Question

an ideal gas undergoes isothermal expansion at constant pressure. during the process ,what happens to enthalpy and entropy,

Do they remain constant???

Or one of them increases??

Answer 1

change of internal energy is zero (ΔU=0) since, for an ideal gas, the internal energy U only depends on amount of substance nn and temperature T, and in a closed system nn is constant (Δn=0)and during an isothermal process also T remain constant (ΔT=0).

Enthalpy H is defined as

H=U+pV

and the ideal gas law states that

pV=nRT

Thus

H=U+nRT

Since n(closed system) and T(isothermal process) are constant, the product nRT is constant, and therefore, according to the ideal gas law, also the product pV is constant.

Furthermore, since U is constant during the given process, the sum H=U+nRT as well asH=U+pV has to remain unchanged

Note that your assumption

ΔH=ΔU+W

(where W=pΔVis the pressure-volume work) only applies to processes under constant pressure p, since enthalpy is defined as

H=U+pV

and thus

dH=dU+d(pV)=dU+Vdp+pdV

which simplifies to

dH=dU+pdV

at constant pressure (dp=0).

However, the pressure does not remain constant during the process given in the question.