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

For isothermal expansion of an ideal gas, the correct combination of t

1.	For isothermal expansion of an ideal gas, the correct combination of thermodynamic parameters will be
Answer» `DeltaU = 0, Q = 0, w ne 0 and DeltaH ne 0` `DeltaU ne 0 q ne 0, w ne 0 and DeltaH = 0` `DeltaU = 0, q ne 0, w = 0 and DeltaH ne 0` `DeltaU = 0 , q ne 0, w ne 0 and DeltaH = 0` SOLUTION :For ISOTHERMAL process, `DeltaT = 0` `DeltaU = nC_(V)DeltaT = 0` `DeltaH = nC_(p)DeltaT = 0` ACCORDING to first law of thermodynamics `DeltaU = q + w` Since `DeltaU = 0, q ne w ne 0`.