Given an arbitry small (infinitesimal) step in a $P,V$-diagram as shown below, how would you calculate the change in all the thermodynamic variables and the resultant work and heat?

enter image description here

I.e. how do all the variables $P_0, V_0, T_0$ change? My first guess would be the following:

$$ P_1 = P_0 + dP$$ $$ V_1 = V_0 + dV $$ $$ T_0 = \frac{P_0 V_0 }{nR}$$ $$ T_1 = \frac{P_1 V_1 }{nR} = \frac{(P_0+dP)(V_0+dV) }{nR} = \frac{1}{nR} (P_0 V_0 + P_0 dV + V_0 dP + dVdP) \ \ \ (1) $$

$$ T_1 = T_0 + W(\text{?})+V_0dP $$

Where $P_0 dV = W$ and $dVdP = 0$. Does this all make any sense? Or should I be using something like:

$$ T_1 = T_0 + dT $$ $$ dT = \frac{\partial T}{\partial P}|_V dP + \frac{\partial T}{\partial V}|_P dV $$ $$ T_1 = T_0 + dT = T_0 + \frac{\partial T}{\partial P}|_V dP + \frac{\partial T}{\partial V}|_P dV $$

$$ \frac{\partial T}{\partial P}|_V = \frac{V}{nR}, \frac{\partial T}{\partial V}|_P = \frac{P}{nR} $$

$$ T_1 = T_0 + \frac{V}{nR} dP + \frac{P}{nR} dV \ \ \ (2) $$

Equation $(2)$ seems similar to $(1)$ however $(1)$ is keeping volume/pressure constant even though the premise is that they both vary like shown in the graph. Are they equivalent or is there important (conceptual) difference between the approaches?

And how would you calculate $Q$ and $W$ for such a step?

How can you expand this to all thermodynamic variables? In other words, how does the following work for an infinitesimal step in the $P,V$-diagram? $$ (P_0, V_0, T_0, U_0, S_0, n_0, ... ) \rightarrow (P_1, V_1, T_1, U_1, S_1, n_1, W, Q ...)$$

I have the feeling that I am confusing different stuff here.

Could you derive anything that makes it easy to compute the thermodanimic quantities and the resultant Work $W$ and heat $Q$ produced along a parametric curve that is divided up in the small steps as shown in the graph?

  • $\begingroup$ Nice, what app did you ues to draw? $\endgroup$ Commented May 5, 2022 at 11:45
  • $\begingroup$ Is the change carried out reversibly, or is this part of an irreversible process. Are dP and dV infinitesimal, or are they finite changes $\Delta P$ and $\Delta V$? $\endgroup$ Commented May 5, 2022 at 11:45
  • $\begingroup$ @Buraian its actually just Word $\endgroup$ Commented May 5, 2022 at 15:48
  • $\begingroup$ @Chet Miller it is infinitesimal $\endgroup$ Commented May 5, 2022 at 15:48

1 Answer 1


There's not a single answer. You can move in any direction from point 0 to point 1, and depending on the slope of that line you will have different relationships between $P$ and $V$, and between all the other properties. That's why you typically specify a process as "isobaric" $dP=0$, isothermal $dT=0$, isentropic $dS=0$... etc. In the isentropic case, you can then use the ideal gas law to derive the relationship $P_0V_0^k=P_1V_1^k$. In contrast, an isothermal process (for an ideal case) will yield $$PV=nRT=const$$ Thus $$P_0V_0=P_1V_1$$

So I'm not sure your attempt to derive a completely generic case will end up with anything meaningful.


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