Steam in a piston-cylinder assembly undergoes a polytropic process, with n = 2, from an initial state where p1 = 500 psi, v1 = 1.701 ft3/lbm, u1 = 1363.3 btu/lbm to a final state where u2 = 990.58 btu/lbm. during the process, there is a heat transfer from the steam of magnitude 342.9 btu. the mass of steam is 1.2 lbm. neglecting changes in kinetic and potential energy:

solution:

ΔU = q + W = 1.2 lbm* [(990.58 Btu/lbm) - (1363.3 Btu/lbm)] =-447.26400 Btu

W = ΔU - q= -447.26400 Btu - (-342.9 Btu)= -104.36500 Btu =-104 Btu

W = -integral[P*dV]           P*V2 = constant   (polytropic)

Steam table gives you that at T = 1000.5 F, u(1) = 1363.3Btu/lbm

T = 811.205556 K

P = 500 psi = 3447378.65 Pa

Vi/n = RT/Pi = (8.314 J/mol/K *811.206 K)/(3 447 378.65 Pa) =0.00195637537 m3/mol

Vi/(n*molar mass) = 0.00195637537 m3/mol*(100cm/1m)3 *(1 in/2.54 cm)3*(1 ft/12 in)3 *(1mol/18g)*(454 g/lbm)

Vi/mass = 1.74257166 ft3/lbm

Vi = 1.2 lbm*(1.74257166 ft3/lbm)= 2.09108599ft3

P*V2 = constant   (polytropic)

constant = (500 psi)*(2.09108599 ft3)2 =2186.32031 psi*ft6

W = -integral [P*dV] = -integral [constant*dV/V2] =constant*[1/Vf - 1/Vi] = -104 Btu (from part a)

1/Vf = -104 Btu/constant + 1/Vi

1/Vf = -104 Btu/ (2186.32031 psi*ft6) + 1/[2.09108599ft3]

answer; the correct answer is technician a; ///100%good luck

newton's first law of motion