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.

Question

Chemistry

Howell

23 February, 22:02

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.

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Saul · Answer 1

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