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I have a question in isothermal process. Since
$$Q = nCdT$$ and
$$dU = nCdT,$$ where $T$ = temperature and $U$ = internal energy. In isothermal process as $$ dT = 0 .$$ Therefore
$$ dU =0. $$ But why is $Q$ not equal to zero?

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  • $\begingroup$ I myself found the answer as in isothermal process C is infinite so we cannot calculate Q by this formula. $\endgroup$ Feb 3, 2020 at 9:30
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    $\begingroup$ I don't think you can say that $C$ is infinite... Why would you think that? $\endgroup$ Feb 3, 2020 at 23:25
  • $\begingroup$ I said that because in isothermal process dT = 0 and C = Q/(n*0) = undefined. $\endgroup$ Feb 4, 2020 at 0:55
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    $\begingroup$ My understanding is that heat capacity ($C_V$ or $C_p$) is not a property of a process, but of the state of the system ($V$ and $T$, for example). For a monoatomic ideal gas, they are always $C_V = 3/2 R$ and $C_p = 5/2 R$, never infinite. $\endgroup$ Feb 4, 2020 at 2:06
  • $\begingroup$ $C_V$ , $C_p$ are calculated while fixed volume and pressure respectively and in isothermal process neither pressure is constant nor volume. If you don't believe that C is infinite then check it out:. quora.com/… $\endgroup$ Feb 4, 2020 at 9:34

3 Answers 3

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In thermodynamics, the heat capacity is not defined in terms of heat Q (i.e., the way they taught you in freshman physics). It is defined as a physical property of the material being processed rather than as a feature of a process, in terms in terms of internal energy or enthalpy:$$nC_v=\left(\frac{\partial U}{\partial T}\right)_V$$and$$nC_p=\left(\frac{\partial H}{\partial T}\right)_P$$The answer is as simple as that.

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  • $\begingroup$ You can equally well define it as $\frac{dQ}{dT}$ for an isochoric process, since $\delta Q = dU + pdV$ (or, more generally, for a process without external work). $\endgroup$ Feb 3, 2020 at 23:29
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    $\begingroup$ @EliasRiedelGårding P-V work is not the only work that can be done. Consider the case of a fluid contained in a constant-volume insulated vessel that is being agitated by a stirrer. The stirrer does work on the fluid (either liquid or gas), and the temperature of the fluid increases. The first law applied to this system is $$\Delta U=nC_v\Delta T = W$$where W is the work done by the stirrer on the fluid. So there is a temperature rise $\Delta T$ at constant volume without any heat being added (Q=0). $\endgroup$ Feb 4, 2020 at 16:46
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    $\begingroup$ In my judgment, you are definitely on the wrong track. It is not adding heat to operate a stirrer on the fluid. The stirrer applies forces to the fluid and these forces are applied through displacements. So work is definitely being done on the fluid. There is no heat transferred across the boundary of the fluid, which is what Q is supposed to represent. $\endgroup$ Feb 5, 2020 at 0:42
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    $\begingroup$ I've seen the new thread that you started. Unfortunately, my background in statistical thermodynamics is very minimal. My experience is with classical thermodynamics. In classical thermodynamics, mechanical work is the result of exerting forces and corresponding displacements at the boundary of a systeml; in the stirrer example, the system is the fluid and work is occurring at its boundary with the stirrer. And heat Q represents energy transferred to the system through its boundary with its surroundings as a result of a temperature difference. $\endgroup$ Feb 5, 2020 at 3:48
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    $\begingroup$ I understand. It would be strange if there were a difference between classical and statistical... Many thanks! $\endgroup$ Feb 5, 2020 at 11:45
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$Q = nCdT$ does not hold in general. Imagine that an ideal gas is under isothermal compression, $dU=0$ and $W=nRTln(V_2/V_1)$, then $Q$ is definitely not zero in this case.

In general, $Q$ depends on the specific process that the system undergoes.

You might have seen $Q = nC_VdT$ in textbooks, but this holds only for ideal gas with constant volume (constant volume so there is no work exchange)

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  • $\begingroup$ In your last equation shouldn't $Q$ be $dU$? $\endgroup$
    – Bob D
    Feb 3, 2020 at 14:07
  • $\begingroup$ And by the way, $dU=nC_{V}dT$ for an ideal gas for ALL processes, not just constant volume processes. $\endgroup$
    – Bob D
    Feb 3, 2020 at 14:08
  • $\begingroup$ @Bob D $C_V$ denotes the heat capacity at constant $V$. For and ideal gas, the internal energy depends only on temperature, that is the origin of relation like $dU=cdV$, but the heat capacity depends on different processes like constant volume we have $C_V$, constant pressure we have $C_P$ $\endgroup$
    – FaDA
    Feb 4, 2020 at 1:21
  • $\begingroup$ I know $C_V$ is the specific heat at constant volume. But for an ideal gas, $dU=C_{V}dT$ for ANY process, not just a constant volume process.Do you not understand that? I suggest you research it and find out. $\endgroup$
    – Bob D
    Feb 4, 2020 at 3:45
  • $\begingroup$ @BobD you should carefully look at my post before trying to make any comments. What i said was "Q depends on the specific process that the system undergoes." I didnt deny that $dU=nC_{V}dT$ $\endgroup$
    – FaDA
    Feb 4, 2020 at 4:31
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You are leaving out one term of the first law: $$ \delta Q = dU + \delta W. $$ Adding heat $\delta Q$ means adding that amount of energy to your system. That energy has to go somewhere. It can either stay in the system as internal energy ($dU$) or it can go out of the system by doing work $\delta W$ on its surroundings.

For an ideal gas, $dU = n C_V dT$ and $\delta W = p dV = nRT \frac{dV}{V}$. So in an isothermal process, adding heat must increase the volume of the gas: $Q = W = nRT \ln\left(\frac{V_\mathrm{final}}{V_\mathrm{initial}}\right)$.

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