`3.0` moles of an ideal gas at `27^(@)C` is compressed at constant temp. reversibly form a volume of `20L` to `10L` slowely increasing the external pressure calculate work done on the gas `w,q,DeltaU` and `DeltaH`.

Explanation

Process is reversible, since the pressure is changed slowely.
a. `w_("rev'isothermal")-=- nRT xx 2.303 "log"(P_(1))/(P_(2))`
`=- nRT xx 2.303 "log"(V_(2))/(V_(1))`
`=- 3xx 8.314 xx 300 xx 2.303 "log"(10)/(20) = 5163.5J`
b. For an isothermal process `DeltaU =0`
c. `DeltaU = q+w`
`0 = q+ 5163.5`
`:.q =- 5163.5`
d. `DeltaH = DeltaU +DeltaPV`
`= DeltaU +Delta(nRT)`
Since `R` and `T` are constant.
`DeltaH = 0 + 0 = 0`