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The derivation that enthalpy equals heat at constant pressure goes like:

$$\begin{align} H &= U + P_{\mathrm{int}}V \\ \Delta H &= \Delta U + \Delta (P_{\mathrm{int}}V) \end{align}$$

If $P_{\mathrm{int}}$ is constant, then

$$\Delta H = \Delta U + P_{\mathrm{int}}\Delta V$$

We also have

$$q = \Delta U - w$$

If external pressure is constant, $w = -P_{\mathrm{ext}}\Delta V$. Then

$$q = \Delta U + P_{\mathrm{ext}}\Delta V$$

So in order for $\Delta H$ to equal $q$, we need $P_{\mathrm{int}} = P_{\mathrm{ext}} = \mathrm{constant}$. This is true when the system is open to a constant external pressure and no PV work is done. But when there's PV work done (and the temperature is still constant), $P_{\mathrm{int}}$ clearly changes, so does $\Delta H$ still equal $q$?

carbenoid
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2 Answers2

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I shall start off by mentioning that your question is not so much about p-V work being done, but rather about the pressure not being constant. It is perfectly possible for p-V work to be done when the pressure is constant, and indeed under these circumstances, the equation $\Delta H = q$ is still obeyed. However, this would necessitate a change in temperature.

Well, you can always try to come up with a counterexample.

Your hypothesis would be that if pressure is not constant and p-V work is done, then $\Delta H = q$. So, let's try calculating those two quantities for a reversible isothermal expansion of $1\mathrm{~mol}$ of an ideal gas, from an initial state of $(p_1, V_1)$ to a final state of $(p_2, V_2)$. Let's just arbitrarily set $V_2 = 2V_1$. Of course, since this is an ideal gas we are talking about, $p_2$ will be equal to $p_1/2$. And since this is a reversible process, $p_\mathrm{int} = p_\mathrm{ext}$ throughout, so let's drop the int and ext subscripts.

Let's calculate the work done (on the system) first.

$$\begin{align} w &= -\int \! p\,\mathrm{d}V \\ &= -\int \! \frac{nRT}{V}\,\mathrm{d}V \\ &= nRT \ln{\left(\frac{V_1}{V_2}\right)} \\ &= -RT \ln 2 \end{align}$$

Now the internal energy change of the system has to be zero, since this is a closed system with an ideal gas and therefore $U = U(T)$. $\Delta T = 0$ therefore implies $\Delta U = 0$. So:

$$\begin{align} q &= \Delta U - w \\ &= RT \ln 2 \end{align}$$

However, the enthalpy $H$ is also only a function of temperature: $H = H(T)$. Therefore,

$$\Delta H = 0 \neq q$$

Hypothesis: disproved!

orthocresol
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  • So if internal pressure changes but external pressure is constant, then enthalpy change can't equal heat right? – carbenoid Oct 31 '15 at 19:01
  • @swenger In general, it will not be equal. – orthocresol Oct 31 '15 at 19:02
  • Then I don't know why in my textbook, it says that when pv work is done at constant external pressure and constant temperature, enthalpy change equals heat? – carbenoid Oct 31 '15 at 19:11
  • That makes no sense, especially if you consider an ideal gas system. Regardless of whether the process is reversible or irreversible, constant temperature of the system means $\Delta H = 0$ and $\Delta U = 0$. And, if $w \neq 0$ then $q = -w \neq 0$ (regardless of what the expression for the work is). – orthocresol Oct 31 '15 at 19:15
  • Actually, in the example in my textbook, there's a reaction occurring which causes the system to expand under a constant temperature and external pressure. So in this case, $\Delta H \neq 0$. But can we say that $\Delta H=q$? – carbenoid Oct 31 '15 at 20:19
  • I am not 100% sure, but I don't think so. As far as I know, there is no reason for $\Delta H$ to be equal to $q$ if the conditions you laid out in your question aren't met. – orthocresol Oct 31 '15 at 20:24
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The external pressure applied by the surroundings on the system at the interface between the system and the surroundings is always equal to the force per unit area exerted by the system on the surroundings at the interface, even if the process is irreversible. It's just that, in the irreversible case, the pressure within the system is not uniform (so there is no single unique value for the pressure) and the force per unit area exerted by the system on the surroundings at the interface also includes viscous stresses. So, if the external pressure applied by the surroundings is constant, the work done by the system on the surroundings is equal to $p_{ext}\Delta V$. If the constant external pressure applied by the surroundings is also equal to the initial pressure p exhibited by the system in its initial thermodynamic equilibrium state, then the enthalpy change is $\Delta H=\Delta U + p_{ext}\Delta V=\Delta U + p\Delta V$. But, by the first law, this is just equal to the heat added to the system. So, irrespective of what the thermal conditions are (say, isothermal or adiabatic) or whether chemical reaction is occurring, if the external pressure is held constant at the initial pressure of the system in its initial thermodynamic equilibrium state, the heat transferred is equal to $\Delta H$. If the process is such that the initial and final temperatures of the system are the same, we call the corresponding $Q = \Delta H$ the heat of reaction (at least for ideal gases for which the heat of mixing is zero).

Chet Miller
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  • So since $\Delta H=\Delta U+\Delta(p_{int}V)$, don't we need $p_{int}$ to be held constant as well (in addition to the external pressure being held constant at the initial pressure of the system in its initial thermodynamic equilibrium state)? – carbenoid Nov 01 '15 at 16:09
  • No. It isn't correct to even talk about $p_{int}$ for a system experiencing an irreversible deformation since the pressure isn't even constant spatially within the system, plus, viscous stresses in the system contribute to the force of the system on the boundary with the surroundings. However, the force per unit area that the system exerts on the surroundings will be constant at the boundary and equal to $p_{ext}$ throughout the deformation. Only at the initial and final states will $p_{int}$ be relevant and equal to $p_{ext}$ throughout the system. – Chet Miller Nov 01 '15 at 20:18