Since y=-= i+r i+1-14, so we get F=2.5 (a) Using the equation of state for an ideal gas: PV=nRT We have the original volume of the gas at the beginning and end of the cycle h=R711×8315×(27315+20)=241×10(m2) P 1×105×1013 b) For isochoric process P Pa The pressure of the gas at the completion of path(1)is P=P2=1×1013027315+150=146×0(a) 273.15+20 For isothermal process: PBB=PVC The volume of the gas at the completion of path(2) P1.46×10°×241×102 1×105×1013 =347×10-(m) P The temperature of the gas at the completion of path(2)is TB=273+150=423K (c)(1)The work Path 1: WI=OJ Path2:W2= RTIn -c=8315×(273+150)×n 3.47×10 2.41×10 2=1282.13(J) Path 3 W3=-PV-V)=-1×1013×103×(347×102-241×10-2)=-107378(J) The total work done by the gas Wu=W+W2+W=0+128213-1073.78=208.35(J) (2)The heat Path1:g1=-C1(T-7A)=1×2.5×8.315×(423-293)=2702.38(J)Since 1.4 1 2 2 1 = + = + = = i i Ri R i C C V P γ , so we get i=2.5 (a) Using the equation of state for an ideal gas: PV = nRT We have the original volume of the gas at the beginning and end of the cycle 2.41 10 (m ) 1 10 1.013 1 8.315 (273.15 20) 2 3 5 − = × × × × × + = = A A A P nRT V (b) For isochoric process: B B A A T P T P = The pressure of the gas at the completion of path (1) is 1.46 10 (Pa) 273.15 20 1 1.013 10 (273.15 150) 5 5 = × + × × × + = = A A B B T P T P For isothermal process: PBVB = PCVC The volume of the gas at the completion of path (2) is 3.47 10 (m ) 1 10 1.013 1.46 10 2.41 10 2 3 5 5 2 − − = × × × × × × = = C B B C P P V V The temperature of the gas at the completion of path (2) is TB = 273 +150 = 423K (c) (1) The work Path 1: 0J W1 = Path 2: 1282.13 (J) 2.41 10 3.47 10 ln 8.315 (273 150) ln 2 2 2 = × × = = × + × − − B C V V W RT Path 3: ( ) 1 1.013 10 (3.47 10 2.41 10 ) 1073.78(J) 5 2 2 3 = − − = − × × × × − × = − − − W PA VC VA The total work done by the gas is 0 1282.13 1073.78 208.35 (J) Wtotal = W1 +W2 +W3 = + − = (2) The heat Path 1: ( ) 1 2.5 8.315 (423 293) 2702.38(J) 1 = CV TB − TA = × × × − = M Q µ