Question

Which process involves an increase in enthalpy due to the system absorbing heat from the surroundings?

• A. Adiabatic expansion
• B. Isothermal compression
• C. Constant pressure heating
• D. Adiabatic compression

Hint:

Enthalpy Change and Heat. At constant pressure, the enthalpy change equals the heat absorbed or released: \(\Delta H = Q_p\). Heating implies \(Q_p>0\), thus \(\Delta H>0\).

Answer 1

The Correct Option is C

Enthalpy change (\(\Delta H\)) is related to heat (\(Q\)) and work (\(W\)) by \(\Delta H = \Delta U + \Delta(PV)\).
For a process occurring at constant pressure (\(P\)), \(\Delta H = Q_p\), where \(Q_p\) is the heat absorbed by the system at constant pressure.
The question asks for a process where enthalpy increases (\(\Delta H>0\)) because the system absorbs heat from the surroundings (\(Q>0\)).
- Adiabatic processes (1, 4): By definition, \(Q=0\).
Enthalpy can change due to work done, but not due to heat absorption.
- Isothermal compression (2): Work is done *on* the system (\(W<0\)).
To keep temperature constant, heat must be *removed* from the system (\(Q = W<0\)).
Enthalpy change for an ideal gas is zero (\(\Delta H = nC_p \Delta T = 0\)).
- Constant pressure heating (3): Heat is added to the system (\(Q_p>0\)).
Since \(\Delta H = Q_p\), the enthalpy increases due to absorbing heat.
Therefore, constant pressure heating fits the description.