InterviewSolution
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InterviewSolution
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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 `"mole"^(-1)K^(-1)`B. 0C. `-1.728 KJ "mole"^(-1)K^(-1)`D. `0.75 KJ "mole"^(-1)K_(1)` |
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Answer» Correct Answer - A `DeltaG=DeltaH - Delta(TS)` `=DeltaH-TDeltaS " "`(isothermal) `0=T DeltaS=-T(int (dq_("rev"))/(T))` `-int dq_("rev")=- q_("rev")=W_("rev")` as process is isothermal so `DeltaE=0=q_("rev")+ W_("rev")` sp `" " DeltaG=- nRT "In" ((V_(f))/(V_(l))) =- RT "In" 2 k=- 8.314 xx300 xx0.693 xx 10^(-3) KJ mol^(-1) K^(-1)` `=1.728 KJ mol^(-1) K^(-1)` |
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