Calculate`q, W, DeltaU` and `DeltaH` for the isothermal reversible expansion of one mole of an ideal gas from an initial pressure of `1.0` bar to a final pressure of `0.1` bar at a constant temperature of `273K.`

Explanation

For isothermal reversible expansion:
`W=-2.303nRT"log"(P_(1))/(P_(2))`
`=-2.303xx1xx8.314xx273 "log"(1.0)/(0.1)`
`W=-5227.2 J`
At constant temperature, for expansion `DeltaT=0` Therefore, `DeltaU=0`
Also, `DeltaU=1xxC_(v)xxDeltaT=0... `(for `1` mole)
and `DeltaU=q+W`
`:. q=-W=5227.2J`
Also when temperature is constant `P_(1)V_(1)=P_(2)V_(2)` or `PV` is constant.
Thus, `DeltaH=DeltaU+Delta(PV) :. DeltaH=0`