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heat capacity of liquid benzene is 1 3 3 and Tm = 1 5 0 0 , a = - 3 3 . 8 9

heat capacity of liquid benzene is 133 and Tm =1500, a =-33.893, b *10^3=471.793, c*10^6=-298.294, d*10^9=70.823
Please do the following before coming to class on Monday to help us solve the problem using different methods:
1. Heat capacity of liquid benzene is 133 J/g K.
Calculate h4-h1 in KJ/Mol.
2. Calculate heat of vaporization of benzene at T4.
3. Calculate van Der waals constants 'a' and 'b'.(Tc and Pc can be obtained from relevant table sent to you.)
4. Calculate vapor volume at state 2 and 3.
5. Integrate Cp(vapor) dT and determine the value.
The Cp of benzene vapor can be obtained from table below.
EManePropanen-Butanen-Pentanen-Hexanen-Heptanen-OctaneAlkenesEthylenePropylene1-Butene1-Pentene1-Hexene1-Heptene1-OcteneAcetylenesAcetylenePropyne1-Butyne1-Pentyne1-Hexyne1-Heptyne1-OctyneMiscellaneousAcetaldehydeAmmonatBenzene216CsHCH10CsH2CH4C,H16C,HsChH,CsHeC,H,CsH10CHi2C,H4CgH6ChHaCjH,CAHsC&HsCeHhoCH2CH4NH,CeHs1000C.H.10001000100010001000100010001000100010001000100015001500150015001500CH,CHO 10001500180015004.493182.22y-4.798307.2591000-160.13232.7430.469385.311-198.84939.9591.443476.394-250.366 S1.2293.0835.020653.6496.907741.7704.196.154.5653.305235.8212.5397.4808.6309.806565.691-300.31962.05112.15718.591344.871424.413514.56350.96916.227605.750150021.402528.087692.18113.458269.990-33.893-2.95819.017442.250348.39423.45685.753-58.33223.431615.81121.879161.19129.74725.108-348.67872.313-397.20482.629471.793-81.07616.813-117.58022.673-191.25241.657-231.86849.593-281.95960.256-334.09571.869-380.986-149.172-188.165-243.460-288.710-337.270-85.883-298.29481.62915.86732.53839.83852.55561.50772.03218.28170.823340.027-223.65256.521-10.797-1.546
532
Chemical engineering thermodynamics
Table A.3 Isobaric molar heat capacities of some selected gases in the ideal gas state*
Cp0=a+bT+cT2+dT3+eT-2;Cp0 is in JmolK and T is in K(from 298.16K to Tm)
\table[[Compound,Formula,Tm,a,b103,c106,d109,e10-5
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