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The ratio of the heat capacities $\frac{C_\mathrm p}{C_\mathrm v}$ for one mole of a gas is $1.67$. The gas is:

a) $\ce{He}$
b) $\ce{H2}$
c) $\ce{CO2}$
d) $\ce{CH4}$

I know how to answer this by analysing the degrees of freedom of the gas molecule. Is there any other way to derive the ratio?

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    $\begingroup$ This question does not make sense without giving the temperature of the gases. $\endgroup$
    – Georg
    Aug 25 '13 at 17:39
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I'd go about such a question looking at the easiest thing first: ideal gases. So, what do you know about the relations between $C_{p}$ and $C_{v}$ (both being the molar heat capacities at constant pressure and constant volume, respectively) for ideal gases? For one thing, it is well known that

\begin{equation} C_{p} - C_{v} = R \end{equation}

with $R$ is the universal gas constant. You can rearrange this equation for the ratio you are after:

\begin{equation} \frac{C_{p} - C_{v}}{C_{v}} = \frac{R}{C_{v}} \\ \frac{C_{p}}{C_{v}} - 1 = \frac{R}{C_{v}} \\ \frac{C_{p}}{C_{v}} = \frac{R}{C_{v}} + 1 \ . \end{equation}

Now, you need to find $C_{v}$. Here, the degrees of freedom that you mentioned come into the game, since the equation that helps us out is the equipartition law, stating that each excited quadratic degree of freedom contributes $\frac{1}{2} R T$ to the molar internal energy $U$

\begin{equation} U = \frac{1}{2} R T (f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}}) \end{equation}

with $f_{\text{trans}}$, $f_{\text{rot}}$ and $f_{\text{vib}}$ being the quadratic degrees of freedom for translation, rotation and vibration respectively. It is important to note that only excited degrees of freedom contribute to $U$. Which ones are active depends heavily on temperatur. Vibrations are usually not excited at room temperatur (there are exceptions, e.g. iodine) - they usually need temperatures in the order of $1000 \, \text{K}$ but that varies greatly. Rotations are much easier to excite - I know of no example where they aren't excited at room temperature. Translations take very little energy to excite and can always be considered active. If all degrees of freedom are active you get: For linear molecules $f_{\text{trans}} = 3$, $f_{\text{rot}} = 2$ and $f_{\text{vib}} = 3N - 5$, where $N$ is the number of atoms in the molecule. For non-linear molecules $f_{\text{trans}} = 3$, $f_{\text{rot}} = 3$ and $f_{\text{vib}} = 3N - 6$. Since $C_{v} = \frac{dU}{dT}$ you get

\begin{equation} C_{v} = \frac{1}{2} R (f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}}) \ . \end{equation}

Substituting this into the equation for the ratio between $C_{p}$ and $C_{v}$ yields

\begin{equation} \frac{C_{p}}{C_{v}} = \frac{2}{(f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}})} + 1 \ . \end{equation}

Now, you only have to know how many degrees of freedom each of the molecules has and plug them into the equation.

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  • $\begingroup$ Hey Philipp, I like your answer, but I think you were a bit to fast with what $n$ stands for (and whether $U$ is molar internal energy or internal energy)… in any case, your statement below the first equation (“$n$ is the number of atoms in the molecule”) is wrong. $\endgroup$
    – F'x
    Aug 25 '13 at 10:14
  • $\begingroup$ @F'x You are right. It's been a while since I took the thermodynamics class and my memory seems to be worse than I hoped. Thanks for pointing that out. Usually I put a little more thought into my answers (as my longer ones hopefully show) but this time I thought I'd remember it correctly and thus didn't look it up again. I corrected my answer and hope everything is all right, now. If my answer still lacks something, please tell me. $\endgroup$
    – Philipp
    Aug 25 '13 at 14:48
  • $\begingroup$ Your answer only is correct for high enough temperatures. rotational and vibrational degrees are not active at low temperatures. So, the question does not make sense without giving the temperatures of the gases. $\endgroup$
    – Georg
    Aug 25 '13 at 17:38
  • $\begingroup$ @Georg Of course, you are right that rotational and vibrational degrees of freedom are not always active. I will add that to my answer. But rotations are usually excited at quite low temperatures and since no temperature was given in the question I assumed room temperature. And for room temperature the choice is clear: $\ce{He}$ since it is the only gas that cannot possibly have rotational or vibrational degrees of freedom and thus $\frac{C_{p}}{C_{v}} = \frac{2}{3} + 1 \approx 1.67$. $\endgroup$
    – Philipp
    Aug 25 '13 at 17:48
  • $\begingroup$ @Philipp what I like to see is mentioning the problem at least. Whether rotation is active in Hydrogen at room temperature or not is not the important detail. $\endgroup$
    – Georg
    Aug 25 '13 at 18:04

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