Calculate the maximum work obtained when 0.75 mol of an ideal gas expands isothermally and reversible at `27^(@)C` from 10 L to 20 L Also calculate value of q and `Delta U` accompanying the process.

Explanation

For isothermal reversible expansion of an ideal gas.
`W=-2.303 "nRT log"(V_(2))/(V_(1))`
`W=-2.303xx0.75xx8.314xx300 log.(20)/(10)` or `W=-1296.7 J`
For isothermal process, `Delta U =0` and `Delta U=q+W rArr 0 = q + (-1296.7)` or `q=+1296.7 J`
It means the system absorbs 1296.7 J heat from surroundings and performs 1296.7 J work surroundings.