Can `DeltaU and DeltaS` discriminate between irreversible and reversible process? Explain.

Explanation

`DeltaU` does not discriminate between irreversible and reversible process. For isothermal process involving ideal gas T is constant. Hence `DeltaU=0` for both reversible and irreversible process.
`DeltaS` discriminates the irreversible and reversible process.
In an isothermal reversible process if the amount of heat Q is absorbed from the surroundings at a temperature T, the increase in the entropy of the system will be
`Delta_("sys")=+(Q)/(T)`
On the other hand surroundings lose the same amount of heat at the same temperature. The decrease in entropy of the surroundings will
`DeltaS_("surr")-(Q)/(T)`
`DeltaS_("Total")=Delta_("sys")+DeltaS_("surr")=(Q)/(T)-(Q)/(T)=0`
In the irreversible process the system is at higher temperature `T_(1)` and its surroundings at lower temperature `T_(2).` 'Q' amount of heat goes irreversibly from system to surroundings
`therefore DeltaS_("sys")=(-Q)/(T_(1))`
`DeltaS_("sur")=(+O)/(T_(2))`
`Delta_("process")=Delta_("sys")+Delta_("sur")`
`=(-Q_1)/(T_(1))+(Q_2)/(T_(2))Q[(T_(1)-T_(2))/(T_(1)T_(2))]`
But `T_(1) gt T_(2)`
`therefore T_(1)-T_(2)= +ve or DeltaS_("process") gt 0`
Hence entropy increases in an irreversible process.