The correct options are
A ΔH=525
B ΔS=5 ln373298+2ln10
C ΔE=375
D ΔG of the process can not be calculated using the given information
The process can be divided into two steps:
(i) Isochoric process in which volume is kept constant at 1L and temperature is raised from 250C (298 K) to 1000C (373K).
(ii) Isothermal process in which the temperature is kept constant at 1000C (373K) and volume is raised from 1L to 10L.
For isothermal process, the entropy change is ΔS=RlnV2V1=2ln101
For isochoric process, the entropy change is ΔS=CVlnT2T1=5ln373298
Hence, the entropy change for the overall process is ΔS=5ln373298+2ln10
Thus, option B is correct.
For the isothermal process, the change in the internal energy is zero. For the isochoric process, the change in internal energy is equal to the heat transferred.
ΔE=Q=CVΔT=5×75=375. This is also equal to the change in the internal energy for the overall process.
Thus, option C is correct.
The expression for the enthalpy change is ΔH=ΔE+(PFVF−PIVI)=ΔE+R(TF−TI)=375+2(393−298)=525.
Thus, option A is correct.
The free energy change for the process cannot be calculated by using the given information as the expression ΔG=ΔH+TΔS is applicable at a given temperature.
Thus, option D is correct.