Enthalpy and PV Work in Constant-Pressure Processes

Key idea: Enthalpy and PV work under constant pressure

• Enthalpy is the heat exchanged with the surroundings under constant pressure.
• Thought experiment setup:
  • Two flasks start in the same state with the same amount of dry ice (solid CO₂) at room temperature (25 °C).
  • Flask A is closed with a stopper (constant volume). Flask B has a deflated plastic bag over the mouth (initially constant pressure, as volume can expand).
  • The same amount of heat q enters each system at the same rate as the dry ice warms from −78 °C toward room temperature and sublimates to gas.
• Dry ice warming: solid CO₂ sublimates to gas as it absorbs heat; temperature of the dry ice rises from −78 °C toward 25 °C.
• Internal energy behavior:
  • In Flask A (constant volume): the sample cannot expand, so no P–V work is done; its internal energy increases as heat is absorbed.
  • In Flask B (constant pressure initially): volume increases as gas forms and expands; the system does work against ambient pressure P while absorbing heat.
• Visual descriptions from Fig. F01-5-1:
  • (a) Constant-volume flask: heat is absorbed, no work is done; cork eventually pops due to rising pressure.
  • (b) Constant-pressure flask: the bag inflates as gas forms; the system performs PV work on the surroundings during inflation.
  • After enough heat intake, the cork pops in (a) and the bag finally inflates and breaks in (b) due to increasing CO₂ pressure.
• Conclusion from the thought experiment: In chemical/physical changes at constant pressure, the primary work the system can do is PV work (ΔV). This PV-work term is especially important when there is a significant volume change, as with gases.
• PV work and state functions:
  • Pressure (P) and volume (V) are state functions; PV work is a state function when considered as part of energy changes.
  • Under constant pressure, when only heat and PV work contribute to energy changes, the energy balance can be written as:
  • $\Delta E = q_p + w = \Delta H - P\,\Delta V$
• Enthalpy and its practical meaning:
  • The heat exchanged with the surroundings under constant pressure is called enthalpy change, defined as $q = \Delta H.$ (In many reactions in liquids/solids with small or zero change in volume, enthalpy change closely tracks the internal energy change.)
• Internal energy vs enthalpy:
  • Both are extensive properties; they depend on the quantity of matter present.
  • In constant-pressure processes where PV work is the only work term, the relation between internal energy change and enthalpy change is:
  • $\Delta E = \Delta H - P\,\Delta V.$
• Sign conventions for heat and work:
  • General first law: $\Delta E = q + w.$
  • Work on the surroundings by the system (PV work) is typically written as $w = -P_{\text{ext}}\,\Delta V$ (negative when the system does work on the surroundings).
  • If the heat flow to the system at constant pressure is considered: $q_p = \Delta H.$
  • Therefore, at constant pressure: $\Delta E = q<em>p - P</em>{\text{ext}}\,\Delta V = \Delta H - P\,\Delta V.$
• Physical interpretations and implications:
  • When ΔH > 0 (heat flows into the system), the process is endothermic (endo = into).
  • When ΔH < 0 (heat flows to the surroundings), the process is exothermic (exo = out of).
  • Endothermic/Exothermic definitions rely on the sign of ΔH, not ΔE.
• Examples and practical relevance:
  • In gas-forming or gas-expanding processes, PV work can be substantial and can significantly affect internal energy even as heat is added.
  • Enthalpy (ΔH) is a convenient measure since, at constant pressure, the heat exchanged equals ΔH and can often be measured calorimetrically.
• Summary takeaways:
  • Under constant-volume conditions, no PV work is done; all absorbed heat increases internal energy.
  • Under constant-pressure conditions, PV work is done as the system expands; this reduces the net increase in internal energy for a given heat input.
  • Enthalpy is a state function representing heat transfer at constant pressure; it is related to internal energy by $H = E + P V$, with corresponding differential relation $\Delta H = \Delta E + P\,\Delta V + V\,\Delta P$; at constant pressure, this reduces to $\Delta H = \Delta E + P\,\Delta V\Rightarrow \Delta E = \Delta H - P\,\Delta V.$