n engineer is regyvired to compress nitrogen isothermall fron latm 7 atm. He kno
ID: 3307902 • Letter: N
Question
n engineer is regyvired to compress nitrogen isothermall fron latm 7 atm. He knows that he cannot achieve this practically, so he instead 7 Carries Ou t the conpression through step process so barica compresses the gas and then isockoricaly compresses it to the isotherm that passe th rough the nilial state of the gas . rror here ,be iso barically conpresses the jes once more and then iseckoricall reaches 7 atm! What are the inter mediate voumes and presseres of -the gas so that the least work is exerted on it? What is the r tio of the engineer's wor k to the ideal isotlermal work?Explanation / Answer
Initial Pressure P1 = 1 atm
initial volume V1 = 1 cu.m
Final pressure P2 = 7 atm
Final volume on the isothermal V2 = 1/7 cu.m ( P1V1 = P2V2 , isothermal process)
Let Pi , Vi are the intermediate pressure and volume on the isothermal
P1V1 = P2V2 = PiVi
Step-1 , isobaric process const pressure P1 = 1 atm
Work done W1 = P1(V1-Vi)
step-2 , isbaric process, const pressure Pi
inital volume - Vi
final volume V2
Work done W2 = Pi(Vi-V2)
Total work done W = W1 + W2 = P1(V1-Vi)+Pi(Vi-V2)
= 2P1V1 - (P1Vi + P1V1V2/Vi) - - - ( replace PiVi = P1V1=P2V2 )
To minimise W set it first derivative wrt Vi to 0
dW/dVi = P1 - P1V1V2/Vi2 =0
Vi = (V1V2)1/2 = 0.378 cu.m
Pi = 1/0.378 = 2.646 atm
work done W = 2P1V1 - (P1Vi + P1Vi) ( put v1v2 = Vi2 )
= 2*101000(1 - 0.378 )
= 125644 J
work done in isothermal process = nRT ln(V2/V1) = P1V1 *ln(V2/V1) = 101000*1 * ln(1/7) = -196537 J
ratio of work done = 125644/196537 = 0.639
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