In his paper clausius describes a Carnot cycle and he continue arguing like below:

In order to make an analytical application of the method...we will assume that all the changes which the gas undergoes are infinitely small. We may then treat the curves as straight lines as they are represented in the accompanying figure. (...) the magnitude $dk$ is the increase on the pressure, while the gas at the constant volume $of$ has its temperature raised from $r$ to $t$ that is by the differential $t-r=dt$. (...) the quantity of heat which must be communicated to a gas while it brought from any former condition in a definite way to that condition in which its volume $=v$ and its temperature its $=t$, may be called $Q$ and the changes of volume in the above process [the figures], which must here be considered seperately may be represented as follows: $ef$ by $dv$,$hg$ by $d'v$, $eh$ by $δv$, $fg$ by $δ'v$.During expansion from volume $v$ to $v+dv$ at constant temperature $t$ the gas must recieve the quantity of heat: $$\frac{dQ}{dv}dv$$

Correspondingly, during expansion from $v+δv$ to $v+δv+d'v$ at temperature $t-dt$ the quantity of heat :


My question is about the last equation. How it is derived?

the paper can be found here:https://archive.org/details/secondlawthermo01kelvgoog/page/n86
(page 76)

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(a) I think your first round bracket is misplaced. It should be immediately before $\frac{d}{dv} \frac{dQ}{dv}$

(a) The stuff in the square brackets is a first order Taylor expansion of $\frac{dQ}{dv}$ regarded as a function of $v$ and $t$, in order to find the appropriate value of $\frac{dQ}{dv}$ for the leg 'de', in terms of the value for the leg 'ab'.

  • $\begingroup$ Could you please explain more I can't see how the terms δv and dt come out through a taylor series of two variables... $\endgroup$ – Vaggelis Kyrilas Apr 27 '19 at 11:06
  • 1
    $\begingroup$ Put $\frac{dQ}{dv}=X.$ Then $$X(v+\Delta v, t+\Delta t) = X(v, t) + \frac{dX}{dv} \Delta v + \frac{dX}{dt} \Delta t.$$Like much of his notation, Clausius's sign convention for temperature change is odd. $\endgroup$ – Philip Wood Apr 27 '19 at 14:11
  • $\begingroup$ Please have a look in a question about the same given paper. physics.stackexchange.com/questions/480325/… $\endgroup$ – Vaggelis Kyrilas May 17 '19 at 12:54

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