Question

a. Using only P-v-T data, estimate the enthalpy of vaporization of water at T oC where, T is T= 38 and so on. b. If the enthalpy (h), as an unmeasured thermodynamic property is related to the temperature (T) and Pressure (P) as measured thermodynamic properties according to the following relation: h = f (T,P). Show that: dh = Cp dT +[ v – T (dv/dT)p] dP Where: Cp is the specific heat at constant pressure, v is the specific volume.

Answer 1

a)

When vaporisation takes place; the liquid and vapour phases are in equilibrium. At equilibrium ;

dP / dT = ΔH^vap / ( T^vap ΔV )

or

ΔP / ΔT = ΔH^vap / ( T^vap ΔV )

which on rearranging gives :

ΔH^vap =  T^vap ΔP ΔV / ΔT

where;

ΔH^vap = Enthalpy of vaporisation

ΔP = Pressure difference

ΔT = Temperature difference

T^vap = Vaporisation temperature

ΔV = V^vap - V^liq

Now;

We can obtain the PVT data for water from the steam table and then calculate the enthalpy of vaporisation at 38 °C.

Now;

At T = 36 °C ; P = 5947.47 Pa

At T = 40 °C ; P = 7384.43 Pa

Then;

ΔP = 7384.43 - 5947.47 = 1436.96 Pa

ΔT = 40 - 36 = 4 °C = 4 K

At T = 38 °C :

V^vap = 21.5954 m3 / kg

V^liq = 0.001 m3 / kg

ΔV = 21.5954 - 0.001 = 21.5944 kg / m3

T^vap = 38 + 273.15 = 311.15 K

Therefore;

ΔH^vap = 311.15 X 1436.96 X 21.5944 / 4

ΔH^vap = 2413769 J / kg

ΔH^vap = 2413.8 kJ / kg

Enthalpy of vaporisation of water = 2413.8 kJ / kg

b)

From thermodynamics we know ;

dh = T ds + v dP

So;

(δh / δP)_T = T (δs / δP)_T + v

Now ;

h = f (T,P)

Then;

dh = (δh / δT)_P dT + (δh / δP)_T dP

We know that ;

(δh / δT)_P = C_p

and

(δh / δP)_T = T (δs / δP)_T + v

Substituting these in the expression :

dh = C_p dT + ( T (δs / δP)_T + v ) dP

From Maxwell's Relations :

(δs / δP)_T = - (δv / δT)_P

Therefore;

dh = Cp dT + ( - T (δv / δT)_P + v ) dP

which is the required expression.