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The molar entropy content of `1 ` mole of oxygen `(O_(2))` gas at `300 K` and `1 atm` is `250 J mol e^(-1)K^(-1)`. Calculate `DeltaG` when 1 mole of oxygen is expanded reversibility and isothermally from `300K`, 1 atm to double its volume ( Take `R=8.314J mol e^(-1)K^(-1),log e=2.303)`
A. `1.728 KJ mol e^(-1)K^(-1)`
B. 0
C. `-1.728 KJ mol e^(-1)K^(-1)`
D. `0.75KJ mol e^(-1)K^(-1)`

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Correct Answer - A
`DeltaG=DeltaH-Delta(TS)`
`DeltaH-T DeltaS" "("isothermal")`
`=0-T DeltaS=-T(int(dq_(rev))/(T))`
`=-intdq_(rev)=-q_(rev)=W_(rev)`
as process is isotheraml so `Delta E=0=q_(rev)+W_(rev)`
so `" "DeltaG=-nRTln((V_(r))/(V_(l)))`
`=-RTln2=-8.314xx300xx0.693xx10^(-3)KJmol^(-1)K^(-1)`
`=1.728KJmol^(-1)K^(-1)`

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