CBSE Class 11 Physics Notes : Thermodynamics part 2

7. Thermodynamic Processes

a) Quasi static Processes

:

• In Quasi static process deviation of system from it's thermodynamic equilibrium is infinitesimally small.
• All the states through which system passed during a quasi static process may be regarded as equilibrium states.
• Consider a system in which gas is contained in a cylinder fitted with a movable piston then if the piston is pushed in a infinitely slow rate, the system will be in quiscent all the time and the process can be considered as quasi-static process.
• Vanishingly slowness of the process is an essential feature of quasi-static process.
• During quasi-static process system at every moment is infinitesimally near the state of thermodynamic equilibrium.
• Quasi static process is an idealized concept and its conditions can never be rigoursly satisfied in practice.

(b) Isothermal Process

:

• In isothermal process temperature of the system remains constant throughout the process.
• For an iso-thermal process equation connecting P, V and T gives.
       PV = constant
  i.e., pressure of given mass of gas varies inversly with its volume this is nothing but the Boyle's law.
• In iso thermal process there is no change in temperature, since internal energy for an ideal gas depends only on temperature hence in iso thermal process there is no change in internal energy.
  Thus, 
       ΔU=0
  therefore,     ΔQ =ΔW
• Thus during iso thermal process
       Heat added (or substacted) from the system = wok done by (or on) the system

(c) Adiabatic Process

:

• Process in which no heat enters or leaves a system is called an adiabatic process
• For every adiabatic process Q=0
• Prevention of heat flow can be acomplished by surrounding system with a thick layer of heat insulating material like cork, asbestos etc.
• Flow of heat requires finite time so if a process is perfomed very quickly then process will be pratically adiabatic.
• On applying first law to adiabatic process we get
       ΔU=U₂ - U₁= - ΔW               (adiabatic process)
• In adiabatic process change in internal energy of a system is equal in magnitade to the work by the system.
• If work is done on the system contracts i.e. ΔW is nagative then.
       ΔU = ΔW
  and internal energy of system increases by an amount equal to the work done on it and temperature of system increases.
• If work is done by the system i.e., ΔW is negative
       ΔU = -Δ W
  here internal energy of systems decreases resulting a drop in temperature.

(d) Isochoric process v:

• In an isochoric process volume of the system remain uncharged throughout i.e. ΔV = O.
• When volume does not change no work is done ; ΔW = 0 and therefore from first law
       U₂ - U₁ = ΔU =ΔQ
• All the heat given to the system has been used to increase the intenal energy of the system.

(e) Isobaric Process

:

• A process taking place at constant pressure is called isobaric process.
• From equation (3) we see that work done in isobaric process is
       W = P(V₂ - V₁) nR (T₂-T₁)
  where pressure is kept constant.
• Here in this process the amount of heat given to the system is partly used in increasing temperature and partly used in doing work.

8. Work done in Isothermal process

• In an isothermal process temperature remains constant.
• Consider pressure and volume of ideal gas changes from (P₁, V₁) to (P₂, V₂) then, from first law of thermodynamics
       ΔW = PΔV
  Now taking ΔV aproaching zero i.e. ΔV𔾴 and suming ΔW over entire process we get total work done by gas so we have
       W = ∫PdV
  where limits of integration goes from V₁ to V₂
  as PV = nRT we have P = nRT / V
       W = ∫(nRT/V)dV
  where limits of integration goes from V₁ to V₂
  on integrating we get,
       W=nRT ln(V₂/V₁)               (3)
  Where n is number of moles in sample of gas taken.

9. Work done in an Adiabatic process

• For an adiabatic process of ideal gas equation we have
       PV^γ = K (Constant)               (14)
  Where γ is the ratio of specific heat (ordinary or molar) at constant pressure and at constant voluume
       γ = C_p/C_v
• Suppose in an adiabatic process pressure and volume of a sample of gas changs from (P₁, V₁) to (P₂, V₂) then we have
       P₁(V₁)^γ=P₂(V₂)^γ=K 
  Thus, P = K/V^γ
• Work done by gas in this process is
       W = ∫PdV
  where limits of integration goes from V₁ to V₂
  Putting for P=K/V^γ, and integrating we get,
       W = (P₁V₁-P₂V₂)/(γ-1)          (16)
• In and adiabatic process if W>0 i.e., work is done by the gas then T₂< T₁
• If work is done on the gas (W<0) then T₂ > T₁ i.e., temperature of gas rises.

10. Heat Engine and efficiency

• Any device which convents heat continously into mechenical work is called a heat engine.
• For any heat engine there are three essential requirements.
  (i) SOURCE : A hot body at fixed temperature T₁ from which heat engine can draw heat
  (ii) Sink : A cold body, at a fixed lower temprature T₂, to which any amount of heat can be rijectd.
  (iii) WOEKING SUBTANCE : The material, which on being supplied with heat will do mechanical work.
• In heat engine, working substances, could be gas in cylinder with a moving piston.
• In heat engine working substance takes heat from the sorce, convents a part of it into mechanical work, gives out rest to the sink and returns to the initial state. This series of operations constitutes a cycle.
• This cycle is represented in fig below
  
   
• Work from heat engine can be continously obtained by performing same cycle again and again.
• Consider,
       Q₁ - heat absorbed by working substance from sorce
       Q₂ - heat rejected to the since
       W - net amount of work done by working substance
       Q₁-Q₂ - net amount of heat absorbed by working substance.
       ΔU = 0 since in the cycle Working Substance returns to its initial condition.
  So on application of first law of thermodynamics
       Q₁-Q₂ = W
• Thermal efficiency of heat engine
       η= work output in energy units / Heat input in same energy units
        = W / Q₁ = (Q₁-Q₂ )/ Q₁
  Or, η = 1-(Q₂/Q₁)                    (17)
  from this equation it is clear that 
       Q = 1 for Q₂=0
  and there would be 100% conversion of heat absorbed into work but such ideal engines are not possible in practice

                                                                 

CBSE Class 11 Physics Notes : Thermodynamics part 3

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