Second Law of Thermodynamics

The second law of Thermodynamics helps us to determine the direction in which energy can be transformed. It also helps us to predict whether a given process or chemical reaction can occur spontaneously or not.  

According to Kelvin: “It is impossible to use a cyclic process to extract heat from a reservoir and to convert it into work without transferring at the same time a certain amount of heat from a hotter to colder part of the body”.  

Entropy Change: Entropy change is the state function and it is the ratio of heat change in a reversible process by the temperature.  

ΔS = q_rev / T  

Thermodynamically irreversible process is always accompanied by an increase in the entropy of the system and its surroundings taken together while in a thermodynamically reversible process, the entropy of the system and its surroundings taken together remains unaltered.    

Solved example:  Calculate entropy change for vaporization of 1 mole of liquid water to steam at 100°C if ΔH_V = 40.8 kJmol^-1.    

Solution: For entropy change of vaporization  

                        ΔS_v = ΔH_v / T  

                        ΔS_v = 40.8 × 10³ / 373 = 109.38 Jk^–1 mol^–1 

Solved example: A system changes its state irreversibly at 300 K in which it absorbs 300 cals of heat. When the same change is carried out reversibly the amount of heat absorbed is 900 cals. The change in entropy of the system is equal to  

                        (A) 1 cal K^–1                                       (B) 3 cals K^–1  

                        (C) 2 cal K^–1                                       (D) 1.5 cals K^–1      

Solution: (B) 

Physical Significance of Entropy: Entropy is the measure of disorderness because spontaneous processes are accompanied by increase in entropy as well as increase in the disorder of the system. Thus, increase in entropy implies increase in disorder.    

Solution: Which of the following statement is true?  

                        (i) A closed system shows exchange of mass and not energy with surroundings.     

                        (ii) Entropy change for fusion reaction is positive.           

                        (iii) Heat is a measure of quantity of energy whereas temperature is a measure of intensity of energy.      

Solution: (i) False          (ii) True          (iii) True  

Some Other State Function: For a spontaneous process entropy change is positive and if it is zero, the system remains in a state of equilibrium. Two other functions are also there to decide the feasibility of the reactions like work function A and free energy change G.  

A = E – TS…….(i)  

G = H – TS…….(ii)  

And ΔA = ΔE - TΔS……(iii)  

ΔG = ΔH - TΔS………...(iv)     (for a finite change at constant temperature)  

Since, ΔS = q_rev./T Hence from eq. (i)  

ΔA = ΔE – q_rev………………..(v)  

and according to first law of Thermodynamics  

ΔE - q_rev =w_rev. …………….(vi)  

If during the change, work is done by the system, it would carry a negative sign,  

-w_rev = ΔE – q_rev…………….(vii)  

Comparing the equation (v) and (vii)  

-ΔA = w_rev  

Since the process is carried out reversibly where w represents the maximum work. It is thus clear that decrease in function A gives maximum work done that can be done by the system during the given change. The work function A is also called as Helmholtz function.  

From equation (iv)  

ΔG = ΔH - TΔS  

and ΔH = ΔE + PΔV  

ΔG = ΔE + PΔV - TΔS  

Comparing it with eq. (iii)  

ΔG = ΔA + PΔV  

Since, ΔA is equal to – w, hence.  

ΔG = - w + PΔV.  

- ΔG = w- PΔV  

Hence decrease in free energy gives maximum work obtainable from a system other than that due to change of volume at constant temperature and pressure. This is called as Net Work.  

Net Work = w-PΔV = -ΔG  

The Net Work may be electrical work or chemical work.

Criterion of spontaneity: For a spontaneous process ΔG should be -ve