Question

For an isobaric process, the heat transferred is equal to the change in ________ of the system.

• A. enthalpy
• B. entropy
• C. Helmholtz free energy
• D. Gibbs free energy

Hint:

Remember: Isochoric process → \(\Delta U\); Isobaric → \(\Delta H\); Isothermal reversible (ideal gas) → \(\Delta U = 0\). Keep these process–potential mappings clear to avoid confusion.

Answer 1

The Correct Option is A

Step 1: Recall the first law of thermodynamics.
\[ dU = \delta Q - p\,dV \] where \(U\) is internal energy, \(p\,dV\) is expansion work. Step 2: Define enthalpy.
\[ H = U + pV \quad \Rightarrow \quad dH = dU + p\,dV + V\,dp \] At constant pressure, \(dp = 0\). Hence, \[ dH = dU + p\,dV. \] Step 3: Connect with heat transfer.
From first law: \[ dU = \delta Q - p\,dV. \] Substituting, \[ dH = (\delta Q - p\,dV) + p\,dV = \delta Q. \] Step 4: Conclude.
Thus, in an isobaric process, heat added is equal to the enthalpy change. \[ \boxed{\Delta Q_p = \Delta H} \] Step 5: Example check (ideal gas).
For 1 mole ideal gas at constant \(p\): \[ \Delta H = nC_p\Delta T \quad \Rightarrow \quad Q_p = nC_p\Delta T. \] This matches the standard thermodynamic relation.