a) Determine the specific enthalpy (kJ/mol) of n-hexane vapor at 200 C and 2.0 atm relative to n-hexane liquid at 20 C and 1.0 atm, assuming ideal gas behavior for the vapor. Show clearly the process path you construct for this calculation and give the enthalpy changes for each step. State where you used the ideal gas assumption.b) What is the enthalpy of n-hexane liquid at 20 C and 1.0 atm relative to n-hexane vapor at 200 C and 2.0 atm?c) Beginning with the value of H calculated in part (a) and still assuming ideal gas behavior, determine the specific internal energy of the vapor at 200 C and 2.0 atm. Once again, state where you used the ideal gas assumption.

(a) Given: n hexane vapor at (Tf) 200 °C and (Pf)2 atm n hexane liquid at (Ti) 20 °C and (Pi) 1 atm Boiling point of hexane (Tb)= 68.73 oC It is required to find the specific enthalpy from liquid to vapor for the given compound ? Process path:∆H=Enthalpy from liquid hexane at 20 oC and 1 atm to vapor hexane 200 oC and 2 atm Since ∆H is path independent ∆H=∆H1+∆H2+∆H3+∆H4where ∆H4= enthalpy change due to increase in pressure fromn 1 atm to 2 atmAssuming ideal gas∆H4=0 (enthalpy is independent of pressure for an ideal gas)⇒∆H=∆H1+∆H2+∆H3 Equation (1) ∆H1 involves sensible heat from 20 °C to 68.73 °C(no phase change)∆H1=∫20+27368.73+273CpdTAssuming Cp=0.2163 kJmol.K in the range 273-400 K∆H1=0.2163 kJmol.K*(341.73-293)K∆H1=10.5403 kJmol ∆H2 involves enthalpy of vaporization of hexane liquid to vapor ∆H2=∆Hvap=28.85 kJmol ∆H3 involves sensible heat from 68.73 °C to 200 °C (no phase change)∆H3=∫68.73+273200+273CpdTAssuming Cp=0.13744+40.85*10-5T-23.72*10-8T2+57.66*10-9T3 kJmol.K in the range 300-500 K∆H3=0.13744*(473-341.73)+40.85*10-52(4732-341.732)-23.72*10-83(4733-341.733)+57.66*10-94(4734-341.734) ∆H3=24.6561 kJmolSubstituting the calculated values in Equation (1) we get ∆H=10.5403+28.85+24.6561⇒∆H=64.0464 kJmol (b) It is required to find the specific enthalpy from vapor to liquid for the given compound ? From part (a) we have find the enthalpy following same path but in different direction. Therefore the magnitude of enthalpy will be same as calculated in part (a) with negative sign ∆H=-10.5403+-28.85+-24.6561⇒∆H=-64.0464 kJmol(c) It is required to find specific internal energy of the vapor at 200 °C and 2.0 atm We know that H=U+PV∆H=∆U+P∆V+V∆Pfrom part (a) ∆H=64.0464 kJmolassuming ideal gas behaviorV1=RT1P1=8.314*2931.013*105=0.024047 m3 (initial volume)V2=RTbP1=8.314*2*1.013*105=0.028046 m3 (volume increases due to vaporization) H=U+PV∆H=∆U+P∆V+V∆P64046.4 =∆U+1.013*105*(0.028046-0.024047)+0.024047*(1*1.013*105)⇒∆U=61.236 kJmol