Thermodynamic 2.8 parts A&C? assuming (ii) temperature \"50C\" Steam undergoes a

Question

Thermodynamic 2.8 parts A&C? assuming (ii) temperature "50C"

Steam undergoes a state change from 450 degree C and 3.5 MPa to 150 degree C and 0.3 MPa. Determine Delta H and Delta U using the following: Steam table data. Ideal gas assumptions.(be sure to use the ideal gas heat capacity for water.) Five grams of the specified pure solvent is placed in a variable volume piston. What are the molar enthalpy and total enthalpy of the pure system when 50% and 75% have been evaporated at: 30 degree C 50 degree C? Use liquid at 25 degree C as a reference state. Benzene (p^L = 0.88 g/cm^3) Ethanol (p^L = 0.79 g/cm^3) Water without using the steam tables (p^L = 1 g/cm^3) Water using the steam tables Create a table of T, U, H for the specified solvent using a reference state of H = 0 for liquid at 25 degree C and l bar. Calculate the table for: liquid at 25 degree C and 1 bar; saturated liquid at 1 bar: saturated vapor at 1 bar; vapor at 110 degree C and 1 bar. Use the Antoine equation (Appendix E) to relate the saturation temperature and saturation pressure. Use the ideal gas law to model the vapor phase. Benzene Ethanol

Explanation / Answer

Q = mc (t2-t1)

DT= -Q

Convention: heat is comming out

m= 5 grams

cp=82.44 J/(mol K)

t2= 50 degree celcius

t1=25 degree celcius

when 50% evaporated

mass=2.5 grams

Benzene= 78 gram/ mol

molar enthalpy= -(m*cp*DT)

= -5.1525 kJ/mol Answer

number of moles = 2.5 / 78 = 0.03205

Total Enthalpy = Molar enthalpy * number of moles = 0.165144231 kJ = -165.144231 J Answer

  

when 75% evaporated

mass= 5*0.75=3.795 grams

molar enthalpy = (-m*cp*Dt)

   = -3.795*82.44*25

   = -7.821495 kJ/ mol Answer

number of moles =3.795/78 = 0.048654

Total Enthalpy =-7.821495*0.048654 = 380.55 kJ Answer

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.