Question

When water changes from the liquid phase to the gaseous (vapor) phase, energy is required for the H2O molecules to escape their mutual molecular attraction (the internal energy change ufg in the steam tables) and for the volume expansion (P·vfg). The total heat of vaporization for the phase change is represented by the enthalpy change hfg in the steam tables. Use the steam tables to calculate the fractions of energy for the volume expansion and for the molecules to escape their mutual attraction at P = 0.1 and 0.2 MPa respectively. What conclusion may be drawn with respect to the effect of the pressure?

Answer 1

The energy required for vaporization of water is expressed as the enthalpy of vaporization (ΔH_v). According to the definition of enthalpy:

Where:

H is the enthalpy

U is the internal energy

P is the pressure of the system

V is the volume of the system

For a change of enthalpy in a vaporization process, equation (1) is rewritten as:

$\Delta H_v = \Delta U_v +\Delta (PV)_v$

At constant pressure, we have;

$\Delta H_v = \Delta U_v +P\Delta V_v\;\;\;\;(2)$

Using steam tables for a given pressure value, we can know the values of the enthalpy of vaporization, the change in internal energy and the change in volume between the gas and liquid phase during a vaporization process.

At pressure of P = 0.1 MPa, we have:

H,-2258.0 kJ/kg

AU,' = 2088.7 kJ kg

$\Delta V_{v} = (1.6940-0.001043)\,m^3/kg = 1.692957\,m^3/kg$

The change in internal energy is required for the molecules to escape to their mutual molecular attraction. The fraction of this energy respect to the heat of vaporization is given by the quotient:

2088.7 k.J k 0.925 2258.0 kJ/kg ΔΗ,

If we divide equation (2) by the heat of vaporization:

$\frac{\Delta H_v}{\Delta H_v}=\frac{\Delta U_v}{\Delta H_v}+\frac{P\Delta V_v}{\Delta H_v}$

0.925 + 1

Then, the fraction of energy required by the volume expansion is:

1-0.925 0.075

At pressure of P = 0.2 MPa, we have:

$\Delta H_v = 2202.0\,kJ/kg$

,-2025.0 kJ/kg

▲Vİ_ (0.8857-0.001061) Tn3/kg-0.884639 7723/kg

The fraction of energy required by the change in internal energy is:

2025,0kJ/kg=0.920 2202.0 kJ/kg My △H,

Then, the fraction of energy required by the volume expansion is:

$\frac{P\Delta V_v}{\Delta H_v}= 1-0.920 = 0.080$

As a conclusion, we can see that pressure increases the fraction of energy required in the volume expansion respect to the total vaporization heat. It is clearly visible in equation (2), where the value of this product is directly proportional to the value of PΔV