Is Delta Enthalpy Zero in Isothermal Expansion with Heat Absorption?

[sachin123]

Is delta enthalpy zero for...

Homework Statement

1 mol of ideal gas expands isothermally at 274 K ,from pressure 1.1 atm to 1 atm.It absorbs 8.314 J of heat.Find free energy change.

Homework Equations

dG=dH-T dS...1
dS=dq/T
dH=dU+Work done
Work done=nRTln(P1/P2)...2

The Attempt at a Solution

dU=0
dH=Work done
dH=whatever u get in 2
dG=whatever u get by substituing values in 1.

But my book says dH=0 because the process is adiabatic.Isn't that wrong?

 

[nate808]

I would like to clarify the statement made in the forum post. The delta enthalpy is not necessarily zero for an ideal gas expanding isothermally. It depends on the conditions and assumptions of the system.

In this case, the process is described as isothermal, meaning that the temperature remains constant throughout the expansion. However, the gas is also said to absorb heat, which suggests that the process is not truly isothermal and there is some heat transfer involved.

Furthermore, the statement that dH=0 because the process is adiabatic is incorrect. Adiabatic means that there is no heat transfer, but it does not necessarily mean that there is no change in enthalpy. Enthalpy is a state function and can change even in adiabatic processes.

To accurately determine the free energy change, the conditions and assumptions of the system must be clearly stated and taken into account. This includes the type of process (isothermal, adiabatic, etc.), the heat transfer, and any work done. Once these factors are determined, the appropriate equations can be used to calculate the free energy change.

 

[Blueshift5]

I would say that the book is incorrect. The process described is not adiabatic, as heat is being absorbed. Adiabatic processes do not involve the transfer of heat. Therefore, the change in enthalpy, or heat content, is not zero. The correct approach would be to use the equation dG=dH-TdS, as you have done, and substitute the values given to find the free energy change. This will give a more accurate result.