation.

Question 3

For a gas that expands adiabatically in a piston-cylinder device (no heat transfer), the internal energy of the gas decreases by exactly the amount of work done by the gas on the piston.

• T. True
• F. False

Answer: True

For an adiabatic process Q = 0, so the first law ΔU = Q − W becomes ΔU = −W. The gas does positive work W on the piston (the gas pushes the piston outward), and its internal energy falls by exactly that amount. Energy is conserved: the gas's thermal energy converts to mechanical work. This is the operating principle of an adiabatic expansion turbine — the working fluid's internal energy decreases and that energy appears as shaft work output.

Question 4

Enthalpy is useful for constant-pressure processes because it represents the total 'heat content' stored in a substance at any given state.

• T. True
• F. False

Answer: False

Enthalpy H = U + PV is a state function, but calling it 'heat content' is a misconception — heat is a path function (energy in transit), not a property stored in a substance. Enthalpy is useful for constant-pressure processes for a specific reason: at constant pressure, ΔH = Q_P, meaning the change in enthalpy equals the heat transferred during a constant-pressure process. This identity absorbs the P ΔV boundary work into H, making the energy balance simpler. But outside of constant-pressure conditions, ΔH does not generally equal heat transferred, and enthalpy has no meaning as a 'stored heat.'

Question 5

Why do engineers typically use enthalpy rather than internal energy when analyzing constant-pressure processes, and what does the enthalpy function absorb that makes it convenient?

Model answer: At constant pressure, the first law is ΔU = Q − PΔV, which requires tracking both internal energy change and the boundary work PΔV separately. Enthalpy is defined as H = U + PV, so ΔH = ΔU + Δ(PV) = ΔU + PΔV at constant pressure. Substituting into the first law gives ΔH = Q_P — the heat transferred at constant pressure equals simply the enthalpy change, with no need to calculate boundary work separately. Enthalpy absorbs the P ΔV expansion work into a single composite state property. This is why thermodynamic tables list H values: most practical engineering devices (boilers, condensers, turbines, nozzles at steady state) operate at approximately constant pressure, and tracking enthalpy reduces every energy balance to a single property difference.

The deeper point is that enthalpy is a convenient accounting fiction tailored to constant-pressure conditions. It is not physically more fundamental than internal energy — it is a bundle designed to simplify a specific class of calculations. Recognizing which process constraint applies (isochoric, isobaric, adiabatic) before writing the energy balance is the central problem-solving skill: the right choice of variables follows directly from the constraint.