Calculating change in state function following different paths

Question

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]

Answer 1

You wrote, in one of your comments, "My biggest question is how to demonstrate the proper use of $\Delta H = n C_p \Delta T$irrespective of the path...".

So:

In a closed system (no flow of matter across the boundaries) of constant composition (no chemical reactions),

$$dH = \left(\frac{\partial H}{\partial T}\right)_ p dT + \left(\frac{\partial H}{\partial p}\right)_ T dp$$

$$= C_ p dT + \left(\frac{\partial H}{\partial p}\right)_ T dp$$

$$= C_ p dT + \left(V - \alpha T V \right) dp$$

$$= C_ p dT + \left(V - T \left(\frac{\partial V}{\partial T}\right)_ p \right) dp$$

Thus, at constant $p$, $dp =0$, and

$$dH_p = C_ p dT \textit{, for all substances.*}$$

It is worth noting, as Chet Miller mentioned in the comments, that $C_p \equiv C_p(T,p)$ (i.e., $C_p \text{ is a function of both } T \text{and } p$ for real substances).

If there are no inter-particle interactions (e.g., an ideal gas), then $V = T \left(\frac{\partial V}{\partial T}\right)_ p $, and thus

$$\left(\frac{\partial H}{\partial p}\right)_ T = 0$$

You can confirm this yourself for an ideal gas by substituting $V = \frac{n R T}{p}$ for the second occurrence of $V$ in: $\left(V - T \left(\frac{\partial V}{\partial T}\right)_ p \right)$.

Hence, for an ideal gas,

$$dH = C_ p dT \text{, always,} \textit{ even if the pressure isn't constant.}$$

In summary, if the substance is an ideal gas, $dH = C_ p dT$ regardless of whether there is a change in pressure. If the substance is real, $dH = C_p dT$ only if the pressure is constant.

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*Note that this relationship holds for all types of work. If we add the constraint that the only work is $pV\text{-work}$, then we have the additional nice relationship that:

$$dH_p = C_ p dT = q_p \textit{, for all substances.}$$

If we keep the constant-$p$, closed system, and $pV\text{-work}$-only constraints, but remove the constant-composition constraint (allowing chemical reactions), then it will no longer be the case that $dH_p = C_ p dT$, but it will be the case that:

$$dH_p = q_p \textit{, for all substances.}$$

Answer 2

You have teen talking about two different processes with two different final states, two different amounts of heat, and two different amounts of work. The only thing that's the same is the temperature change, which, for an ideal gas makes $\Delta U$ and $\Delta H$ the same. If it were not an ideal gas, these two would not even be the same either.

Your mistake is associating the amount of heat q with the heat capacity C. In thermodynamics, q is dependent on path, and can vary between two different paths between the same two end states, while C is a state function, defined in terms of the path-independent state functions U and H: $$C_v=\frac{1}{n}\left(\frac{\partial U}{\partial T}\right)_V$$and$$C_p=\frac{1}{n}\left(\frac{\partial H}{\partial T}\right)_P$$ For an ideal gas, U and H are functions only of T, but this is not generally true.

In the specific process you analyzed, once you changed the temperature at constant volume, there was no isobaric path between these same two end states. The equation of state requires that P=P(V,T), and, once you said that P is constant along your alternate path, for the same temperature change, V would have to change.