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We know that:

$$\begin{align} H &= U + pV \\ \Delta H &= \Delta U + \Delta (pV) \end{align}$$

Now,

  1. If $p$ is constant,

$$\begin{align} \Delta H &= \Delta U + p\Delta V \\ &= q_p + w + p\Delta V \\ &= q_p \qquad\qquad \text{(since } w = -p\Delta V\text{)} \end{align}$$

  1. If $V$ is constant, $$\begin{align} \Delta H &= \Delta U + V\Delta p \\ &= q_V + w + V\Delta p \\ &= q_V + V\Delta p \qquad \qquad \text{(since } w = -p\Delta V = 0\text{)} \end{align}$$

  2. We also know that at constant $V$,

$$\Delta U = q_V$$

Now my question is that: my book says that at constant $V$, $q = \Delta H = \Delta U$, which I think is false, since I obtained $\Delta U = q_V$, but not $\Delta H = q_v$.

Please tell me where am I going wrong?


Here is a link to the chapter.

Please refer to Pg 161, right column, first paragraph.

Link to chapter: Thermodynamics

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  • $\begingroup$ Just a note: from what I've seen, $\LaTeX$ in titles is discouraged. $\endgroup$ Commented Mar 5, 2016 at 7:54
  • $\begingroup$ @DGS Removed latex! $\endgroup$
    – Max Payne
    Commented Mar 5, 2016 at 7:57
  • $\begingroup$ Apparently, it's mostly the "ce" that is a problem. But here are the specifics, if you're interested. Oh, and it looks funky on search engines. $\endgroup$ Commented Mar 5, 2016 at 7:59
  • $\begingroup$ @DGS Thanks ill make a note of it. although at other sites, I can't do without latex in titles ( eg math SE) ! $\endgroup$
    – Max Payne
    Commented Mar 5, 2016 at 8:01
  • $\begingroup$ I think in the book they take both pressure and volume to be constant when showing you this result since they refer to $\Delta{H} = \Delta{U} + p\Delta{V}$ (eq 6.8) in their derivation which is enthalpy at constant pressure. I agree it's bit confusing. $\endgroup$
    – wuschi
    Commented Mar 5, 2016 at 9:00

1 Answer 1

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No the book is not completely wrong.If you read the next 6 lines then you will find that it wants to point that this approximately true only for solids and liquids, not for gases.

From the First Law,

$$\Delta U = q + w$$ and since work done is $0$ in an isochoric process (constant $V$),

$$\Delta U = q_V$$

Furthermore,

$$\begin{align} \Delta H &= \Delta U + \Delta (pV) \\ &= q_V + \Delta (pV) \end{align}$$

Since we are considering solids and liquids, the changes in the volume of these matter upon changing pressure are negligible (as they are incompressible matter) when compared to gases, so the $\Delta (pV)$ term can be considered negligible.

Therefore

$$\boxed{\Delta H \approx \Delta U = q_V}$$

for these incompressible solids and liquids. However this fails when we consider gases as the term $\Delta (pV)$ will no longer be negligible.

Comparison between the $\Delta (pV)$ values for different states of matter, being heated from $20\ ^\circ\mathrm{C}$ to $30\ ^\circ\mathrm{C}$:

$$\begin{array}{c|c} \text{Substance} & \Delta (pV)\text{ / J} \\ \hline \text{Air} & 2850 \\ \text{Water} & 0.1 \\ \text{Iron} & 0.0004 \end{array}$$

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  • $\begingroup$ You are right, its true for solids and liquids, but it wasn't mentioned specifically so I got confused. I think your answer is very relevant to this question. +1 $\endgroup$
    – Max Payne
    Commented Mar 11, 2016 at 11:00

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