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Calculate the amount of work done by 2 mole of an ideal gas at 298 K in reversible isothermal expansion from 10 litre to 20 litre. |
Answer» Amount of work done in reversible isothermal expansion, `w=-2.303nRTlog((V_(2))/(V_(1)))` given, `n=2,R=8.314K^(-1)mol^(-1),T=298K,V_(2)=20L and V_(1)=10L`. ltBrgt Substituting the values in above equation, `w=-2.303xx2xx8.314xx298log((20)/(10))` `=-2.303xx2xx8.314xx298xx0.3010=-3434.9J` i.e., work is done by the system. |
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