Thermodynamics/Mass Transfer: Vapor-Liquid Composition

When the two phases are in equilibrium, their chemical potential or fugacity will be same. fi(L) = fi(V)

The simplest case is that the liquid phase is ideal solution (i.e. similar chemical nature molecules) and vapour phase is ideal gas (i.e. low pressure) which is the Raoult’s Law; i.e. 

fi(ideal solution) = fi(ideal gas)  

Xi.Pi(sat) = yi.P  

for the calculation of composition from this we have to know the saturation pressure of the molecule at a given temperature, which is generally found from Antoine equation but the condition is that the temperature should below its critical temperature. So for a given temperature/Pressure and composition in a phase we can find the composition in other phase.

For a system of air water we can not apply this law to find the composition of air in liquid because the critical temperature of air is less than room temperature and the vapour pressure or saturation pressure of air can not be find. The amount of water in vapour phase can be found by assuming no air is dissolved in water i.e. water is pure but the amount of air in the liquid phase can not be calculated from Raoult’s law. 

To get the composition of air in liquid we can use Henry's law. Henry's law can be applied for pressure low enough that the vapor phase may be assumed an ideal gas and a species present as a very dilute solute in the liquid phase. Henry's law yi.P = Hi.xi and the value of Henry's constant come from experiment. for the air water system the composition of air is calculated from Henry's Law and composition of water is calculated by Raoult's Law. 
For a real solution we can apply Henry Law in dilute region and Lewis Randall Rule at pure species composition for approximate values.

Thermodynamics: Fugacity

Fugacity (fi) is a thermodynamic property defined to correlate behavior of real gases with ideal gases.  It is used to evaluate the most useful property Gibbs Free energy (chemical potential) for real species/component either in pure or in solution.

If we have pure species, for ideal gas fugacity is equal to pressure of the gas, and for ideal gas solution fugacity of the species/component is equal to partial pressure of the species/component and in ideal liquid solution it is equal to fuacity x mole fraction (fi.xi) (Lewis/Randall Rule)

The fugacity can be calculated by comparing the Gibbs free energy for real conditions to its ideal value. The gas will behave like an ideal gas at lowest possible pressure at a given temperature. At ideal conditions the fugacity can be calculated from its conditions for pure species or in solution. The difference between the Gibbs free energy can be calculated from its fundamental equation of (∆H-T∆S).
The fugacity can be calculated using equation of state or compressiblility factor versus pressure data.
In case of steam we can use saturated steam table and super heated steam table to find fugacity.

Industry-Institute Interaction

I attended the industry-academic interaction, in which it was discussed how we can enhance the quality of education of B. Tech students. A teacher should learn everyday. In my views, the faculty should have a knowledge of the current processes going on in the industry. For that there should be the provision for the industrial exposure for the faculty. there should be at least one week industrial visit for the faculty in his concerned company and then he should get the industrial problems related to the subject. the students will get a chance to work on real problems and see what is the importance of the subject in industry.   

Thermodynamics: Partial molar property

property value in pure state and in solution differs. for example if we take the molar volume of pure water is 18.069 ml/mol but if 1 mol of water is added to 1 mol of ethanol, the partial molar volume of water is about 16.9 ml/mol. if we add 1017 ml of methanol and 1053 ml of water, we expect to get 2070 ml solution, but we get 2000 ml solution. this shows that the volume of solution decreases or volume change of mixing is -70 ml.
Like volume the different properties like enthalpy, entropy, Gibbs free energy also differs in the pure state and in solution. the solution property like U, H S, G, V can be calculated form partial properties by multiplying with the mole fraction or number of moles and summing it.
the difference between the solution property and sum of the partial molar enthalpy is called the change of property of mixing. i.e. M-∑yiMi, where Mi is the molar property in its pure state.
the enthalpy change of mixing of ideal gases is zero, but the entropy change of mixing of ideal gases is greater than zero, as it is an irreversible process which has entropy change positive.

Thermodynamic properties for real gases

The thermodynamic properties like internal energy, enthalpy, entropy for real gases can be calculated from the fundamental relations like
ΔU=Q+W,      H=U+PV,             A= U-TS,                 G=H-TS,      
dW=-PdV,       dQr = TdS
so dU=TdS-PdV like that all these are state functions and independent of path whether reversible or irrversible processes. from the differential equations of U, H, A & G, we get the Maxwell's equations applicable to ideal as well as real gases. by combining the differential equations and Maxwell's equations we can get general differential equations for the calculation of change in internal energy, enthalpy and entropy.
For pure component degree of freedom is 2. So, a property will be a function of 2 variables. for example U (T,V), H (T,P), S(T,V) or S(T,P). U is taken as a function of T & V because heat supplied at constant volume is change in internal energy and heat supplied at constant pressure is change in enthalpy.
dU =  CvdT + [T(dP/dT)v - P]dV

dH =  CpdT + [V - T(dV/dT)p]dP

dS  =  dQ/T = (dU + PdV)/T  = (Cv/T)dT  + [(dP/dT)v]dV
                    =  (dH - VdP)/T  = (Cp/T)dT  -  [(dV/dT)p]dP

Now to remember the expressions of dU & dH
at constant volume dU =  CvdT true for any gas
for any process  dU =  CvdT + [T(dP/dT)v - P]dV, the whole expression should have the units of energy i.e. PdV and the changes in other properties P & T keeping V constant.
Similarly dH =  CpdT + [V - T(dV/dT)p]dP

Chemical Engineering Thermodynamics questions

Here are some question from first & second laws of thermodynamics and thermodynamic properties.

1.    Two moles of an ideal gas from an initial state of 150°C and 1 bar is compressed isobarically to 70°C reversibly. Calculate ΔU, ΔH, Q and W for the process. If the process is carried out irreversibly but so as to accomplish exactly the same changes of state, calculate ΔU, ΔH, Q and W if it is carried out with an efficiency of 80%.  Take C_P=(5/2)R

2.  Entropy is a state function, should not depend on the path taken, yet entropy change is different for reversible and irreversible paths (Clausius inequality). Give the proper justification of these statements.

3. A heat engine absorbs 250 kJ of heat from a source at 350 K and rejects heat to two cold reservoirs at 300 K and 275 K. If the heat rejected to 300 K is 100 kJ, calculate the maximum work done by the engine.                                       

4.  One mole of an ideal gas Cp=7/2R is compressed adiabatically in a piston-cylinder device from 2 bar and 25°C to 8 bar. The process is irreversible and requires 35% more work than a reversible, adiabatic compression from the same initial state to the final pressure. What is the entropy change of the gas?       

5.    The equation of state of a certain substance is given by the expression V= (RT/P - C/T3)and the specific heat is given by the relation Cp=A+BT, where A, B and C are constants. Derive the expression for the changes in internal energy, enthalpy and entropy for constant volume process. 

6.  A gas obeying the following equation of state P(V − b) = RT is subjected to a Joule-Thomson expansion. Will the temperature increase, decrease, or remain the same? The Joule-Thomson coefficient is given by  (dT/dP)h

Applications of the various Subjects in Chemical Engineering

1.      Process Calculation:

This subject deals with material balance, energy balance. In any chemical industry, we have to find the amount of raw materials needed to get the required desired product and how much heat is needed to get the desired operating conditions

      2.      Fluid Flow and Mechanical Operation(Fluid and Particle Mechanics):

In any chemical industry, the fluid and solid has to be transported from one place to another. These subjects deal with properties of fluid and solid, their transportation (devices, energy), solid particles size reduction and separation.

       3.      Heat Transfer:

 In any chemical industry, in order to get the desired operating temperature, we have to analyse the amount of heat transfer. In this subject, we study the different modes of heat transfer and various devices like heat exchanger, evaporators.
4. Chemical Engineering Thermodynamics
In this subject, we study about the various properties of the components, like its Gibbs Free energy, Work, Enthalpy, lost work, conditions to optimise the process.