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Entropy changes in irreversible Processes

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Presentation on theme: "Entropy changes in irreversible Processes"— Presentation transcript:

1 Entropy changes in irreversible Processes
To obtain the change in entropy in an irreversible process we have to calculate DS along a reversible path between the initial state and the final state. Freezing of water below its freezing point Irrev H2O( l , -10 °C) H2O( s , -10 °C) H2O( l , 0°C) H2O( s , 0 °C)

2 A sample of 1. 00 mole of monoatomic perfect gas with Cv,m = 1
A sample of 1.00 mole of monoatomic perfect gas with Cv,m = 1.5 R, initially at 298 K and 10 L, is expanded, with the surroundings maintained at 298 K, to a final volume of 20 L, in three ways (a) isothermally and reversibly (b) isothermally against a constant external pressure of 0.5 atm (c) adiabatically against a constant external pressure of 0.5 atm . Calculate ΔS and ΔSsurr, ΔH and ΔT for every path.

3 Question? A 50 gm mass of Cu at a temperature of 393 K is placed in contact with a 100 g mass of Cu at a temperature of 303 K in a thermally insulated container. Calculate q and ΔStotal for the reversible process. Use a value of Jg-1K-1 for specific heat capacity of Cu.

4 Absolute entropy of a substance
Third law of thermodynamics: The entropy of each pure element or substance in a perfectly crystalline form is zero at absolute zero.

5 Spontaneous process

6

7 Enthalpy/ Entropy driven process
A-Enthalpy driven B-Entropy driven

8 Hydrophobic Effect in Protein Folding
HOH + S=+ Folded Unfolded More Hydrocarbon-Water Interfacial Area Less Hydrocarbon-Water

9 Hydrophobic forces Hydrophobic interactions are considered to make a major contribution to stabilizing the native structures of proteins in aqueous environment. The hydrophobic effect is a unique organizing force based on repulsion by the solvent instead of attractive forces at the site of organization. Thermodynamics of protein unfolding can be explained on the basis of transfer of non-polar groups from organic solvent to water.

10 Enthalpy and Entropy Go = Ho - TSo
Typically, G and H are measurable and S calculated H and S Provide Mechanistic Insight In very rough generalities: H related to bond formation/breaking S related to configurational freedom and water ordering dH = CpdT Gives Thermal Dependence of K

11 Extreme ordering of waters caused by hydrophobic molecule
Reason for negative S O H Hydrophobic bond Water molecules seek favorable positions and partially freeze their thermal motion. Extreme ordering of waters caused by hydrophobic molecule Net effect is entropic rather than energetic in nature just because the energy of H-bonds is extremely high and waters would prefer to become frozen and sacrifice a part of their freedom than to lose the large energy of a hydrogen bond.

12 A simple binding process
Binding can be enthalpy or entropy driven + +

13 Enthalpic contribution to Free energy of binding
Ionic and hydrogen bonds. Electrostatic interaction. Van der Waal interaction

14 entropy of binding Binding is favoured if it leads to a net increase in disorder or entropy. This includes entropy of the system (interacting molecules and solvent) represented as change in entropy or S the environment (everything else) as the system releases or absorbs heat it changes the entropy of the surroundings heat release is measured as change in enthalpy or H 14

15 Combining the First and Second law

16 S H P U G T V A Good people have studied under very able teachers.

17

18 The Thermodynamic Square
P U G T V A

19 Thermodynamic equation of state

20 Prove it Derive the expression for for a gas following the virial equation with Z=1+B/V

21 True or False ?? No heat transfer occurs when liquid water is reversibly and isothermally compressed. During an adiabatic, reversible process at constant volume the internal energy can never increase. The internal energy of a system and its surroundings is never conserved during an irreversible process, but is conserved for reversible processes. The work done by a closed system can exceed the decrease in the system’s internal energy

22 When extra (useful) work is involved.

23 Important Thermodynamic equations

24 Helmholtz free energy Maximum work done (including expansion work) can be calculated by measuring the change in A.

25 Gibb’s Free Energy

26 E.M.F. work

27 Displacement of metal from metal salts
A more active metal can "displace" a less active one from a solution of its salt. “Active" metals are all "attacked by acids“. Zn(s) + Cu2+ → Zn2+ + Cu(s) .

28 Pressure as a function of height
An increase in height will correspond to a decrease in pressure. Boiling point of water is raised if the pressure above water is increased; it is lowered if the pressure is reduced. This explains why water boils at 70 °C up in the Himalaya.

29 The Thermodynamics of a Rubber band

30 Q.Experiments show that the retractive force f of polymeric elastomers as a function of temperature and expansion L is given by f(T,L) = aT(L-L0) where a and L0 are constants. (a)Use Maxwell relations to determine the entropy and enthalpy at constant T and p. (b) If you adiabatically stretch a rubber band by small amount, its temperature increases but volume does not change. Derive an expression for its temperature as a function of L, L0, a and C (heat capacity).

31 The variation of Gibb’s energy with temperature.

32 Variation of the Gibb’s energy with pressure

33 Question ? Calculate ΔG for the conversion of 3 mol of liquid benzene at 80 C (normal boiling point) to vapour at same temperature and a pressure of 0.66 bar? Consider the vapour as an ideal gas. In the transition CaCO3 (aragonite) to CaCO3 (calcite), ΔGm (298) = -800 J and DVm = 2.75 cm3. At what pressure would aragonite become the stable form at 298 K?.


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