in thermodynamics, we learn that change in state function is independent of the path/process, and thus we can conveniently pick any arbitrary path that allows easier calculation. I'm TA'ing a class and found that it would be great to actually show students an example where that a simpler path can indeed be constructed to calculate the change of state function. However, I feel that there must be some blind spot that I myself am struggling with. And would like to get everyone's feedback.
Example and detail
We are asked to calculate $\Delta H$ for an isochoric process where a known amount of energy is transferred into the system (of a monatomic ideal gas) as heat. Assuming only P-V work is allowed in the system.
I can first calculate change of internal energy ($\Delta U$) using the first law, and subsequently $\Delta T$ with the known heat capacity (over constant volume) $c_V = \frac{3R}{2}$. Then to calcuate $\Delta H$, I'll follow an isobaric process instead (with $c_P = \frac{5R}{2}$), but it will give the same result. Here's the detail:
$w = -\int_i^f P_\text{ext} dV = 0$ (no work is done at constant volume)
$\Delta U = w + q = 0 + q = q$
$\Delta U (= q_V) = n c_V \Delta T \implies \Delta T = \frac{\Delta U}{n c_V} = \frac{q}{n c_V}$
$\Delta H = q_P = n c_P \Delta T = n c_P \frac{q}{n c_V} = \frac{c_P}{c_V} q$
Question
My question - under the context of this example - is how to make up a path connecting the same endpoints (the initial and final states) that is under constant pressure?
Without loss of generality, assume that $q > 0$, and thus the temperature at the end increases as the internal energy increases (no work is done due to the constant volume constraint). This suggests that the pressure also increases. But I just said that I'd like to construct an isobaric path that connects the initial and final state..., which now seems to be contradicting the fact.
My thought is - that instead of a single isobaric path (which connects the initial state and potentially an intermediate state), maybe I need another segment that connects the intermediate state and the final state? Such that overall the volume stays the same? If this is the approach, what would the strategy be for picking the intermediate state and the two segments making up the simpler path?
Would the following work?
1st segment: isobaric expansion to an intermediate state whose temperature is the same as the final state. The relevant heat is $q_P$ which will contribute to the overall change in enthalpy with its contribution $\Delta H_1 = q_P$.
2nd segment: isothermal compression to the final state (i.e., need to volume to go back to its initial/final state volume). For this part - there is going to be additional heat and work involved, but internal energy and enthalpy should remain unchanged (as same as the intermediate state), resulting in $\Delta H_2 = 0$.
Overall, $\Delta H = \Delta H_1 + \Delta H_2 = \Delta H_1 = q_P$.
I'm not very sure about this as it invokes that enthalpy change is zero for an isothermal process -- which I don't know if it will hold for non-ideal gas? And if not - this suggests my proposal doesn't work universally...
[Edits for fixing typo]