So recently I've come up with this thought

In an isochoric process, many books state that the work done is zero because there is no change in volume i.e. the gas isn't expanding nor it is contracting. So by knowing that we could write an equation;

$$\delta Q = dU$$

But since by the definition of heat capacity, we have that;

$$\delta Q = C_pdT$$

$$dU = C_vdT$$

Substituting, and now our new equation becomes;

$$C_pdT = C_vdT$$

$$C_p = C_v$$

Which is clearly not true because $C_p$ can never be equal to $C_v$ (knowing that $C_p$ $-$ $C_v$ = $R$). Am I missing something here? What I think that I'm missing is that even though there is no change of volume in the system, there still is a change in pressure and we know that $W$ is a function of $P$ and $V$. Therefore in an isochoric process, there should exist a work that is being done by the system (or an external work?) that is proportional to the change of pressure. Because from what I think is true;

$$\delta Q = dU + \delta W$$

And since the integral of a derivative is basically the function itself; hence,

$$\int d(\delta W) = \delta W$$

Substituting for $d(\delta W)$ where $P$ and $V$ are both not constants,

$$\int (P \delta V + V \delta P) = \delta W$$

$$\delta Q = dU + \int_1^2 (P \delta V + V \delta P)$$

Again, no change in volume,

$$\delta Q = dU + \int_1^2 V \delta P$$

And this is when the problem comes. I've tried finding the integral and it became,

$$\int_1^2 V \delta P = nR \delta T \ln \frac{P_2}{P_1}$$

Change in temperature is not zero in an isochoric processes, hence the $\delta T$. I'm stuck in this equation. But then I read about isentropic processes and I found a similar equation where,

$$W = VdP$$

My intuition tells me that the value of the integral(I'm talking about the $\int_1^2\ V \delta P$ one) is basically the product of the volume $V$, which is a constant, times all of the infinitesimally small $\delta P$s, hence it leads to the equation above(could be wrong though). So to sum up my question; is work done by the change of pressure?

Just for you to know, I'm still new to this calculus notation. So apologies for the incorrection(s). Oh and quick question, does $dT$ means the change of $T$ with respect to a constant or does it simply means $\Delta T$?


I've did the integral and found a mistake, here is what I did,

$$\delta W = \int_1^2 V\delta P$$

Since $P = \frac{nRT}{V}$

and $\delta P = \frac{nR \delta T}{V}$ (accounting that $V$ is constant and $T$ is not).

substituting we get,

$$\delta W = \int_1^2 nR \delta T$$

$$\delta W = nRdT$$

Knowing that $T = \frac {PV}{nR}$

we get,

$$\delta W = VdP$$(is this correct?)

  • $\begingroup$ Ignore the questions about $dT$s and $ \Delta T$s, there's hundreds of questions asking the same. $\endgroup$
    – Felix
    Jan 19, 2020 at 14:19
  • 1
    $\begingroup$ A differential (say $\text{d}x$) times some scalar is still a differential: e.g. $\text{d}A=A\text{d}x$. $\endgroup$
    – Gert
    Jan 19, 2020 at 14:48
  • 2
    $\begingroup$ Why $\delta Q = C_p dT$ ? Why you have equated heat given is equal to the heat capacity at constant pressure times change in temperature? For isochoric process, $\delta Q = C_v dT$. Heat is given and no work is done by the system (I.e. the constant volume) $\endgroup$
    – user240696
    Jan 19, 2020 at 17:06
  • $\begingroup$ Yes I thought from the definition that it applies to any processes, which isn't true. Thanks for pointing out though. $\endgroup$
    – Felix
    Jan 19, 2020 at 17:13

1 Answer 1


$Q = C_pdT$ is valid when the pressure is constant and the volume is changing (see here), so it is inconsistent that you equate the $dU$ and the $Q$ of two different processes.

  • $\begingroup$ If the equation is only valid when the system's pressure is constant but the volume is changing, then why does this says that it is true in any process? $\endgroup$
    – Felix
    Jan 19, 2020 at 14:53
  • $\begingroup$ sorry it was a typo, I meant $c_p$, I corrected it $\endgroup$
    – user65081
    Jan 19, 2020 at 15:03
  • $\begingroup$ Ahh i see. So going back to the question, does that mean the change of pressure doesn't mean in any way that work is done? If so then what is with this isentropic process where work is defined as $VdP$? $\endgroup$
    – Felix
    Jan 19, 2020 at 16:32
  • $\begingroup$ I never seen work defined that way $\endgroup$
    – user65081
    Jan 19, 2020 at 16:48
  • 2
    $\begingroup$ An isentropic process where work is defined as VdP is for steady flow in an open system (not a closed system), and the VdP is only the shaft work portion of the work, not the total work. The work on each material parcel of fluid passing through the system is still PdV. $\endgroup$ Jan 19, 2020 at 17:12

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