Module 5 : Electrochemistry
Lecture 21 : Review Of Thermodynamics
 
  Ex 21.1
Calculate the work done in the reversible and irreversible expansion of Fig 21.2, if p1 = 4 atm, pC = 3 atm, PD = 2 atm, PE = 1 atm, V1 = 10 lit, VC= 13.33 lit, VD = 20 lit and V2 = 40 lit. Take T = 298 K and assume ideal gas behaviour. The number of moles of gas n = 1.635, 1atm = 1.01325 bar, 1 bar =105 N/m 2, R = 8.314 J/(mol K) = 0.08025 lit atm/ (mol K)
  
Solution
For an ideal gas, PV = nRT

Work done on the surroundings = Pdv

If the system is in equilibrium throughout (reversible process)
P = nRT / V
And w sur = nRT/ V dv = nRT ln V 2 / V 1 = 55.45 lit atm
              = 5.62 kJ (Because 1 lit atm = 101.325 J or 1 lit atm = 1.01325 kJ)
In the irreversible process, using eq (21.3),
w irr = 3(13.33-10) + 2(20-13.33) + 1(40-20)
       = 43.33 lit.atm
       = 4.39 kJ (Because 1 lit atm = 101.325 J or 1 lit atm = 1.01325 kJ)

as

It is seen that the work done in an irreversible process is less than that in a reversible process. Note that the work done on the system = - w sur . In an expansion process, work done on the system is negative.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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