Thermodynamics

Statement IDuring isothermal expansion of an ideal gas its enthalpy decreases.For an ideal gas, enthalpy$H=U+pV$.For an ideal gas:$U=nC_VT$and since internal energy$U$depends only on temperature for an ideal gas,$dU=0 \;\;$if temperature is constant (isothermal process).Also,$pV=nRT$.Therefore,$H=U+pV=nC_VT+nRT=n(C_V+R)T=nC_PT$Since$T$is constant (isothermal), enthalpy$H$is also constant.Hence, enthalpy does not decrease, it remains the same.Statement I is incorrect.Statement IIWhen 2.0 L of an ideal gas expands isothermally into vacuum,$\Delta U=0$.For free expansion into vacuum:External pressure$=0$So$w=0$No heat exchange ($q=0$) because the system is isolated (expanding into vacuum)Thus,$\Delta U=q+w=0+0=0$Also, for an ideal gas,$\Delta U$depends only on temperature, and since it is isothermal, $\Delta U=0$. Statement II is correct. Final AnswerStatement I is not correct but Statement II is correct.Correct option: D