Question

For liquid benzene at 25 ̊C, the specific heat at constant pressure is 1.78 J K-1 g-1, the density is 0.88 g cm-3, the coefficient of thermal expansion is 0.00124 K-1, and the isothermal compressibility is 9.8x10-5 atm-1. Calculate the molar heat capacity of benzene at 25 ̊C, at constant pressure and at constant volume.

Answer 1

Specific heat at constant pressure =1.78 J K-1 g-1

molar mass of benzene=78.11 g/mol

So, molar heat capacity of benzene at 25 ̊C, at constant pressure=Cp=sp.heat /78.11g/mol=1.78 J K-1 g-1 * 78.11 g/mol=139.036 J K-1 mol-1

Cp=139.036 J K-1 mol-1

Equation for relation between Cp ,Cv, coefficient of thermal expansion( ), and the isothermal compressibility ():

Cp-Cv=T^2/d

d=density of substance,T=temperature=25+273=298K

Cp-Cv=T^2/d

or, 139.036 J K-1 mol-1 -Cv=298K*(0.00124K-1)^2/(0.88g/cm3)*(9.8*10^-5 atm-1)

or, 139.036 J K-1 mol-1 -Cv=5.313 K-1 Latm/kg =5.313 K-1* 101.267 J =538.030 J/K kg [see unit conversion below]

or, 139.036 J K-1 mol-1 -Cv =538.030 J/K kg=538.030 J/K kg *78.11 g/mol=42025.541 /1000 J/K mol=42.025 J/K.mol

or Cv= 139.036 J K-1 mol-1 - 42.025 J/K.mol=97.010 J/K mol

Cv=97.010 J/K mol=molar heat capacity of benzene at 25 ̊C at constant volume

[Unit conversion: g /cm3=g/ml=g*1000/ml*1000=kg/L]

[kg/Latm=kg (Latm)^-1]

[R=8.314 J/K.mol=0.0821 L atm/K mol, So 1 Latm=8.314 J/K.mol/0.0821/K.mol=101.267 J]