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What work is to be done on `2` mole of a perfect gas at `27^@C` if it is compressed reversibly and isothermally from a pressure of `1.01 xx 10^5 Nm^(-2)` to `5.05 xx 10^6 Nm^(-2)`? Also calculate the free energy change. |
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Answer» For reversible process: `W_(rev)=-2.303 nRT"log"_(10)(P_(1))/(P_(2))` `=-2.303xx2xx8.314xx300 "log"_(10)(1.01xx10^(5))/(5.05xx10^(6))` `=+1.9518xx10^(4)` joule Since, `W_(rev)` is a measure of free energy change `:. -DeltaG=-W_(rev)=-W_(max)` or `DeltaG=1.9518xx10^(4)` joule |
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