# Applicability of Enthalpy-Heat relation for a non-isobaric process

Consider a process in which state variables changes from $$(p_1, V_1, T_1)$$ to $$(p_2, V_2, T_3)$$ . I've learned that to calculate the change in enthalpy for any process, we consider a hypothetical isobaric process (Since enthalpy is a state function, it doesn't depend on the path).

Should the hypothetical process have the same initial and final state variable or just the same initial and final temperature? Because if we consider the same initial and final state variables, then we can't form a hypothetical isobaric process since the pressure changes in the actual process and the formula $$\Delta H = q_p = n \cdot c_p \cdot \Delta T$$ fails here.

• For an ideal gas, H is a function only of T. So it doesn’t matter if P and V change. Oct 9, 2020 at 15:29
• Of course, unless P is constant, q is not equal to delta H. Oct 9, 2020 at 16:10
• Could you clarify if you were asking about ideal gas or general?
– Babu
Oct 10, 2020 at 11:01

Disclaimer: The below discussion only holds to be completely true for reversible processes and for ideal gases.

I think the problem is you're thinking of it the wrong way, that is, you first think that $$q= nC_p \Delta T$$ and for constant pressure process $$q_p= \Delta H$$ , and hence, $$\Delta H = nC_p \Delta T$$

In actuality, it is the other way around. Similar to how internal energy is always $$nC_v \Delta T$$, the enthalpy is always $$nC_p \Delta T$$ for an ideal gas. A simple proof is shown below:

$$\Delta H = \Delta U + \Delta (PV)$$

For $$n$$ fixed moles of an ideal gas,

$$U = n C_v \Delta T$$

and,

$$\Delta (PV) = nR \Delta T$$

Hence,

$$\Delta H =n\Delta T [ R + C_v]$$

Now, from Mayers relation,

$$C_p = C_v + R$$

And hence,

$$\Delta H = n \Delta T C_p$$

So, as seen above, if you can accept that $$\Delta U= nC_v \Delta T$$ for an ideal gas always then you must be able to accept a similar statement for enthalpy

Regarding state variables:

Yes, there are indeed many states for an ideal gas which correspond to the same enthalpy since enthalpy is a function of the only temperature. The state variables of pressure and volume are irrelevant for enthalpy calculations unless you don't have the final and initial temperatures. In such a case, you can use the ideal gas law to find expression for the temperature at different states.

Note:

A discussion of proving $$q_p =\Delta H$$ can be found in this post

You may think the above proof is circular reasoning since regularly Mayers relation is proven by using $$\Delta H = nC_p \Delta T$$. However, there exists another way to prove it by considering path functions. Have a look at the first twenty minutes of this lecture by MIT OCW.

• If you suddenly drop the external pressure on the gas to a new lower value and then hold it constant throughout the process, would you consider that a constant pressure expansion? Oct 10, 2020 at 3:48
• Didn't I ask you this question on physics stack exchange? xD I'm not sure if the constant variable process means constant variable for state or constant variable for external. However, for reversible it's simple because both are at equilibrium for each point in the process.
– Babu
Oct 10, 2020 at 5:07
• Is that Joule-Thomson expansion? Oct 10, 2020 at 9:27
• No, not necessarily. Oct 10, 2020 at 10:09

If your process is not isobaric, there is no way you can consider it as only one hypothetical isobaric step. So you need to 'decompose' your process into more than one steps, and among those several steps, there can be isobaric step.

For example, your process of interest is

$$(p_1, V_1, T_1) {\rightarrow} (p_2, V_2, T_2)$$

where $$p_1 ≠ p_2, V_1 ≠ V_2, T_1 ≠ T_2$$.

You can consider it as two steps:

$$(p_1, V_1, T_1) {\rightarrow} (p_1, V_a, T_2) {\rightarrow} (p_2, V_2, T_2)$$

where your first step is isobaric heating/ cooling and your second step is isothermal expansion/ compression.

Your formula applies only to the first step which is isobaric. If your system made up of perfect gas, $$ΔH$$ for second step is zero as commented by Sir Chet Miller so you can apply that formula to get final $$ΔH$$ directly. Else, you will need to find $$ΔH$$ for second step and sum up $$ΔH$$ of both steps to get $$ΔH$$ of process of interest.

In short, when comparing process of interest and hypothetical steps, the initial states for both must be the same, and the final states for both must be the same. If your process is not isobaric, and you want to 'decompose' it into only one step which is isobaric as illustrated by your question, the first sentence of this paragraph will not be true, this cannot be done!