Consider an absolutely incompressible body, which does not change its volume $V$ under pressure $p$ but can expand or shrink when its temperature $T$ changes. Then its equation of state has a simple form $$ T=T(V). $$

Then imagine we are performing the Brayton cycle with this body consisting of two isobaric processes at the pressures $p_1$ and $p_2$, and two adiabatic processes. The latter two processes are also isothermal and isochoric since the body is incompressible.

The efficiency of this cycle is $\eta=A/Q_1$, where $A=(p_1-p_2)(V_1-V_2)$ is the performed work and $Q_1$ is the heat received from the hot reservoir. To calculate $Q_1$, we can use the first law of thermodynamics: $\delta Q=c_VdT+pdV$. The Maxwell's relations imply the infinite heat capacity ratio $\gamma=c_p/c_V\rightarrow\infty$ for incompressible body then $c_V=0$ (it can be also understood considering that $T$ and hence the internal energy do not change at $V=\mathrm{const}$).

So $\delta Q=pdV$, and the heat, which is revieved by the system at the top isobare, is $Q_1=p_1(V_1-V_2)$. The efficiency becomes $$ \eta=1-\frac{p_2}{p_1}. $$

From the other side, the Carnot efficiency $\eta_\mathrm{Carnot}$ depends on the ratio of the highest $T_1=T(V_1)$ to the lowest $T_2=T(V_2)$ temperatures during the cycle, and $$ \eta_\mathrm{Carnot}=1-\frac{T(V_2)}{T(V_1)}. $$ Taking the pressure difference $p_1-p_2$ large enough while keeping $V_1-V_2$ constant, we can achieve $\eta>\eta_\mathrm{Carnot}$.

Why the Carnot's theorem is violated in this case? Does it mean that the Carnot's theorem prohibits existence of absolutely incompressible materials? Or something in my calculations is wrong?


Indeed, as noted in the comments, the Carnot's theorem is applicable when only two heat reservoirs with fixed temperatures are present. However, it can be easily generalized to the case of variable temperatures. If $T_1$ is the maximum temperature achieved by the reservoir which transfers heat to the system and $T_2$ is the minimum temperature of the second reservoir which absorbs heat then the efficiency of any cycle cannot exceed $$ \eta_\mathrm{max}=1-\frac{T_2}{T_1} $$ (see, e.g., R. Kubo, Thermodynamics (1968), problem 4 on page 85 and its solution on pp. 97-98).

Moreover, I have found even stranger thing when applied the proof of this relation to incompressible body. Assume "a" and "b" are, respectively, the top and bottom isobares where the system absorbs and gives off amounts of heat $Q_1$ and $Q_2$. Then, according to the Clausius' inequality, $$ \int\limits_a\frac{\delta Q_1}{T}\leqslant\int\limits_b\frac{\delta Q_2}{T}. $$ As stated above, for incompressible body $\delta Q=pdV$ then $$ \int\limits_a\frac{\delta Q_1}{T}=\int\limits_{V_2}^{V_1}\frac{p_1dV}{T(V)}, \qquad\int\limits_b\frac{\delta Q_2}{T}=\int\limits_{V_2}^{V_1}\frac{p_2dV}{T(V)}. $$ Canceling $\int_{V_2}^{V_1}dV/T(V)$ on both sides of the inequality, we get $$ p_1\leqslant p_2 $$ in contradiction with the initial assumption $p_1>p_2$ about the shape of the cycle.

So, now I suppose that even the Clausius' inequality prohibits existence of incompressible bodies. Is this true and how it can be understood?


I tried to state the problem anew. Assume the system at given state variables $T$ and $p$, it is characterized by the Gibbs potential $G$ with $$ dG=-SdT+Vdp,\qquad\left(\frac{\partial G}{\partial p}\right)_T=V,\qquad\left(\frac{\partial G}{\partial T}\right)_p=-S, $$ so incompressibility $\partial V/\partial p=0$ means $V$ depends only on $T$: $(\partial G/\partial p)_T=V(T)$. Integrating this relation, we get $$ G=pV(T)+A(T), $$ where $A(T)$ is some unknown function. Accordingly, we get the entropy $$ S=-pV'(T)-A'(T) $$ and the internal energy $U=G+TS-pV$: $$ U=-pTV'(T)-TA'(T)+A(T). $$

However, it this case the calculations of the cycle efficiency should be reconsidered because $$ \delta Q=TdS=-TV'dp-(pTV''+TA'')dT $$ implies heat exchange during isothermal/isochoric processes! Thus my assumption about adiabatic nature of isochoric processes is incorrect.

For example, assuming $V=\alpha T$, $A=0$, we get $\Delta Q=0$ at the both isobares, $\Delta Q=(p_1-p_2)T_1\alpha=Q_1$ at the right isochore and $\Delta Q=-(p_1-p_2)T_2\alpha=-Q_2$ at the left isochore. Thus the efficiency $$ \eta=1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1} $$ matches the Carnot's limit.

So what is wrong in this case? I see that some results seem strange (e.g., dependence of $U$ on $p$ at $T=\mathrm{const}$), but are there any fundamental problems (with stability, existence of thermodynamic equilibrium etc.)?

  • $\begingroup$ The temperature is not constant during isobaric processes during which heat exchange occurs. So how are you going to compare it with Carnot cycle which strictly works between two heat reservoirs? $\endgroup$
    – Deep
    Commented Apr 26, 2017 at 5:30
  • $\begingroup$ @Deep See the update. $\endgroup$ Commented May 2, 2017 at 15:40
  • $\begingroup$ Are you completely sure that $C_V = 0$? Without having done the math, I feel like $C_V$ should be infinite instead. $\endgroup$
    – knzhou
    Commented May 2, 2017 at 18:39
  • $\begingroup$ @knzhou I'm not completely sure, but the Maxwell's relations state that incompressibility implies $c_p/c_V=\infty$, so either $c_p=\infty$ of $c_V=0$. In other words, incompressibility implies vertical adiabats, which means infinitely large $\gamma=c_p/c_V$. $\endgroup$ Commented May 2, 2017 at 19:25

2 Answers 2


Perfectly incompressible bodies are actually prohibited by stability of thermodynamic equilibrium (entropy maximization). Section 21 of Landau & Lifshitz Vol 5: Statistical Physics has a nice discussion and derivation of the general thermodynamic inequality $C_V > 0$.

So the equation of state $T = T(V)$ is pathological, and lets you derive all sorts of silly results:

Assume that $T = T(V)$, and imagine going from $(P_1,V_1) \to (P_1,V_2)$ in two different ways. On one hand, you can just go from $V_1 \to V_2$ at fixed $P = P_1$. The change in internal energy is $\Delta U = Q_1 + P_1 (V_1 - V_2)$ by the first law. On the other hand, you can also go along the path $(P_1, V_1) \to (0,V_1) \to (0,V_2) \to (P_1,V_2)$. The first and third legs must have $\Delta U = 0$ because nothing depends on $P$. The second leg will have $\Delta U = Q_0$. If internal energy is a state variable, then these should be equal: $$ Q_1 + P_1(V_1 - V_2) = Q_0 $$ Here $Q_1$ is the amount of heat needed to change the temperature of the body from $T(V_1)$ to $T(V_2)$ at $P = P_1$, and $Q_0$ is the amount of heat needed to change the temperature of the body from $T(V_1)$ to $T(V_2)$ at $P = 0$. The equation above says that these must differ. But this contradicts the original assumption that the equation of state $T = T(V)$ is independent of $P$!

  • $\begingroup$ Thanks for the reference about $c_V>0$, I need to think about it, but the derivation does not seem to me very clear... Also the the second update to my question, I think the calculations of $\delta Q$ in isochoric processes need to be reconsidered. $\endgroup$ Commented May 5, 2017 at 23:01

This is what I think: A true thermodynamic variable must be a function of two others. Therefore $T=T(V)$ is not a thermodynamic variable. However we have encountered such a pathology before: for ideal gas we have $U=U(T)$. In your very first equation you are piling both of these pathologies together to get $\delta Q=dU+p~dV=C_vdT+p~dV$. I suspect nothing good will come out of this! In this equation $T$ and $V$ are not independent variables. But in writing down $\delta Q=p~dV$ you have assumed that they are independent. More correctly: $\delta Q=\left( C_v\frac{dT}{dV}+p \right)dV$.

If you are not convinced, the same equation may be derived more formally, by avoiding writing in terms of $\delta Q$. Say the fundamental equation of the system is given by $S=f(U,V)$ (see Thermodynamics by Callen). But since $U=U(T(V))$, $S$ really depends on only one variable, $V$. Now: \begin{align} dS & =\frac{\partial f}{\partial U}dU+\frac{\partial f}{\partial V}dV \\ & = \left[ \frac{\partial f}{\partial U}\frac{dU}{dT}\frac{dT}{dV}+\frac{\partial f}{\partial V} \right] dV \\ & = \left[ \frac{C_v}{T}\frac{dT}{dV}+\frac{p}{T} \right] dV \end{align} which is the same equation as before (when multiplied by $T$ throughout).

Anyway, you will be happy to know that when Clausius inequality is applied, the absurd conclusion $p_1\leq p_2$ results nevertheless. Therefore thermodynamics seems to prohibit a body with a fundamental equation that depends on a single extensive thermodynamic variable such as $V$. Two independent extensive thermodynamic variables are absolutely required (such as $U,V$).

P.S. I am not sure why you keep calling the body in your question, characterized by $T=T(V)$, an incompressible body which actually requires $\frac{\partial V}{\partial p}=0$.

  • $\begingroup$ Thanks! Although I do not understand the point about necessary existence of at least two extensive thermodynamic variables. I always though that even one state variable $T$ (microscopic system, no volume and pressure, no external fields) is sufficient to build consistent, although very simple thermodynamics. Also see the second update to the question... $\endgroup$ Commented May 5, 2017 at 23:07

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