# Calculating change in state function following different paths

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]

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.

*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:

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

• I think it is worth mentioning that, for real gases, Cp is a function not only of temperature, but also of pressure. Only in the ideal gas region, in the limit of low pressures, is Cp a function only of temperature. So, at higher pressures, the ideal gas heat capacity will not be an accurate representation of Cp. Also, usually, heat capacity data is only available in the ideal gas limit. So the equation you developed for the effect of pressure on enthalpy needs to be used to calculate and apply Cp at higher pressures. Oct 28 '20 at 10:37
• @ChetMiller Good point. I added that as an explicit note to my answer. Oct 28 '20 at 22:42

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.

• Thanks for your reply! I'm aware of work/heat being path-dependent. My biggest question is how to demonstrate the proper use of $\Delta H = n c_P \Delta T$ irrespective of the path, other than proofing the equation is valid under the assumption (only P-V work is allowed)? This led to my thought of wanting to construct an additional path consist of one or more segments with the net heat equal to $n c_P \Delta T$. Oct 27 '20 at 15:43
• It is correct only for an ideal gas. Is that what you are trying to prove? Oct 27 '20 at 15:50
• $H = U + PV$ so $(dH)_P = (dU)_P + (d(PV))_P = -PdV + \cancel{d}q_P + PdV = \cancel{d}q_P$. Does this somehow suggest $\Delta H = q_P$ is valid only for an ideal gas? I don't see it, could you point out during which step did I inexplicitly assume ideal gas? (PS - I don't know how to properly make distinction between exact and inexact differential on this platform and thus the weird \cancel command... sorry) Oct 27 '20 at 15:57
• The derivation starts with $dH=TdS+VdP$ and ends with $$\left(\frac{\partial H}{\partial P}\right)_T=V-T\left(\frac{\partial V}{\partial T}\right)_P$$ The right hand side of this equation is equal to zero for an ideal gas. This derivation is in every thermodynamics book. Oct 27 '20 at 16:48
• I'll need to think a little bit more about what you mentioned. But are you saying that $\Delta H = q_P$ is only valid for ideal gases? This sounds a bit confusing to me - because in practice we conduct most experiments under constant pressure and report $\Delta _r H$ for various chemical reactions. It would not be as useful if $\Delta H = q_P$ is indeed true only for ideal gases...? Oct 27 '20 at 21:41