Problem 3 (30 points Water at 140 °C and 15 bar pressure is initially held in pl
ID: 718162 • Letter: P
Question
Problem 3 (30 points Water at 140 °C and 15 bar pressure is initially held in place by a piston that is latched, as indicated in the diagram (the diagram is not to scale). The piston is weightless, and there is a vacuum above the piston. The system (including the piston) is well insulated. When the latch is removed, the final pressure in the system is 1.0 bar. You may assume that the entropy and internal energy of liquid water depend only on temperature. vacuum insulation latch water 8 points - Choose your system to be the water: what is the internal energy change of this system? Show your work, and justify all conclusions that you draw. 22 points - What are the entropy changes of the system (per kg of water) and of the surroundings? a) b)Explanation / Answer
Given:
Initial temperature (T1) = 140 °C =140+273 K = 413 K
Initial pressure (P1) = 15 bar
Well insulated system. This represents flow of heat (Q) = 0
Water is expanding against vacuum.
Final pressure (P2) = 1 bar
a) Water at 140 °C and 15 bar pressure will be in the form of steam. On removing the latch, steam is free to expand against vacuum, as there is nothing to oppose the expansion.
Therefore, the workdone in expanding (W) will be zero
W= 0
From First Law of Thermodynamics, dU = Q + W
Since, Q and W are zero, dU=0
Therefore change in internal energy of the system (dU) = 0
b) Since dU=0, The process is an isothermal irreversible process.
that means. Q = -W = RTln(P1/P2) =8.314*413*ln(15/1) = 9298.583 J/mole
Change in entropy of system (Ss) = Q/T = 9298.583/413 = 22.51 J/mole K
Change in entropy of system per kg of water = 22.51/18 = 1.25 J/K
Change in entropy of surroundings is zero, since there is no interaction between system and surroundings, the state of the surroundings does not change, and all thermodynamic functions of the surroundings remain constant, including entropy.
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