Question

Why is the change in enthalpy zero for isothermal processes?

Answer 1

Hint: An isothermal process may be a thermodynamic process during which the temperature of the system remains constant. Thus, the change in internal energy for an isothermal process is zero. For ideal gases, enthalpy is simply a function of temperature. As the temperature is constant in an isothermal process, change in enthalpy of the process is zero.

Complete answer:
An isothermal process is a thermodynamic process in which the temperature of the system remains constant. Thus, For an isothermal process \[PV = {\text{Constant}}\]. For ideal gases, the change in internal energy is zero for an isothermal process since an ideal gas has no interactions between particles, no intermolecular forces, so pressure change at constant temperature does not change internal energy.

For ideal gases, enthalpy is a function of the only temperature. Thus, in an isothermal process involving only ideal gases, the change in enthalpy is zero.
From the Maxwell Relation for the enthalpy for a reversible process in a thermodynamically-closed system,
\[{\text{dH = TdS + VdP}} - - - - - - \left( 1 \right)\]
 Now, on infinitesimally varying the pressure at constant temperature, we get
 \[{\left( {\dfrac{{\partial H}}{{\partial P}}} \right)_T} = T{\left( {\dfrac{{\partial S}}{{\partial P}}} \right)_T} + V{\left( {\dfrac{{\partial P}}{{\partial P}}} \right)_T} - - - - - - \left( 2 \right)\]
\[\left( {\dfrac{{\partial P}}{{\partial P}}} \right) = 1\]

Now we will examine the entropy term, the Gibbs' free energy is a function of temperature and pressure from its Maxwell Relation for a reversible process in a thermodynamically closed system:
\[dG = - SdT + VdP - - - - - \left( 3 \right)\]
Since the Gibbs' free energy (as with any thermodynamic function) is a state function, its cross-derivatives are equal
\[{\left( {\dfrac{{\partial S}}{{\partial P}}} \right)_T} = - {\left( {\dfrac{{\partial V}}{{\partial T}}} \right)_P} - - - - - \left( 4 \right)\]
Substituting \[\left( 4 \right)\] in \[\left( 2 \right)\] we get,
\[ \Rightarrow {\left( {\dfrac{{\partial H}}{{\partial P}}} \right)_T} = - T{\left( {\dfrac{{\partial V}}{{\partial T}}} \right)_P} + V\]
Using the ideal gas equation, we can say \[V = \dfrac{{nRT}}{P}\]
Substituting this in the above equation we get,
\[{\left( {\dfrac{{\partial H}}{{\partial P}}} \right)_T} = - T\dfrac{\partial }{{\partial T}}{\left[ {\dfrac{{nRT}}{P}} \right]_P} + \dfrac{{nRT}}{P}\]
\[ \Rightarrow {\left( {\dfrac{{\partial H}}{{\partial P}}} \right)_T} = - \dfrac{{nRT}}{P}\dfrac{{\partial \left[ T \right]}}{{\partial T}} + \dfrac{{nRT}}{P}\]
\[\Rightarrow {\left( {\dfrac{{\partial H}}{{\partial P}}} \right)_T} = - \dfrac{{nRT}}{P} + \dfrac{{nRT}}{P}\]
\[\therefore {\left( {\dfrac{{\partial H}}{{\partial P}}} \right)_T} = 0\]

Thus, for ideal gases enthalpy does not change for an isothermal process.

Note: When a process takes place at constant pressure, the heat absorbed or released is equal to the Enthalpy change. Enthalpy is the sum of the internal energy and the product of pressure and volume.Enthalpy can also be written as \[\Delta H = \Delta U + \Delta \left( {PV} \right)\].