ENTROPY (S) INTRODUCTION

ENTHALPY (H) INTRODUCTION

We known 1^st law of thermodynamics   

                                        dU=dq+dW 

  If P_ext=constant

                Then dW= -PdV

     and     dU=dq- PdV

                dq=dU-PdV

                dq=(U₂-U₁)+P(V₂-V₁)

                dq=(U2+PV2)-(U1+PV1)

                dQ=H2-H1                    ( H₂=U₂+PV₂  and H₁=U₁+PV₁)

                dq=dH

hence

The enthalpy of a system is defined as:

                  H = U + PV

                  DH = DU + D(PV)

Where

H is the enthalpy of the system

U is the internal energy of the system

P is the pressure at the boundary of the system and its environment.

(1) In thermodynamics the quantity U + PV is a new state function and known as the    enthalpy of the system and is denoted by H=U+PV. It represents the total energy stored in the system.

(2) It may be noted that change in enthalpy is equal to heat exchange at constant pressure.(3) Enthalpy is also an extensive property as well as a state function.                                 (4)The absolute value of enthalpy cannot be determined, however the change in enthalpy can  be experimentally determined.

               DH = DU + D(PV)

(5) Change in enthalpy is a more useful quantity than its absolute value.

(6) The unit of measurement for enthalpy (SI) is joule.

(7)The enthalpy is the preferred expression of system energy changes in many chemical and physical measurements, because it simplifies certain descriptions of energy transfer. This is because a change in enthalpy takes account of energy transferred to the environment through the expansion of the system under study.

(8)The change dH is positive in endothermic reactions, and negative in exothermic processes. dH of a system is equal to the sum of non-mechanical work done on it and the heat supplied to it.

(9) For quasi static processes under constant pressure, dH is equal to the change in the internal energy of the system, plus the work that the system has done on its surroundings. This means that the change in enthalpy under such conditions is the heat absorbed (or released) by a chemical reaction.

NOTE:

Transfer of heat at constant volume brings about a change in the internal energy(DU) of the system whereas that at constant pressure brings about a change in the enthalpy (DH) of the system.

For Ideal gas

                     H=U+PV   and U=f(T)

                     PV=nRT

                    H=U+nRT   and H=f(T) only for ideal gas

For other substance and real gas

                     H=U+PV  

                     U=f(P,V,T) and H=f(,PV,T)

       So         H=f(P,T)/ f(V,T)/ f(P,V)

H=f(T,P)

           dH=(dH/dT)_p dT+(dH/dP)_T dP------------------------------------- (1)

H=f(V,T)

           dH=(dH/dV)_T dV+(dH/dT)_V dT------------------------------------- (2)

H=f(P,V)

           dH=(dH/dV)_P dV+(dH/dP)_V dP------------------------------------- (3)

Out of the above three relation H as function on of (T,P) Has a greater significance. The above differential equation simplified for different substance for different condition.

For isobaric process : dP = 0 (Molar Heat capacity at constant Pressure)

We known

                 Q_P=nC_pmdT (Molar Heat capacity at constant Pressure)

                 C_pm= (dQ/dT)_P   for 1 mole of gas

                            dQ=dH  at dP=0

   then         C_pm= (dH/dT)_P  

         

For an ideal gas: change in enthalpy at constant temperature with change in

pressure is zero. i.e.

Continue...........

RELATIONSHIP BETWEEN DH AND DU

THIRD LAW OF THERMODYNAMICS (TLOT)

We know that on increasing the entropy of a pure crystalline substance increases because molecular motion increases with increase of temperature and decreases on decreasing temperature. Or we can say that the entropy of a perfectly crystalline solid approaches zero as the absolute zero of temperature is approached. Which means that at absolute zero every crystalline solid is in a state of perfect order and its entropy should be zero. This is third law of thermodynamics.

We can calculate absolute value of entropy of any substance in any state (solid ,liquid, gas) at any temperature by calculating dS for the processes in going from the initial state to the state of the substance for which entropy is to be calculated.

EXAMPLE: Find the entropy of change when 90 g of H₂O at 10^C was converted into steam at 100^C.( Given Cp(H₂O)=75.29 JK^-1 and dH_Vap=43.932 k JK^-1  mol^-1)

SOLUTION:  

EXAMPLE:  Calculate dG for

 (i) H₂O (l, 1 atm, 300 K ) ------à H₂O (g, 1 atm, 300 K)

 (ii) H₂O (l, 2 atm, 373 K ) -----à H₂O (g, 2 atm, 373 K)

 Given: dH₃₇₃ = 40 kJ ; C_P(H₂O,l)= 75 J / mol /K ; C_P(H₂O,g) = 35 J / mol / K

SOLUTION:  (1)

(2)

WORK DONE (PV-WORK ANALYSIS )

Energy that is transmitted from one system to another in such a way that difference of temperature is not directly involved. This definition is consistent with our understanding of work as dw= Fdx. The force (F) can arise from electrical, magnetic, gravitational & other sources. It is a path function.

PV-Work analysis: Consider a cylinder fitted with a friction less piston, which enclosed no more of an ideal gas. Let an external force F pushes the piston inside producing displacement in piston.

Let distance of piston from a fixed point is x and distance of bottom of piston at the same fixed point is l. This means the volume of cylinder = (l – x) A where A is area of cross section of piston.

For a small displacement dx due to force F, work done on the system.

                        dw = F.dx

                                                  Also  F = PA 

                        dW = PA.dx

                                                    V = (l – x)A

                         dV = –A . dx

                        dW = –Pext. dV

             

Note :

(1): Litre atmosphere term is unit of energy. It is useful to remember the conversion: 

         1 litre atm= 101.3 Joules = 24.206 Cal.

(2): During expansion dV is positive and hence sign of w is negative since work is done by the system and negative sign representing decease in energy content of system. During compression, the sign of dV is negative which gives positive value of w representing the increase in energy content of system during compression.

EXAMPLE.1 mole of ideal monatomic gas at 27°C expands adiabatically against a constant external pressure of 1.5 atm from a volume of 4dm³ to 16 dm³.            Calculate (i) q (ii) w and (iii) DU

SOLUTION:   (i) Since process is adiabatic  \ q = 0

                       (ii) As the gas expands against the constant external pressure.

                             W = - PVd=-P(V2-V1)

                             W =-1.5(16-4)

                             W= - 18 dm³

                            (iii) DU = q + w = 0 + (-18) = -18 atm dm³

(1) Work done in Isothermal Irreversible Process:
(2) Work done in Isothermal Reversible Process:
(3) Work done in Adiabatic Irreversible Process:
(4) Work done in Adiabatic Reversible Process:

SECOND LAW OF THERMODYNAMICS (SLOT)

(1) FLOT is law of conservation of energy, and according to it all chemical and physical processes take place in a such way that energy remain constant

(2) In FLOT we introduce enthalpy and internal energy.

(3)When two body at different temperatures are kept closed toeach other there will be transfer of heat while the transfer of heat take place called not explain by FLOT.

(4) we can not explain spontaneous and irreversible behaviour of processes by using FLOT.

In 2nd law of thermodynamics we will introduce entropy as criteria of spontaneity.

STATEMENT OF 2ND LAW :. "It is impossible to construct the heat engine which can take heat from a source and completely convert into work without creating any disturbance in the surrounding.