Question

During a constant pressure process, the heat added equals:

• A. Change in internal energy
• B. Work done
• C. Change in enthalpy
• D. Zero

Hint:

At constant pressure, the heat added to the system equals the change in enthalpy, which combines both the internal energy change and the work done due to volume expansion.

Answer 1

The Correct Option is C

For a thermodynamic process occurring at constant pressure, the first law of thermodynamics states that the heat added to the system is related to the change in internal energy and the work done by the system: \[ Q = \Delta U + W \] where \( Q \) is the heat added, \( \Delta U \) is the change in internal energy, and \( W \) is the work done by the system.
At constant pressure, the work done is \( W = P \Delta V \), where \( P \) is the pressure and \( \Delta V \) is the change in volume. Thus, the heat added becomes: \[ Q = \Delta U + P \Delta V \] This expression can be rewritten as: \[ Q = \Delta H \] where \( \Delta H \) is the change in enthalpy. Therefore, during a constant pressure process, the heat added equals the change in enthalpy.
Thus, the correct answer is Change in enthalpy.