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Explain reversible and irreversible processes . Describe the working of Carnot engine . Obtain an expression for the efficiency .

Answer»

Solution :Reversible process : In reversible process a thermodynamic system can be retraced back in opposite direction to the changes that take place in the direct process or in forward process .
A reversible process is only an ideal concept .
Examples for reversible process :
1) Peltier effect and Seebeck effect .
2) Fusion of ice and vapourisation of water .
Irreversible process : A thermodynamic process that cannot be taken back in oposite direction is called "irreversible process."
Examples :
1) Work done against frition
2) Magnetization of materials .
Carnot's Engine : Carnot's engine works on the PRINCIPLE of reversible process with-in the temperatures`T_(1) and T_(2)`
It consists of four continuous PROCESSES .
Total process is known as Carnot Cycle .
Step 1 : In carnot cycle the 1st step consists ofisothermalexpansionof gases . So TEMPERATURE T is constant , P ,V changes areshown as`P_(1)V_(1)T_(1)overset("to")rarrP_(2)V_(2)T_(2)`
Work done in isothermal process
`W_(1rarr2)=Q_(1)-muRT_(1)log_(e)((V_(2))/(V_(1)))""....(1)`
Step : 2 In this stage gases will expand adiabatically . So energy to the system Q is constant .
So , P,V relation is `P_(2)V_(2)T_(1)overset("to")rarrP_(3)V_(3)T_(2)`
Work done in adiabaticallyprocess
`W_(2rarr3)=(muR(T_(1)-T_(2)))/(gamma-1)"" ...(2)`

Step 3 : In this stage gases will be compressed isothermally , So`P_(1)V` changes are `P_(3)V_(3)T_(2)overset("to")rarrP_(4)V_(4)T_(2)`
Work done in isothermal compression
`W_(3rarr4)=Q_(1)-muRTlog_(e)((V_(3))/(V_(4)))""....(3)`
Step 4 : In the fourth stage the gas suffers adiabatic compression and returns to original stage .
So , P,V changes are`P_(4)V_(4)T_(2)overset("to")rarrP_(1)V_(1)T_(1)`
Work done`W_(4rarr1)=(muR(T_(2)-T_(1))/(gamma-1)"" ...(4)`
Total work done `W=muRT_(1)log((V_(2))/(V_(1)))`
`+(muR)/(gamma-1)(T_(1)-T_(2))-muRT_(2)log(V_(3)/(V_(4)))-(muR)/(gamma-1)(T_(1)-T_(2))`
`therefore W=muRT_(1)log((V_(2))/(V_(1)))-muRT_(2)log(V_(3)/(V_(4)))`
The total work done `W=Q_(1)-Q_(2)` i.e., the difference to heat energy ABSORBED from source and heat energy given to sink Efficiency of Carnot engine
`eta=("work done by Carnot engine")/("heat energy supplied")`
`thereforeeta=(Q_(1)-Q_(2))/(Q_(1))" or " eta=1-(Q_(2))/(Q_(1))=1-(T_(2))/(T_(1))`


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