# If $dQ_p = dU_p + pdV = dH_p$, then how can $dQ_p / dT = \partial H_p / \partial T$

In the book of Kondepudi & Prigogine, Modern Theormodynamics, at page 65,

(under constant pressure)

$$dQ_p = dU_p + pdV = dH_p,$$ where $$H_p$$ is the entalpy at the constant pressure $$p$$.

However, then they argue that

If the system consists of 1 mol of a substance, and if the change in temperature due to the exchange of heat is $$dT$$, then it follows that

$$\left(\frac{\mathrm{d} Q}{\mathrm{d} T}\right)_{p}=C_{\mathrm{m} p}=\left(\frac{\partial H_{\mathrm{m}}}{\partial T}\right)_{p},$$ where $$C_{mp}$$ is the molar constant-pressure heat capacity.

However, I'm having hard time understanding how do they go from total derivative to partial derivative.

Mathematically speaking, in order for that to be true, $$Q$$, hence $$H$$, has to be a function of a single variable, $$T$$. However, I don't see why should be the case.

Just to know what we are dealing with here -- LHS: We note that $$Q$$ is not a state function, and the differential $$\delta Q$$ is in-exact. To implement $$\delta Q$$ in our calculations and finally relate it to heat capacity, as inspired from the very same book you mentioned, we interpret it so that $$Q$$ is a function of time,

$$C(t) = \frac{dQ/dt}{dT/dt}$$

and write,

$$dQ=C\mathrm dT.$$

Now, RHS: What does $$\left(\frac{\partial H}{\partial T}\right)_P$$ mean?

Here we have an embedded operation that's done without explicit record: The expression treats $$V$$ itself as a function, $$\hat{V}(P,T)$$. Through such treatment we are essentially requesting the partial derivative of the function $$H(\hat{V}(P,T), T)$$, where

$$H(\hat{V}(P,T), T) \equiv\hat{H}(P,T).$$

This explains why the definition of heat capacity above assumes such dependence as you noticed, there is an implicit consideration of variables themselves as functions of other variables.

• Thanks for your anter @Gulce; however, when $p$ is constant, $dQ$ becomes exact, since the integral $\int pdV = p (V_f - V_i)$ becomes path-independent. – onurcanbektas Aug 25 at 12:58
• But, apart from that, I see your point; since there is no restriction on choosing our macroscopic state variables, choosing them as $P,T$ leads directly to the equality of partial derivative to the total derivative; thanks a lot. – onurcanbektas Aug 25 at 13:03
• Yes, I was just commenting about that -- this also captures my motivation to write the part related to LHS without specifying what is viewing as constant (It is originally a trick introduced in the book, as mentioned). Always happy to share! :) – Gulce Kardes Aug 25 at 13:09

The reason for your confusion is that, in thermodynamics, unbeknownst to you, they have changed the definition of heat capacity without telling you. In thermodynamics, heat capacity (which is supposed to be a physical property of the fluid) is no longer defined in terms of heat Q (because doing work W can also affect the temperature, without transferring heat). Instead, we now define heat capacity in terms of enthalpy and internal energy. Enthalpy per mole is usually expressed as a function of temperature and pressure H(T,P) and internal energy per mole us usually expressed as a function of temperature and specific volume U(T,V). So, differential changes in these physical properties are expressed mathematically by: $$dH=\left(\frac{\partial H}{\partial T}\right)_PdT+\left(\frac{\partial H}{\partial P}\right)_TdP$$and $$dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV$$We then define the heat capacities at constant pressure and at constant volume by:$$C_p\equiv\left(\frac{\partial H}{\partial T}\right)_P$$and$$C_v\equiv\left(\frac{\partial U}{\partial T}\right)_V$$