5
$\begingroup$

It's a fact that $$ Q= nC \Delta T$$

Where $Q$ is the heat of the body and $C$ is the molar heat capacity while $T$ is the temperature.

In an isothermal process we keep the temperature constant which means that internal energy is constant, but still according to the second law there is a net change is heat which is equal to the work done. According to our previous equation, there must be no heat absorbed, then how come the heat change exist according to the second law?

$\endgroup$
  • 2
    $\begingroup$ Note, to be precise there shouldn't be a $\Delta$ on the $\Delta Q$. Heat $Q$ is in itself already a change (an energy transfer process) $\endgroup$ – Steeven Dec 22 '16 at 9:41
  • 2
    $\begingroup$ "Question titles you can't <del>play in a guitar store</del> use on Physics.SE" $\endgroup$ – bright-star Dec 22 '16 at 19:25
  • 1
    $\begingroup$ For future reference, \triangle shouldn't be used, instead \Delta should be used to yield $\Delta$, the correct notation for Difference operator. $\endgroup$ – user36790 Dec 23 '16 at 2:37
13
$\begingroup$

To keep things simple, they fooled us in freshman physics. They told us that $Q=nC\Delta T$. But this equation is not correct when work is being done. If they didn't want to confuse us, they should have introduced the internal energy U, and correctly defined the heat capacity (at least for ideal gases and incompressible solids and liquids) in terms of U by the equation $\Delta U=nC\Delta T$. This equation still gives the correct result for Q when no work is being done. But for cases in which work is being done, we obtain from the 1st law$$\Delta U=nC\Delta T=Q-W$$For an isothermal case, this reduces to Q = W.

$\endgroup$
  • 1
    $\begingroup$ General rule, OP: when you write "it's a fact" in your question, assume your assumption is wrong :P $\endgroup$ – Lightness Races in Orbit Dec 22 '16 at 22:11
  • 1
    $\begingroup$ Don't be so hard on lower level physics classes. It took hundreds of years to build up the mathematical models we have of Physics now, and many of them require quite advanced mathematics. Additionally, many of the "simpler case" based equations are still useful today because that "special case" is common enough. I don't see anything fundamentally wrong with incrementally learning more complete models of increasing complexity; that's how it happened in history, after all. That said, there is of course great value in ensuring that those classes teach the limitations of the models. $\endgroup$ – jpmc26 Dec 22 '16 at 23:37
  • 1
    $\begingroup$ @jpmc26 With respect, I disagree. If teaching it the wrong way to begin with causes confusion among students, then it should not be taught that way. In freshman physics, it would have taken very little additional effort (if any) to teach it the correct way (and avoid all this confusion). I wish I had a nickle for every time I've seen a student ask this exact same question. $\endgroup$ – Chet Miller Dec 22 '16 at 23:49
  • $\begingroup$ @ChesterMiller Extending your argument to it's logical conclusion, we should start with quantum and GR to avoid creating bad intuition that comes from classical physics. Any teaching method is going to result in some students being confused. The root cause here is that asker doesn't realize it's not applicable in all cases, but that's true of nearly every theory we have. I think putting more emphasis on what assumptions/limitations go with an equation is a much more valuable and broadly applicable skill and would allow students to work with simpler theories early on. $\endgroup$ – jpmc26 Dec 22 '16 at 23:56
  • $\begingroup$ Valid point. I guess we both agree it's a matter of striking the right balance. $\endgroup$ – Chet Miller Dec 23 '16 at 1:23
6
$\begingroup$

According to our previous equation, there must be no heat absorbed

The equation $$Q=nC\Delta T$$ doesn't say that no heat is absorbed. Only that no heat absorbed causes a temperature change. Heat can be absorbed and work be done to cancel it at the same time.

$\endgroup$
  • 1
    $\begingroup$ Actually, the equation does say that if $\Delta T =0$, not heat is absorbed. Actually, the equation is incorrect for cases where work is done. See my answer. $\endgroup$ – Chet Miller Dec 22 '16 at 12:30
  • $\begingroup$ @ChesterMiller Yes, I agree that your answer is good. Note, though, that there is no issue in interpreting this equation as the heat (or energy) that would be required to cause a chosen temperature difference. But agreed, the proper formula with $\Delta U$ instead of $Q$ makes more sense. $\endgroup$ – Steeven Dec 22 '16 at 13:02
3
$\begingroup$

Let me jot down about ideal gas, first.

Thermal capacity at constant volume $\textrm C_V$ is defined as

$$\mathrm C_V ~=~\left(\frac{\partial U}{\partial T}\right)_V$$ where $U$ is the internal energy of the system.

For an ideal gas, $U=U(T);$ so, $$\mathrm C_V~\mathrm dT ~=~ \mathrm dU\,.\tag I$$

Substituting $\mathrm{(I)}$ in the First Law of Thermodynamics,

$$\mathrm C_V~\mathrm dT +đw~=~ đQ,\tag{II} $$

$đw = p~\mathrm dV;$ this implies $$\mathrm C_V~\mathrm dT +p~\mathrm dV~=~ đQ.\tag{II.a} $$

Therefore, change in entropy $\mathrm dS_\textrm{ideal gas}$ for our system of ideal gas can be written as: $$\mathrm dS_\textrm{ideal gas} =\frac{đQ}T = \frac{\mathrm C_V}{T}~\mathrm dT + \underbrace{\frac{\mathcal R}V}_{pV~ =~\mathcal RT}~\mathrm dV . \tag{III}$$

When, $\mathrm dT = 0$ for isothermal transformation, then $$\mathrm dS_\textrm{ideal gas} = \frac{đQ}T = \frac{\mathcal R }{V}~\mathrm dV\tag{III.a}$$


Suppose, we choose $T, V$ as the independent variable to define the state of a general system.

So, entropy change $\mathrm dS$ is given by $$\mathrm dS = \frac{đQ}T = \frac1T\left(\frac{\partial U}{\partial T}\right)_V ~\mathrm dT + \frac1T\left[\left(\frac{\partial U}{\partial V}\right)_T + p~\right]~\mathrm dV;\tag{IV}$$ when $\mathrm dT = 0,$ $$\mathrm dS =\frac{đQ}T =\frac1T\left[\left(\frac{\partial U}{\partial V}\right)_T + p~\right]~\mathrm dV.\tag{IV.a} $$

There is nothing contradictory here. $\mathrm dT= 0$ doesn't mean $đQ = 0$ as is evident above in $\mathrm{(II)}$.

$\endgroup$

Your Answer

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