Question

100 kg of steam is enclosed in a piston-cylinder device, initially at 573.15 K and 0.5 MPa. It expands and cools to 473.15 K and 0.1 MPa.

A. What is the change in internal energy of the steam in this process?

B. If the external pressure is constant at 0.1 MPa, how much work was done by the steam on the surroundings?

Answer 1

steam is enclosed in a piston - cylinder device. Initial condition Pressure P_1 = 0.5 MPa Temperature T_1 = 573.15 K Final Conditions P_2 = 0.1 MPA T_2 = 473.15 K A. Change in Internal Energy of Steam in this process: Both initial & Final states are in superheated region. since, at Prat = 0.5 T_rat = 424.55 k T_rat < T_1 material implication superheated at P_rat = 0.1 MPa T_rat = 372.76 K T_rat < T_2 material implication Super heated. Internal energy of steam at P_1 & T_1 U_1 = 2803.3 KJ/Kg Internal energy of steam at P_2, T_2 U_2 = 2658.2 KJ/Kg So, change in internal energy for this process material implication Delta V = m times (U_2 - U_1) Delta U = 100 Kg times (2658.2 - 2803.3) KJ/Kg Delta U = -14510 KJ (a) (b) V_1 at P_1 & T_1 material implication V_1 = 0.523 m^2/kg (volume) V_2 at P_2 & T_2 material implication V_2 = 0.1725 m^3/kg & P = 0.1 MPa = 0.1 times 10^6 Pa

Thank you