Question

ΔH=−μCpΔP+CpΔT

Since for VDW gas μCp=2aRT−b.

The process took place at 300 K which means ΔT=0.

∴ΔH=−μCpΔP+CpΔT=(−2aRT+b)ΔP=0.4399 kJ/mol.

But the answer is −30.5 J/mol
I think my mistake is taking ΔT=0 but if that were the case what should I take as ΔT, I don't even know the process.

Is the author's answer correct ?

Answer 1

The expression ΔH = -μCpΔP + CpΔT represents the enthalpy change for a gas undergoing a process. In the given equation, ΔT is the change in temperature and ΔP is the change in pressure.

Step-by-step explanation:

The expression ΔH = -μCpΔP + CpΔT represents the enthalpy change for a gas undergoing a process. In the given equation, ΔT is the change in temperature and ΔP is the change in pressure.

When ΔT = 0, the equation becomes ΔH = (-2aRT + b)ΔP. However, to obtain the correct answer, you need to know the value of ΔT for the process.

Without knowing the process and the value of ΔT, it is not possible to determine the correct value of ΔH. Therefore, it is not possible to determine if the author's answer of -30.5 J/mol is correct or not.