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Under what condition will the enthalpy change of a process equal the amount of heat transferred into or out of the system? Recall that the definition of enthalpy (H) includes both the internal energy (U) and the product of pressure (P) and volume (V): \(H = U + PV\) For a process occurring in a system, the change in enthalpy (ΔH) is given by: ΔH = ΔU + PΔV We know that heat transfer (Q) is related to internal energy and work done on/by the system (W) by the first law of thermodynamics: ΔU = Q - W Since work is done by/on the system in a process involving volume change, we have: W = -PΔV (the negative sign indicated work done by the system) Now, we can substitute the expression for W into the equation for ΔU: ΔU = Q - (-PΔV) = Q + PΔV Finally, we substitute ΔU from the thermodynamics equation back into the enthalpy change equation, obtaining: ΔH = Q + PΔV The condition for enthalpy change to equal heat transfer is when the volume change is zero (ΔV = 0) in the process. This occurs during a constant-volume (isochoric) process.

During a constant-pressure process, the system absorbs heat from the surroundings. Does the enthalpy of the system increase or decrease during the process? During a constant-pressure process (isobaric process), the work done by/on the system is given by: W = -PΔV In this case, the system absorbs heat (Q > 0) from the surroundings. According to the first law of thermodynamics: ΔU = Q - W Since the system is absorbing heat, the internal energy of the system (U) increases (ΔU > 0). This implies that the work done by the system W is positive, meaning that the system expands (ΔV > 0) as it absorbs heat. Now, we go back to the enthalpy change equation: ΔH = ΔU + PΔV We already established that ΔU > 0 and ΔV > 0 during a constant-pressure heat absorption process. Therefore, the enthalpy of the system (H) increases during the process (ΔH > 0).