The standard molar engthalpy of vaporisation of benzene `Delta_(vap)H^(@)` at 353 K is 30.8 kJ `mol^(-1)`.
If the benzene vapours behave as an ideal gas, the change in internal energy of vaporisation of 78 g of benzene at 353 K in kJ `mol^(-1)` is

Options

A. 37.87
B. 27.87
C. 33.74
D. 17.87

Answer

B

Explanation

Given,
Standard molar enthalpy of benzene
`" "(DeltaH)=30.8kJ" "mol^(-1)`
Temperature (T) = 353 K
Equation
`" "underset("Liquid")(C_(6)H_(6)) overset(Delta)(rarr)underset("Gas")(C_(6)H_(6))`
Herem change in gaseous mole `(Deltan_(g))=1`
Now, `" "DeltaH=DeltaE+Deltan_(g)RT`
`30.8kJ" "mol^(-1)=DeltaE+(1)(0.082) times (101.32)(353)10^(-3)kJ" "mol^(-1)`
`30.8kJ" "mol^(-1)=DeltaE+2934.84 times 10^(-3)kJ" "mol^(-1)`
`" "DeltaE=30.8-2.93`
`" "DeltaE=27.87kJ" "mol^(-1)`