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BSC. -II PHYSICAL CHEMISTRY THERMODYNAMICS-II It does not give information concerning feasibility of a thermodynamic process. NOT EXPLAINED BY FIRST.

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Presentation on theme: "BSC. -II PHYSICAL CHEMISTRY THERMODYNAMICS-II It does not give information concerning feasibility of a thermodynamic process. NOT EXPLAINED BY FIRST."— Presentation transcript:

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2 BSC. -II PHYSICAL CHEMISTRY THERMODYNAMICS-II

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4 It does not give information concerning feasibility of a thermodynamic process. NOT EXPLAINED BY FIRST LAW

5 It tells that work obtained is equal to heat absorbed but in actual heat absorbed can not be completely converted into work. Limitations motivating second law

6 5 The second law of thermodynamics It will arouse changes while the heat transfers from low temp.substance to high temp. one.

7 All spontaneous processes are thermodynamically irreversible..

8 The complete conversion of heat into work is impossible without leaving its effects somewhere. SOME AMOUNT OF HEAT MOVES TO SURROUNDING

9 Without the use of any external agency,heat can not by itself pass from a colder body to a hotter body. HOT BODY COLD BODY HEAT TRANSFER EXTERNAL AGENCY

10 Sadi Carnot devised a technique, called Carnot Cycle. The Carnot Cycle consists of 4 different operations: 1.Isothermal expansion 2.Adiabetic expansion 3.Isothermal compression 4.Adiabetic compression Sadi Carnot devised a technique, called Carnot Cycle. The Carnot Cycle consists of 4 different operations: 1.Isothermal expansion 2.Adiabetic expansion 3.Isothermal compression 4.Adiabetic compression

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12 ISOTHERMAL(T 2 ) ISOTHERMAL(T 1 ) ADIABETIC I II III IV PRESSURE VOLUME A(V 1 ) B(V 2 ) C(V 3 ) D(V 4 ) State 1 T 2,V 1 State 1 T 2,V 1 Pressure –volume diagram (indicator diagram)

13 State 1 T 2,V 1 State 2 T 2,V 1 State 4 T 1 State 3 T 1 isothermal adiabatic isothermal q=0, W 4 =C v  T q=0, w 2 = - C V  T w 1 = -RT 2 ln (V2/V1) q 2 = - w W 3 = RT 1 ln (V4/V3) -q 1 = w 3

14 The net work done by the system (w)=(-w 1 )+(-w 2 )+w 3 +w 4 w=RT2lnV2/V 1 - CV(T2-T1) + RT1lnV4/V3 + CV(T2-T1) w=RT 2 lnV 2 /V 1 +RT 1 ln V 4 /V 3 Since V 1 and V 4 lie on one adiabetic curve and V 3 and V 2 lie on Another, it follows that ( V 4 /V 1 ) =T 2 /T 1 Comparing both the equations, we get: 1 2 3 4

15 V 4 /V 3 = V 1 /V 2 Putting these values in equation 1, we get : The net work done by the system,w = R(T 2 -T 1 ) ln V 2 /V 1 The heat absorbed by the system during isothermal expansion is equal to amount of work done during isothermal expansion. Therefore -w 1 = q 2 =RT 2l lnV 2 /V 1 7 5 6

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88 State 1 T hot State 2 T hot State 4 T cold State 3 T cold isothermal adiabatic isothermal q=0, w=C V  T q=0, w=-C V  T w= -nRT hot ln(V2/V1) q=-w w= -nRT cold ln(V4/V3) q=-w

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115 114 The second law of thermodynamics It will arouse changes while the heat transfers from low temp.substance to high temp. one.

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122 121 The second law of thermodynamics It will arouse changes while the heat transfers from low temp.substance to high temp. one.

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