The specific enthalpy of benzene vapor at 45 c and 0.7 atm absolute pressure, relative to a reference state of benzene vapor at 45 c and 1.27 atm absolute pressure will be 0 kJ/mol.
<h3>What is specific enthalpy and how was it calculated in the question?</h3>
A thermodynamic system has a property called enthalpy (H). It is calculated by the sum of the internal energy (U) of the thermodynamic system and the product of its volume (V) and pressure (p). The SI Unit is Joule (J).
Equation:
H = U+pV
The specific enthalpy of vapor can be defined as the amount of energy spent in order to transform a liquid substance into its vapor or gaseous form. The SI Unit is kJ/mol.
In the above question, the formula to be used is
P1/P2 = (Δ Hvap)/R)(1/T2-1/T1)
T1 & P1 --> the starting temperature & pressure respectively (= 1.27 atm and 45c),
T2 & P2 --> the final temperature & pressure respectively (= 0.7 atm and 45c),
R --> the real gas constant i.e. 8.314kJ/mol and
ΔHvap --> The specific enthalpy of vaporization.
Putting the values in the equation;
1.27/0.7=(ΔHvap/8.314)(1/45-1/45)
Hence as after subtracting the equation becomes 0, our final answer also comes out to be ΔHvap= 0 kJ/mol.
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