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Question

The molar enthalpy of vaporization of benzene at its boiling point (353 K) is 30.84 kJ mol1. What is the molar internal energy change? For how long would a 12 V source need to supply a 0.5 A current in order to vaporises 7.8 g of the sample at its boiling point?

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Solution

C6H6(l)C6H6(g)
Δn=nPnR=10=1
Given that:-
ΔH=30.84KJ/mol=30840J/mol
T=353K
As we know that,
ΔH=ΔE+ΔngRT
30840=ΔE+1×8.314×353
ΔE=308402934.842=27905.158=27.9KJ
Now
Molecular weight of benzene =78g/mol
Weight of benzene =7.8g
No. of moles of benzene =7.878=0.1 mole
Energy provided to evaporate 0.1 mole of benzene (E)=30.84×0.1=3084J
As we know that,
E=V×I×t
Given that,
V=12 volt
t=?
I=0.5A
3084=12×0.5×t
t=30846=514s
Hence the molar internal energy change will be 27.9KJ and for t=514s, a 12 volt source need to supply a 0.5A current in order to vaporise 7.8g of the sample at its boiling point.

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