For an isothermal expansion of an ideal gas, the correct combination of the thermodynamic parameters will be-

Options

A. `DeltaU=0,Q=0,wne0 and DeltaH ne0`
B. `DeltaU ne0,Qne0,wne0 and DeltaH=0`.
C. `DeltaU=0,Qne0,w=0 and DeltaHne0`
D. `DeltaU=0,Qne0,wne0 and DeltaH=0`

Answer

D

Explanation

For an ideal gas undergoing an isothermal process, `DeltaD=0`. When a gas undergoes expansion it absorbs heat from its surroundings and performs external work. So, in such a process, `qne0 and wne0`.
We know, `DeltaH=DeltaU+DeltaU(PV)`. For ideal gas, `PV=nRT`.
`therefore DeltaH=DeltaU+Delta(nRT)=DeltaU+nRDeltaT`
In an isothermal proces of an ideal gas, `DeltaU=0,DeltaT=0 and` hence `DeltaH=0`.