Question

one kilogram of water (V1 = 1003 cm^3*kg-1) in a piston cylinder device at (25°C) and 1 bar is compressed in a mechanically reversible, isothermal process to 1500 bar. Determine Q , W, ΔU , ΔH , and ΔS given that β = 250 x 10-6K-1 and K = 45 x 10^-6 bar-1.

Answer 1

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