Recall $Q = dU + W$ (First Thermodynamic law, energy conservation).
If $Q=0$ (for adiabatic process, either adiabatic expansion or contraction), and supposing $W = pdV$, then $0 = dU + pdV$.

Either by definition, or the thermodynamic identity, $\tau d\sigma = dU + pdV$, then $C_V := \left( \frac{ \partial U}{ \partial \tau} \right)_V$. My question is this: for a manifold of thermodynamic states $M$, $M=(U,V)$, i.e. $U$ is a global coordinate, being the internal energy of the gas system, and $V$ is a ``local'' coordinate. One can make a Legendre transformation such that $M$ is parametrized by $(\tau,V)$, where $\tau$ is the temperature. In general, one should say that $U=U(\tau,V) \in C^{\infty}(M)$, and so $dU = \frac{ \partial U }{ \partial \tau} d\tau + \frac{ \partial U}{ \partial V} dV$, $dU \in \Omega^1(M)$.

However, for this adiabatic process, we want

$Q = 0 = dU + W = dU + pdV = C_V d \tau + pdV$

which implies that $dU = C_V d\tau$. What happened to the $\frac{ \partial U}{ \partial V} dV$? Is it that in this adiabatic process, the internal energy of the gas system goes to either doing work (expansion) or increases due to work being done on it (contraction), and is characterized completely by a drop or increase in its temperature, respectively? And so $dU = C_V \tau$, and $pdV$ completely describes what's going on with work done or work done on it?

Nevertheless, using the (empirical) ideal gas law, $pV = N\tau$,

$0 = C_V d\tau + pdV = C_V d\tau + \frac{N\tau}{V} dV$

Consider a path $\begin{aligned} & \gamma : \mathbb{R} \to M \\ & \gamma(t) = (\tau(t), V(t)) \end{aligned}$

in $M$, so that $\dot{\gamma}(t) = \dot{\tau} \frac{ \partial }{ \partial \tau } + \dot{V} \frac{ \partial }{ \partial V} \in \mathfrak{X}(M)$.


\begin{gathered} 0 = C_v \dot{\tau} + \frac{N \tau}{V} \dot{V} \text{ so } \frac{ \dot{\tau}}{\tau} + \frac{N}{C_V} \frac{ \dot{V}}{V} \\ \xrightarrow{ \int dt } \ln{ \frac{ \tau_f}{ \tau_i } } + \frac{N}{C_V} \ln{ \frac{V_f}{ V_i } } = 0 \text{ or } \ln{\tau V^{\frac{N}{C_V}} } = \text{ const. } \end{gathered}


$\frac{N}{C_V} = \frac{C_P - C_V}{ C_V} = \gamma -1 $

which is true, assuming the (empirical) ideal gas law, the thermodynamic identity, and, surely, for the case of a monatomic gas.


$\tau_f V_f^{\gamma-1} = \tau_i V_i^{\gamma -1}$

Summary: This should also clarify, for undergraduate level thermodynamics, this question: when is it that the change in internal energy, $dU$ (or if you like, $\Delta E$ or $\Delta U$), only changes with temperature, and so one can employ the heat capacity that keeps volume constant $C_V$, in that $dU = C_V d\tau$, knowing that $C_V := \left( \frac{ \partial U}{ \partial \tau} \right)_V$ (by definition)?


  • Ralph Baierlein Thermal Physics
  • Kittel and Kroemer Thermal Physics
  • Bernard Schutz Geometrical Methods of Mathematical Physics Chapter 5 Section A Thermodynamics (very different perspective on thermodynamics from a differential geometry point of view)

1 Answer 1


$dU = C_v dT$ is the case for ideal gases (its actually called Joule's law).

Obviously it's not the case for general substances. A dependency on the total volume can be thought of position constraints between different parts of the substance, which for example appear when the energy has a non-zero potential term. As you may recall, ideal gases are not victims of such devices.

Are there OTHER substances with the property $dU = C_v dT$?? Obviously there is a family of them given by letting $C_v$ take different values... in real life one could expect them to be all sort of complex molecules in a gaseous state at sufficiently high temperature (kinetic term much bigger than potential one). Two issues with this:

  1. Relativistic Thermodynamics should be better suited for this... for example relativistic energy ain't the sum of two terms
  2. The "gaseous at high temperature" non-monoatomic substances will probably become unstable before being a good "ideal gas"

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