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It is known that the specific heat of a gas is process dependent.So it must be theoretically possible to have a negative value for a gas according to the following equation (for polytropic process): $$ C = \frac R{\gamma-1} + \frac R{1-n} $$ where $C$ is molar specific heat and $\gamma$ is adiabatic exponent.

(supposing $\gamma = \frac{5}{3}$ and $n = \frac{4}{3}$, $C$ comes out to be negative)

Is it practically possible and if so what would it signify? As you provide more heat to a gas in such a process, would it lose temperature? Please clarify.

Thomas Fritsch
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evamPUNdit
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  • Is the example for an ideal gas? – Bob D May 02 '19 at 17:06
  • And do you have a reference for this equation? – Bob D May 02 '19 at 17:12
  • I came across this in a material provided by my faculty in a coaching institite – evamPUNdit May 02 '19 at 17:18
  • Which molar specific heat is $C$ supposed to be? $C_{p}$ or $C_{v}$? – Bob D May 02 '19 at 17:22
  • This might be helpful: https://physics.stackexchange.com/questions/142461 – Eagle May 02 '19 at 17:24
  • I did come across this question. But my doubt is specific to gases. – evamPUNdit May 02 '19 at 17:30
  • @ImmortalUchiha Since you appear to be dealing with an ideal gas based on the value of gamma, which $C$ are you talking about, $C_v$ or $C_p$? – Bob D May 02 '19 at 17:33
  • neither. Cp and Cv are for specific processes. An ideal ideal gas can undergo infinitely many processes and C is for some a case. – evamPUNdit May 02 '19 at 17:36
  • OK I can confirm you will can get a negative number for $C$, but that depends on whether the equation is true. You need to do better than simply saying it was in material provided by your faculty. How was it presented? In what context? – Bob D May 02 '19 at 18:00
  • I found the derivation of this equation in this answer:https://physics.stackexchange.com/a/108954/206442 – evamPUNdit May 02 '19 at 18:04

3 Answers3

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I don’t know where that formula comes from, but to the best of my knowledge (this is not to say I’m right) $C$ is a nonnegative quantity. Some extravagant cases do exist where it comes out to be negative indeed, but it is normally (again, to the best of my knowledge) a consequence of not properly considering one or several energy sources.

KernelPanic
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In the OP's equation, $C$ will be negative if $\gamma > 1$ and $1 < n < \gamma$.

That implies a system where mechanical work and heat are flowing in opposite directions. That is physically possible, for example a mechanical compressor is doing work to compress the gas, but the high temperature gas is losing heat to the environment by conduction through the compressor structure.

You could imagine a situation where the gas is losing heat to the environment faster than you do mechanical work to it, in which case its temperature would decrease.

But I can't quite see what you are going to discover about the system from this bit of mathematics that you didn't know already.

alephzero
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  • Yes, but that all depends on whether or not the OP's equation is correct. Have you seen this equation? As far as I know, the specific heat should not be be negative for a gas undergoing a polytropic process.. – Bob D May 02 '19 at 18:05
  • @BobD Its as you say. I do not know the actual validity of this equation. If I'm wrong anywhere please correct me – evamPUNdit May 02 '19 at 18:12
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Assuming ideal gas conditions, The above equation $$C = \frac R{\gamma-1} + \frac R{1-n}$$

You need to know two things:

  1. The first part of the sum that is $ \frac R{\gamma-1}$ refers to the heat required to change the temperature of the gas by $1$K When no work is done by the gas.(Isochoric condition)

  2. The second part of the sum $\frac R{1-n}$ refers to the work that is done by the gas.
    As you change the temperature, the pressure and volume also changes and so the gas does work.

Now your question, when you have such a process which renders the specific heat capacity negative, that process actually decreases the volume of the gas with increase in temperature

Here, look at this graph

Hence work done by the gas is negative and this results in the above sum (the C equation) to become negative.

In short:
Heating the gas would lead to contraction of the gas, and the work done by the gas will be negative and greater than the heat required to changed temperature of the gas in isochoric conditions. This makes the thermal capacity negative

Bhaskar0120
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  • Probably minor point, but the OP used the term "molar specific heat" in describing C. Is that the same as your term "thermal capacity"?. In other words, are you saying that the "molar specific heat' can be negative? Because that's the term the OP used for C. – Bob D May 04 '19 at 22:55
  • @BobD The term molar heat capacity is the heat required to raise temp of 1 mole by 1K, while thermal capacity could be defined as the heat required to increase the temperature of a sample of gas by 1k, this sample could have any number of moles, 1 mole, 10 moles, etc. – Bhaskar0120 May 04 '19 at 23:08
  • @BobD And I am not saying that the molar heat capacity can be negative, But the following equation, the way it has been structured, it takes into account the work done by the gas and change in internal energy of the gas. Because the work term becomea negative here, the equation becomes negative. – Bhaskar0120 May 04 '19 at 23:12
  • @Thanks for the reply. But my question is simply can the "molar specific heat" be negative? – Bob D May 04 '19 at 23:12
  • @Let me put it another way, should the OP have said that C is the "thermal capacity" and not the "molar specific heat"? Because my answer was based on the latter. Thanks. – Bob D May 04 '19 at 23:29