Question

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 1

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.