CBSE Class 11 Physics Notes : Thermodynamics part 3

11. Principle of a Refrigerator

• Refrigerators works in reverse direction of heat engines.
• In refrigerators working substance extracts heat Q₂ from sink at lower temperature T₂
• Some external work is performed by the compressor of refrigerator and then heat Q₁ is rejected to the 
  source, to the radiator of the refrigerator.
  
   
  Coefficent of performance : 
       β= Amount of heat absorbed from the cold reservoir / work done in running the mechinery
  Q₂ - heat absorbed from cold reservoir.
  Q₁ - heat rejected to hot reservoir during one complete cycle
  W = (Q₁-Q₂ ) is the work done in running the machinery
  thus,
       β= Q₂/W =Q₂/(Q₁-Q₂)          (18)
• Like heat engines refrigerators can not work without some external work done on the system. Hence coefficent of performance can not be infinite.

12. Second law of thermodynamics

• First law of thermodynamics states the equivalance of heat and energy.
• It does not state anything about the limitation in the conversion of heat into work or about the condition necessary for such conversion.
• Second law of thermodynamics is generalization of certain experience and observation and is concerned with tine direction in which energy flow takes place.
• This law can be stated in number of ways. Although differently said, they are essentially equvalent.
  (i)Kelvin Plank Statment :
  "It is impossible to construct a device which, operating in a cycle, has a sole effect of extracting heat from a reservoir and performing an equivalent amount of work".
  (ii)Clasius Statement :
  "It is impossible for a self acting machine, unaided by enternal agency, to transfer heat from a colder body to a hotter body".
• It can ne proved that these two statements of second law are completely equivalent and voilation of Kelvin Plank statement leads to voilation of Clasius statement and vice-versa.

13. Reversibility and irreversibility

• Reversible process is the one which can be retraced in opposite order by changing external conditions slightly.
• Those processes which can not be retraced in opposite order by reversing the controling factors are known as irreversible process.
• It is a consequence of second law that all the natural processes are irreversible process.
• Conditions for reversibility of a process are
       (i) Process is performed quasi-statically
       (ii) it is not accompained by any dissipative effects.
• It is impossible to satisfy these two conditions perfectly, thus requessible process is purely an ideal abstraction.

14. Carnot's Heat Engine

• According to second law of thermodynamics, no heat engine can have 100% efficiency
• Carnot�s heat engine is an idealized heat engine that has maximum possible efficiency consistent with the second law.
• Cycle through which working substance passed in Carnot�s engine is known as Carnot�s Cycle.
• Carnot's engine works between two temperatures
       T₁ - temperature of hot reservoir
       T₂ - temperature of cold reservoir
• In a Complete Carnot's Cycle system is taken from temperature T₁ to T₂ and then back from
  temerature T₂ to T₁.
• We have taken ideal gas as the working substance of cornot engine.
• Fig below is an indicator digram for Cornot Cycle of an ideal gas
  
   
  (i) In step b→c iso thermal esepansion of gas taken place and thermodynamic variables of gas changes from (P₁, V₁,T₁) to (P₂,V₂,T₁)
• If Q₁ is the amount of heat absorbed by working substance from the source and W₁ the work done by the gas then from eqn (13)
       Q₁ = W₁ = nRT1 ln (V₂/V₁)     (19)
  as process is iso thermal.
  (ii) Step c→d is an adiabatic expension of gas from (P₂, V₂, T₁) to (P₃,V₃,T₂). Work done by gas in adiabatic esepansion is given by eqn (16)
       W₂ = nR (T₁-T₂)/(γ-1)               (20)
  (iii)Step d→a is iso-thermal compression of gas from (P₃,V₃,T₂) to (P₄,V₄,T₂). Heat Q₂ would be released by the gas to the at temperature T₂
• Work done on the gas by the environment is
       W₃ = Q₂ 
             = nRT₂ln(V₃/ V₄)               (21)
  (iv)Step a→b is adiabatic compression of gas from (P₄, V₄, T₂) to (P₁, V₁, T₁)
• Work done on the gas is
       W₄ =nR (T₁-T₂)/(γ-1)                    (22)
• Now total work done in one complete cycle is
       W = W₁ + W₂ - W₃ - W₄
        = nRT₁ln(V₂/V₁)-nRT₂ln(V₃/V₄)          (23)
  as W₂ = W₄
  
• Efficency of carnot engine
       η=W/Q₁ = 1-(Q₂/Q₁)
        = 1-(T₂/T₁)ln(V₃/V₄)/ln(V₂/V₁)          (24)
  or     η= 1-[T₂ln(V₃/V₄)/T₁ln(V₂/V₁)]          (25)
  Since points b and c lie on same iso thermal
  ⇒ P₁V₁=P₂V₂                    (26)
  also points c and d lie on same adiabatic
  ⇒     P₂(V₂)^γ=P₃(V₃)^γ                    (27)
  also points d and a lie on same iso thermal and points a and b on sum adiabatic thus,
        P₃V₃=P₄V₄                    (28)
       P₄(V₄)^γ=P₂(V₁)^γ                         (29)
  multiplying all the above four eqns me get
       V₃/₄ = V₂/V₁                    (30)
  Putting this in equation (25) we get
       η= 1-(T₂/T₁)                         (31)
  From above eqn we can draw following conclusions that efficency of Carnot engine is
  (i) independent of the nature of working substance
  (ii) depend on temperature of source and sink

15. Carnot Theorem

• Carnot Engine is a reversible engine.
• Carnot�s theorem consists of two parts
  (i) no engine working between two given temperatures can be more efficent than a reversible Carnot engine working between same source and sink.
  (ii) all reversible engines working between same source and sink (same limits or temperature) have the same efficiency irrespective the working substance

                                                                 

                                                                         credit goes to physicscatalyst team

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