LECTURE ONTHE CONCEPT OF SECOND LAW OF THERMODYNAMICS

BY

DR. NNABUK OKON EDDY

Outline

Introduction The need for the second law Concept of entropy Statements of the second law Properties of entropy Derivation of equation for entropy Consequences of the second law

Introduction The second law explains the phenomenon of

irreversibility in nature The need for the second law arises because

the first law failed in some aspects. For example,

It fails to explain why natural processes have a preferred direction

The first law fails to produce thermodynamic functions that can be used to predict the direction of a spontaneous reaction

The second law deals with entropy

Entropy

The key concept for the explanation of phenomenon through the second law is the definition of a physical property called entropy

Entropy is a measure of the degree of disorderliness of a system.

A change in entropy of a system is the infinitesimal transfer of heat to a close system driving a reversible process divided by the equilibrium temperature (T) of the system, i.e dS = dqrev /T

Statements of the second law No process is possible whose sole result is the

transfer of heat from a body of lower temperature to a body of higher temperature (Clausius-Mussoti)

No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work (Kelvin-Plank)

Equivalent ways of stating the laws are i. the entropy of a spontaneous reaction

increases and tends toward a maximum ii. After any spontaneous reaction, work must

be converted to heat in order to restore the system to its initial state

Properties of entropy

Entropy is a state function: its properties depends on the initial and final state of the system

Entropy is additive. i.e ST = S1 + S2 + S3 + -----

Entropy is a probability function

Derivation of expression for entropy change

If the probability of finding a system in state 1 and 2 are W1 and W2, then the probability of finding the system in the two parts is the total probability, W = W1 x W2,

S(W) = S(W1) x S(W2) = S(W1) + S(W2) (1)

Conditions set by Eq 1 can only be fulfilled if entropy is logarithm dependent, i.eS = log(W1 x W2) = logW1 x logW2 (2)

Consider an ideal gas expanding into two systems joined together, the probabilities for the first and second is proportional to their respective volumes, therefore, W1 = aV1, W2 = aV2 and since S is additive, S = S2 – S1

= log(aV2) – log(aV1) = log(V2/V1)

From first law of thermodynamics, it can be shown that the reversible work done = reversible heat absorbed = nRTln(V2/V1) and if we multiply S by the constants, 2.303R, we have, qads = T x S. It therefore follows that S can be expressed as follows

S = qads/T (3)

Consequence of the second law of thermodynamics

We shall consider the following consequences of the 2nd law of thermodynamics,

Entropy change for an ideal gas Entropy of mixing ideal gases Carnot cycle Free energy change

Entropy change for an ideal gasdS = ∫ 𝑑𝑞𝑇 = 𝐶𝑣𝑑𝑇𝑇 + RT 𝑑𝑉𝑉 4

= CVln((𝑇2𝑇1) + Rln(𝑉2𝑉1) 5

From the general gas equation, it can be shown that 𝑉2𝑉1 = (𝑃1𝑃2)( 𝑇2𝑇1) and by substituting for V2/V1, we

have,

S = CVln((𝑇2𝑇1) + R[(ln(𝑃1𝑃2 ) + ln(𝑇2𝑇1)] 6

The relationship between Cp, CV and R can be written as Cp = CV + R, therefore equation 6 can be written as follows

= CPln((𝑇2𝑇1) - Rln ln(𝑃2𝑃1 ) 7

Equation 7 can be applied to three special cases as follows,

i. Isothermal condition, S = 2.303RTlog(𝑉2𝑉1)

ii. Isobaric condition, S = 2.303Cplogn((𝑇2𝑇1)

iii. Isochoric condition, S = 2.303CVlog((𝑇2𝑇1)

Entropy of mixing ideal gasesConsider two gases, A and B mixed together. The work done in mixing the gases can be defined as follows,

dW = WA + WB = PAdVA + PBdVB 8

From the ideal gas equation, P = nRT/V, therefore, PA = nART/VA and PB = nART/VB. substituting for PA and PB in equation 8 yields equation 9 and upon simplication, equation 10 is obtained,

∫dW = nA𝑅𝑇 𝑉𝐴𝑉𝐴𝑉𝐴+𝑉𝐵 𝑑𝑉𝐴 + RT 𝑉𝐵𝑉𝐵

𝑉𝐴+𝑉𝐵 𝑑𝑉𝐵 ) 9

W = RTln(nA + nB)/nA + RTln(nA + nB)/nB 10

= nARTln( 1𝑥𝐴) + nB RTln( 1𝑥𝐵) 11

Also, since W = dqads and S = 𝑑𝑞𝑎𝑑𝑠𝑇 , S can be expresses as follows,

S = -2.303RT(nAlnXA + nBlnXB) 12

Reversible cycle and efficiency: Carnot cycle

A cycle process in which a succession of changes occurs as a results of which the system returns to its original state and all properties assume their original values. This findings was made by Carnot and the cycle is commonly called Carnot cycle. Carnot cycle consist of four major components

i. Reversible isothermal expansion ii. Adiabatic reversible expansion iii. Isothermal reversible compression iv. Adiabatic reversible compression

In step i., the change in internal energy is zero, and S = - 𝑞2𝑇2. In step ii, q = 0 and S = 0. In

step iii, S = - 𝑞1𝑇1 and the change in internal energy = 0. Finally in step iv, q=0 and S = 0

In a carnot cycle, the heat absorbed = q2 – q1 and the efficiency is defined as the fraction of the workdone to the heat absorbed at higher temperature. i.e Efficiency = W/q2. Also the total entropy change in the cycle is q2/T2 – q1/T1 = 0. Therefore, q2/q1 = T2/T1 hence

Efficiency = (q2 – q1)/q2 = (T2 – T1)/T2 13

Free energy Enthalpy and entropy are state functions obtained from the first and second law of thermodynamics respectively. Enthalpy measures the tendency towards orderliness while entropy represents the tendency towards disorderliness. The difference between orderliness and diorderliness leads to the concept of free energy, which can be expressed as follows,

G = H - TS 14

Note, H = E + PV , therefore,

G = E + PV - TS 15

Differentiating both sides of equation 15 yields equation 16

dG = dE + VdP + PdV - TdS + SdT 16

Note, qads = dE + PdV and qads = TdS and by substitution, equation 16 simplifies to equation 17

dG = VdP - SdT 17

Effect of pressure and temperature

At constant temperature,

dG = VdP and dG𝑑𝑃 = V. Also, since PV = RT, V = 𝑅𝑇𝑃 , then

dG = RT𝑑𝑃𝑃 18

integration of equation 18 yields

dG = RTln𝑃2𝑃1 19

At constant pressure, dG = -SdT and (dG𝑑𝑇)p = -S

Also, (dG2𝑑𝑇)p - (dG1𝑑𝑇)p = -S = S2 - S1 20

But G −H𝑇 = -S 21

Then, 𝑑𝑑𝑇(∆G𝑇) = (1𝑇) 𝑑𝑑𝑥 (∆G) - (∆G) 𝑑𝑑𝑇(1𝑇) = 1𝑇(G −H𝑇 ) - ∆G𝑇2 = -∆H𝑇2 22

Since From equation 22, it can be deduced that a plot of ∆G𝑇 versus 1/T should give a

straight line with slope equal to H

Note, 𝑑𝑑𝑇(∆G𝑇 ) = - ∆H𝑇2 and 𝑑𝑑𝑇(1𝑇) = 1/T2, therefore 𝑑𝑑𝑇((G/T1/𝑇) = H 23

S and spontaneousity of a reaction

When S is positive, spontaneous reaction

When S is zero, reaction at equilibrium When S is negative, non spontaneous Limitation is that we who measures the

entropy are part of the environment. Therefore S is not a unique parameter for predicting the direction of a chemical reaction

G and spontaneousity of a reaction

G > 0, non spontaneous (H > TS) G < 0, spontaneous (H < TS) G = 0, reaction at equilibrium (H = TSG is a state function obtained at constant pressure.

At constant volume the state function is work function expressed as

A = E - TSWhen A > 0, spontaneous

When A <0 , non spontaneous

When A = 0, at equilibrium

CONCLUSION

Thermodynamic function obtained from the second law is entropy

Entropy is a measure of disorderliness while enthalpy measures orderliness

Entropy data must be combined with enthalpy (or internal energy data) in order to predict the direction of a chemical reaction