Mechanical Knowledge

 Q1.Prove that PV^Y=C for reversible adiabatic process. Solution:        δQ = dU + PdV Reversible Adiabatic: δQ=0              (:: There is no heat interaction between                                                                   system and surroundings = adiabatic  dU + dV = 0 dU = -PdV m(Cv)dT = -PdV  ----------------------------------------------(1) Now we want 'Y' , so, We know      H = U + PV                  dH = dU + PdV + VdP                  dH =   δQ + VdP (:: There is no heat interaction δQ =0)              dH = VdP       m(Cp)dT = VdP       ------------------------------------------(2) Dividing (2) by (1)     Y = -V/P(dP/dV)                              (:: Y= Gama=Cp/Cv)     Y = -V/dV(dP/P) Integral both side     ∫YdV/V  +  ∫dP/P  = ∫0 Y(lnV)  + lnP = lnC ln PV^Y = lnC PV^Y = C  ------------Since, this has been found for  δQ =0 so, only                                                     apply it for reversible adiabatic 

 Q. A 1 gm nitrogen undergoes a following sequence of process in a piston cylinder arrangement.(1) An adiabatic expansion in which volume double. (2) A constant pressure process in which volume is reduced to its initial state. Represent the cycle on p -v diagram, calculate net work done by the gas. Assuming  t1 = 150°C , P1 = 5 atm, Y = 1.4, R = 297 J/kg K. Solution: t1 = 150°C                   T1 = 150 + 273 = 423 K                   P1 = 5 × 1.01325 = 5.06625 bar Process (1-2): T2/T1 = (V1/V2)^Y-1 T2/423 = (V1/2V1)^1.4-1 T2 = 320.5K Process(2-3): T3/T2 = V1/V2 T3/T2 = V1/2V1 T3 = T2/2 T3=320.5/2=160.25K Now : W(net) = W(1-2) + W(2-3) + W(3-1) W(net) = P1V1 -P2V2/Y-1  + P3V3 -P2V2 + 0 W(net) = mR(T1-T2)/Y-1   + mR(T3-T2) W(net) = 10^-3 × 297(423 - T2)/1.4 -1  + 10^-3×297(T3 - T2) W(net) = 10^-3 × 297(423 - 320.5)/0.4 + 10^-3×297(160.25 - 320.5) W(net) = 28.86 J