5
$\begingroup$

The heat capacity of $\ce{H2}$ at constant $P$ and temperature is ______.

It was explained to me that the answer is infinity with the following reasoning:

Since heat capacity = $\mathrm{d}H/\mathrm{d}T$ and $\mathrm{d}T$ is zero since constant temperature is part of the question, the answer is infinity.

My attempt: Since $\ce{H2}$ is a diatomic gas, its $C_p$ is $7R/2$ which gives $7~\mathrm{cal \cdot mol^{-1} \cdot K^{-1}}$.

Which method is correct?

$\endgroup$
  • 1
    $\begingroup$ I had an elementary school teacher who loved to flip A/THE to make true false questions. This is one of those gotcha type questions. // Look at it another way. No amount of hydrogen was defined in the problem, yet your answer has units of $mol^{-1}$. // So with an infinite amount of hydrogen you could dump an infinite amount of heat and still not get a temperature or pressure rise. There are a lot of these sorts of "fuzzy problems" that get weird if you dissect them too much. $\endgroup$ – MaxW Nov 4 '15 at 17:51
  • $\begingroup$ To me I thought of the hydrogen like a brass cylinder which gets rid of pressure. "The heat capacity of a brass cylinder at constant temperature is ___". Since I don't know how big the brass cylinder is, I can't determine its heat capacity (change in temperature per gram of brass). But if I add any heat then the temperature has to change. So to keep the brass cylinder at the same temperature it can't lose or gain heat. So dH =0. $\endgroup$ – MaxW Nov 4 '15 at 20:44
1
$\begingroup$

If you assume ideal behaviour and a closed system, then since the ideal gas law holds, with pressure and temperature remaining constant volume must also be constant.So that means that nothing is happening to the system, everything is as it is with $dq=0$ and $dw=0$.Since there is no process involved there is no meaning for the term heat capacity.

| improve this answer | |
$\endgroup$
  • $\begingroup$ I'd answered 0. If you add heat then either (1) system does work to keep P and T constant (2) P rises (3) T rises (4) some combination of P and T rising. I never would have deduced infinity. $\endgroup$ – MaxW Nov 4 '15 at 18:45
  • $\begingroup$ Heat capacity is not a property of a process. It is a property of a system: $C_{p,\mathrm{m}} = (\partial H_\mathrm m / \partial T)_p$. $\endgroup$ – orthocresol Nov 4 '15 at 20:16
  • $\begingroup$ @orthocresol but Heat capacity is not independent of the path.Here though the pressure is constant, since temperature is also the same,how should one define heat capacity-we could even claim it to be C_v instead of using C_p.That is what seems ambiguous to me. $\endgroup$ – Aditya Anand Nov 5 '15 at 19:01
1
$\begingroup$

In my humble opinion, your answer is correct and your teacher's answer is incorrect. Heat capacity is an intensive property of a material, defined as you expressed it. If the gas is ideal, then, if dT is equal to zero, dH is also equal to zero, so the ratio dH/dT appears to be indeterminate. However the limit of $\frac{\Delta H}{\Delta T}$ as $\Delta T$ approaches zero is the heat capacity, and that is not indeterminate.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I hate problems like this because you have "read between the lines." "Heat capacity" is ill defined. Does it mean (1) "molar heat capacity" or (2) "specific heat capacity" or (3) "heat capacity" of some undefined system (amount of hydrogen.) I took it to mean (3). So it doesn't matter how much hydrogen gas that the system has, if you add heat then at least one of the two parameters of pressure and temperature must change. $\endgroup$ – MaxW Nov 4 '15 at 20:35
  • $\begingroup$ I agree. The most precise definition of $C_p$ that I am familiar with is $C_p=\left(\frac{\partial H}{\partial T}\right)_P$, but this doesn't specify whether if is per mole or per unit mass. Anyway, in terms of the problem at hand, with a change of phase such as vaporization or melting, the enthalpy can change discontinuously at constant temperature and pressure, in which case the heat capacity of the material at such a point can be regarded as infinite (a Dirac delta function in temperature). $\endgroup$ – Chet Miller Nov 4 '15 at 22:15
1
$\begingroup$

I think the answer infinity is correct.

We know that $c=\frac{\partial H}{\partial T}$. But, in your case, $\partial T$ is not $\to0$, in fact, $\partial T$ is said to be exactly zero. You might be tempted now to say that $c=\frac{\partial H}{\partial T}=\frac{\partial H}0=\infty$ but this is NOT the correct way to interpret this question (and neither is it mathematically valid). The correct way to interpret this question is that "a very large change in enthalpy causes a negligible change in temperature". Hence, $c\to\infty$ (a mathematical way to express "enormously large").

PS: This process is commonly taken to be the phase change process at the boiling point or the melting point. However, it is more complicated at that point than it seems, because the mass of the system also changes. At this point, we may have to thus use the specific heat capacity, $c=\frac 1m \frac{\partial H}{\partial T}$ which is defined per unit mass. But again, the specific heat isn't defined for processes in which mass changes. So, it is really tough to do any reasonable calorimetry at this point.

| improve this answer | |
$\endgroup$
  • $\begingroup$ The question never says that there is a change in enthalpy, let alone "a very large change in enthalpy ". $\endgroup$ – DavePhD Mar 1 '18 at 17:04
  • $\begingroup$ @DavePhD Indeed. I am not saying that there is necessarily an exchange in enthalpy. Certainly, the hydrogen gas can remain wherever it is without gaining or losing any enthalpy. I am saying that if there was a certain change in enthalpy, then the change in temperature wrt that enthalpy change would be almost negligible. $\endgroup$ – Gaurang Tandon Mar 1 '18 at 17:08
  • $\begingroup$ It seems like you're saying that if I have a glass of water sitting in my kitchen, at constant temperature and pressure, then the heat capacity of water becomes zero. $\endgroup$ – DavePhD Mar 1 '18 at 17:32
  • $\begingroup$ @DavePhD No. But if you do supply enthalpy to that glass of water, then its temperature will rise. By calculating $\left(\frac{\partial H}{\partial T}\right)$, you would be able to measure the heat capacity of that glass of water at that temperature. This is also what I am doing in my answer post above. $\endgroup$ – Gaurang Tandon Mar 2 '18 at 2:02
1
$\begingroup$

Heat capacity $C$ is not well-defined for a closed system undergoing constant temperature processes, at least with a reversible process. For a reversible process in a closed system,

$$\delta q_\pu{rev} = C\pu{d}T.$$

If temperature is constant, then $\mathrm{d}T=0$ gives $q_\pu{rev} = 0$ for any finite value of $C$. Other formulae for heat capacity, expressed through internal energy or enthalpy, require additional constraints such as constant volume or pressure, and only expansion work is allowed. Neither of these alternative depictions change the outcome.

This is not a problem because heat capacity does not make sense anyway if we assume a priori that temperature cannot change at all.

My attempt: Since $\ce{H2}$ is a diatomic gas, its $C_P$ is $7R/2$ which gives $\pu{7 cal mol-1 K^-1}$.

I would be fine with this answer as well. If heat is supplied at constant pressure to a closed non-reactive system, then the temperature must change. The alternative would be an change in volume to keep temperature constant but this would alter the pressure of the system.

In other words, in the net-constant pressure and net-constant temperature process dihydrogen had to warm up (according to its heat capacity at constant pressure), and then cool off again. It is not a good idea to equate the heat capacity here to infinity; that would make little physical and mathematical sense in my opinion.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.