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.

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

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i