Problem 1 (i) Calculate the change in enthalpy between the following states of benzene (25) State 1: Temperature =55C, Pressure =900mmHg State 2: Temperature =200C, Pressure =1500mmHg The following data is given (a) The vapor pressure of liquid benzene is given by: log10P(mmHg)=6.88t(C)+219.1611196.76 (b) The substance is known to follow generalized virial equation of state Z=1+RTcBPcPrBTrRTcBPc=B0+B1,Bo=0.083Tr1.60.422,B1=0.139Tr4.20.172 (c) The critical temperature, pressure and acentric factor of benzene are 562.1K,49.24bar and 0.212 respectively. (d) The heat capacity of liquid benzene may be taken as 2kJ/kgK (e) The density of liquid benzene may be assumed constant at 0.77g/cm3 (f) The isobaric molar heat capacity of benzene in ideal gas state can be estimated from Cpo(J/molK)=33.89+0.472T where T is in kelvin The latent heat of vaporization hvn may be calculated using the following equations hvn(kJ/kg)=1.093RTc[Tbr0.93TbrlnlnPc1.013]hv2=hv1(1Tr11Tr2) Sketch the path used for calculation on a P-v diagram