Comparing what to how I did back in September?
Answer:
Q = -18118.5KJ
W = -18118.5KJ
∆U = 0
∆H = 0
∆S = -60.80KJ/KgK
Explanation:
W = RTln(P1/P2)
P1 = 1bar = 100KN/m^2, P2 = 1500bar = 1500×100 = 150000KN/m^2, T = 23°C = 23 + 273K = 298K
W = 8.314×298ln(100/150000) = 8.314×298×-7.313 = -18118.5KJ ( work is negative because the isothermal process involves compression)
∆U = Cv(T2 - T1)
For an isothermal process, temperature is constant, so T2 = T1
∆U = Cv(T1 - T1) = Cv × 0 = 0
Q = ∆U + W = 0 + (-18118.5) = 0 - 18118.5 = -18118.5KJ
∆H = Cp(T2 - T1)
T2 = T1
∆H = Cp(T1 - T1) = Cp × 0 = 0
∆S = Q/T
Mass of water = 1kg
Heat transferred (Q) per kilogram of water = -18118.5KJ/Kg
∆S = (-18118.5KJ/Kg)/298K = -60.80KJ/KgK
6.022*10^23 is the answer
