0
$\begingroup$

Sign Conventions:-

Work if done by system, heat given to system and Increase in Internal Energy implies $+$ change.
Otherwise $-$.

Ok, consider an ideal gas in a container with perfectly conducting walls and a frictionless piston.

In my textbook, specific heat is given as,
$c=\frac{\Delta Q}{m \Delta T}$

If work is done on the gas such that rate of work done to compress the gas is more than the rate at which it loses heat in the surroundings then,

There will be increase in internal energy or in it's temp.

In that case, as per the equations and sign conventions used,

Specific heat of the ideal gas should be negative.

Heat Capacity wikipedia page: Go to Negative Heat Capacity under Measurement

In that section(which I understand little bit), it says

A negative heat capacity can result in a negative temperature.

So, the statement implies that negative specific heat is not something one can observe in ideal gases(because in theory, to be precise, in high school physics theory, there can't be a temperature less than absolute 0).

So,if the following is possible

if work is done on the gas such that rate of work done to compress the gas is more[...]

then, can specific heat be negative, specifically in my case?

$\endgroup$
  • $\begingroup$ "rate at which gas is compressed is more than the rate at which it loses heat in the surroundings" How can you compare these two things? The units are not the same. Do you mean to say the work done in compressing the gas requires more energy than the energy lost to heat? $\endgroup$ – Aaron Stevens Jul 6 '18 at 16:07
  • $\begingroup$ I edited the post. Please have a look at it $\endgroup$ – lakhi Jul 6 '18 at 18:20
  • $\begingroup$ The specific heat capacity of an ideal gas is a physical property of that gas, that depends on temperature. Various inaccuracies and imprecision in experimental results does not change that fact. The Wikipedia article should probably be reworded. $\endgroup$ – David White Dec 31 '18 at 21:34
  • $\begingroup$ @lakhi, in reading the Wikipedia article, I also note that you are taking some of the Wikipedia commentary WAY out of context. The text clearly states that the "negative temperature" situations do not occur under thermodynamic equilibrium conditions, implying that extrapolating to conclusions regarding ideal gases on earth is probably not a good idea. $\endgroup$ – David White Dec 31 '18 at 21:43
0
$\begingroup$

I think your confusion is in understanding the definition of specific heat you have given. The idea is that we supply a known amount of energy as heat to our object and then look at the resulting temperature change. In other words, the specific heat tells us how much energy is needed to change the temperature of an object. You cannot apply this definition to any general process (it is not the specific heat of a process, it is the specific heat of the material).

In other words, just because a process involves transfer of energy as heat and a temperature change does not mean we just plug in those values into the specific heat definition and then say that is the heat capacity of that material. This is just not the case.

$\endgroup$
  • $\begingroup$ Ok, see, I have done some work and it results loss of heat from the gas. Evaluating that heat loss and the resulting temperature change of gas, I calculate the specific heat capacity of the gas and found out it to be negative as per sign conventions. Is that the 2 unknowns I found, loss of heat and temperature change are incorrect to put into that standard formula of specific heat? If you agree with this, then can you explain, why so?? $\endgroup$ – lakhi Jul 7 '18 at 5:54
  • $\begingroup$ @lakhi you are just restating your question in your comment, so I will just refer you to my answer. Since you are doing work on the system, you cannot use the given equation for heat capacity. $\endgroup$ – Aaron Stevens Jul 8 '18 at 3:09
  • $\begingroup$ Ok, agreeing on what you say, what problem arises when work comes into the picture while evaluating specific heat of a gas?? Sorry if I am being too reluctant. $\endgroup$ – lakhi Jul 8 '18 at 6:10
  • $\begingroup$ Are you saying that we cannot supply energy as work as that will contradict the very definition of specific heat?? $\endgroup$ – lakhi Jul 8 '18 at 8:12
  • $\begingroup$ @lakhi For this definition no. Perhaps you should look into the difference between heat capacity at constant volume and constant pressure for an ideal gas. $\endgroup$ – Aaron Stevens Jul 8 '18 at 20:17
0
$\begingroup$

Yes, the specific heat capacity would be negative in that case. Of course it wouldn't be the heat capacity $c_V$ at constant Volume or $c_p$ at constant pressure. These are positive for ideal gases. It would be the heat capacity for some more unusual process, where the system gains more energy through work than it looses as heat (as you describe it in your question).

An example are polytropic processes, which obey

$$ p \cdot V^n = \mathrm{const.} $$

with a constant exponent $n$. The molar heat capacity for such a process is

$$ C_{\mathrm{mol}, n} = R \left[ \frac{1}{\gamma-1} - \frac{1}{n-1} \right], $$

where $\gamma$ is the adiabatic index ($\gamma = 5/3$ for a monoatomic ideal gas) and $R$ the gas constant. (See also this question about its derivation.)

If $1 < n < \gamma$, the heat capacity $C_{\mathrm{mol}, n}$ actually becomes negative.

I don't know where such a process is relevant, but you can make up a mechanical apparatus that realizes it, at least in theory. Maybe something like this:

cylinder with piston and a weight on top, a rope that gets thinner towards its ends is attached to the weight

The weight puts some pressure on the piston, but a heavy rope is attached to it, which is thicker in the middle. The rope runs over a pulley and if the piston goes up part of its weight pulls the piston up and reduces pressure. With the right mass distribution you should get $p \propto V^{-n}$.

However, the negative heat capacity would mean that the system is unstable. If it comes in contact with a hotter reservoir, it absorbs heat, whereupon it expands and cools down so it can absorb even more heat and continue expanding. On the other hand, if the reservoir is cooler, it will continue to contract. I can imagine that, in other systems, this may ultimately lead to states which can be described by negative temperatures, but with an ideal gas this is not possible. Ideal gases can only have positive temperatures. Instead, the above contraption would eventually reach its limits, like the end of the piston or the end of the rope, where the pressure no longer follows a polytropic process.

$\endgroup$

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.