Question

Which of the following relation is not correct?

• A. \( \Delta H = \Delta U - P\Delta V \)
• B. \( \Delta U = q + W \)
• C. \( \Delta S_{{sys}} + \Delta S_{{surr}} \geq 0 \)
• D. \( \Delta G = \Delta H - T \Delta S \)

Hint:

Enthalpy change, \( \Delta H \), is related to internal energy change by the equation \( \Delta H = \Delta U + P\Delta V \), which is crucial for understanding thermodynamic systems at constant pressure.

Answer 1

The Correct Option is A

Step 1: {Understanding the Relations}
Relation (A) is the incorrect one.
The correct expression for enthalpy change is \( \Delta H = \Delta U + P\Delta V \).
The expression provided in option (A) is incorrect because it subtracts \( P\Delta V \) instead of adding it.
Step 2: {Correct Relations}
Relation (B) is correct because the change in internal energy \( \Delta U \) is the sum of heat added to the system \( q \) and the work done on the system \( W \), i.e., \( \Delta U = q + W \).
Relation (C) is correct according to the second law of thermodynamics, which states that the total entropy change (system plus surroundings) for a spontaneous process is greater than or equal to zero.
Relation (D) is the correct form of the Gibbs free energy equation.
Thus, the correct answer is (A).

Answer 2

Step 1: Recall the definitions of enthalpy and internal energy
The relation between enthalpy (ΔH), internal energy (ΔU), pressure (P), and volume change (ΔV) is derived from the definition:
H = U + PV

Step 2: Derive the correct relationship
Taking differential form:
ΔH = ΔU + Δ(PV)
At constant pressure, PV becomes PΔV, so:
ΔH = ΔU + PΔV

Step 3: Identify the incorrect relation
The given option is:
ΔH = ΔU − PΔV
This is incorrect because it has the wrong sign for the pressure-volume work term.
The correct relation is:
ΔH = ΔU + PΔV

Step 4: Conclusion
The relation ΔH = ΔU − PΔV is not correct.

Final Answer: \( \Delta H = \Delta U - P\Delta V \)