Consider an isobaric process. Which of the following statements is correct?

Question

BlackTom AI · Accepted Answer

For an isobaric process (constant pressure), use the first law of thermodynamics and the definition of enthalpy to reach the result.

1) Relevant knowledge points and concepts
- First law (closed system, sign convention Q in, W out): ΔU = Q - W.
- Mechanical work for a quasi-static process: W = ∫P dV. At constant pressure, W = P ΔV.
- Enthalpy: H ≡ U + PV, so ΔH = ΔU + Δ(PV). For constant pressure, ΔH = ΔU + P ΔV.

2) Step-by-step reasoning
- Start with the first law: ΔU = Q - W.
- Substitute W = P ΔV (isobaric): ΔU = Q - P ΔV.
- Rearrange: Q = ΔU + P ΔV.
- Use enthalpy relation at constant pressure: ΔH = ΔU + P ΔV.
- Therefore Q = ΔH.

3) Intermediate steps shown algebraically
- ΔU = Q - W
- W = P ΔV
- Q = ΔU + P ΔV
- ΔH = ΔU + P ΔV
- Hence Q = ΔH

4) Why the correct answer is correct
Because at constant pressure the heat added to the system changes both its internal energy and does work equal to PΔV. Enthalpy precisely accounts for the internal energy plus the PV term, so the heat transfer at constant pressure equals the change in enthalpy.

5) Why the other options are incorrect
- a. "The work done is equal to the change in internal energy." Generally false: W = PΔV, while ΔU = Q - W, so W ≠ ΔU except in special cases (e.g., when Q = 0 and sign conventions permit).
- c. "The work done is equal to the heat transfer." Only true if ΔU = 0 (no change in internal energy), which is not generally true for isobaric processes.
- d. "The heat transfer is equal to the change in internal energy." This is true for isochoric (constant-volume) processes (where W = 0), not for isobaric processes where W = PΔV ≠ 0.
- e. "No work is done." False: at constant pressure, if volume changes, work PΔV is done.

6) Conclusion
Therefore the heat transfer in an isobaric process equals the change in enthalpy, confirming answer (b).

Consider an isobaric process. Which of the following statements is correct?单项选择题

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