Question

The molar enthalpy of vaporization of benzene at its boiling point (353 K) is 30.84 kJmol^-1. What is the molar internal energy change? For how long would a 12 volt source need to supply a 0.5A current in order to vaporise 7.8g of the sample at its boiling point?  

Answer 1

∆H_vap = 30.84 kJ/mol 

V_vap =(1 0.082 x 3533/1)

= 28.9 L

W = – P∆V = +PV_vap

= +0.0821 × 353 = 2.89 kJ

∆E = ∆H – ∆PV

= +30.84 – 2.89

= 27.91 kJ/mol

We know VIT = Q

Q + W = ∆E

∴ Q = ∆E – W

= ∆E + ∆PV = ∆H

∴12 × 0.5 × t = 30.84 × 10³ × 7.8/78

t = 514 sec.