From my past studies I have had the presumption that for an ideal gas the equation:

$du=c_v dT$

( where $du$ is the density of internal energy, $c_v$ is the specific heat capacity at constant volume and $T$ is temperature)

is only valid when the process is isentropic. However searching the internet I'm doubting if this is really true. For example the wikipedia entry for the subject and this MIT blogpost, mention the equation without any assumptions such as being reversible, adiabatic or isentropic. So my question is if the equation above is valid all the time or just unders certain assumptions? if later then what are those assumptions and what is the general form? and if former is the case how we can prove that?


3 Answers 3


I assume by the "density of internal energy" $u$ you mean

$$u = \frac{E_{therm}}{n},$$

where $E_{therm}$ is the thermal energy of the gas, and $n$ is the number of moles.

In this case your equation is true for any ideal gas process. Here is how we can show that this is true.

We begin by considering an isochoric process (V=const). By definition, the amount of heat $Q$ needed to increase the temperature of gas by $\Delta T$ is

$$Q = n C_v \Delta T\tag{1},$$

where $C_v$ is the molar specific heat at constant volume. Since there is no change in volume, the work done is zero. Therefore, from the 1st law of thermodynamics:

$$\Delta E_{therm} = Q + 0 = Q.$$

Combining with Equation (1) we get:

$$\Delta E_{therm} = n C_v \Delta T. \tag{2}$$

Next we apply a key idea from thermodynamics: the change in internal energy $E_{therm}$ of gas is the same for any two processes that results in the same change in temperature $\Delta T$. Therefore, Equation (2) is true for any ideal gas process, and not just the isochoric process.

If we divide both sides by $n$, we get

$$\Delta u = C_v \Delta T,$$

which is what we needed to show.

  • $\begingroup$ there is also another law for the ideal gas stating that the state of the gas can be determined only given two of the independent variables? any idea what is it called? I think that should also be taken into the account. $\endgroup$
    – Foad
    Feb 5, 2018 at 15:42
  • $\begingroup$ I'm not familiar with this law, but it sounds right because pV=nRT has three variables. Therefore, the third one is always determined by the choice of two others. I don't know how this can be incorporated into my proof though. Do you mean it can make it shorter? $\endgroup$
    – Evgenii
    Feb 5, 2018 at 21:21

You're right. $\text{d}U=n\ c_v \text{d}T$ is independent of process: it is the equation linking change in internal energy, U, with change in temperature, T, for a system. But, secondarily, $c_v$ is also the molar heat capacity at constant volume, because at constant volume, no work is done, so the heat inflow is equal to the gain in internal energy! So the notation $c_v$ for the constant in the general equation $\text{d}U=n\ c_v \text{d}T$ is derived from its restricted role in constant volume heating. This causes endless confusion!

You'll have noted that I didn't use your form of the equation. That's because I don't understand your use of 'energy density'. I think the form I've used is more standard (especially when working with gases). n is the number of moles.

Edit: For completeness I should add that, except for ideal gases, U is usually a function not just of temperature but of one more variable for a fluid, and maybe more than one other for other systems. So, rather than $\text{d}U=n\ c_v \text{d}T$, we should really write: $n\ c_v =\left(\frac{\partial U}{\partial T}\right)_{\text{other independent variables}}$!

  • $\begingroup$ thanks for the post. 1. You are right my form is not standard, but I'm trying to write the equations in an Eulerian form so later I can use them in conjunction with my NAvier-Stokes equations. 2. from definition we have $dU=dQ+dW$ where $dW=P.dV$ is the work done by the system. and for an ideal gas we have $PV=nRT$. and second law of entropy is $dS>\frac{dQ}{T}$. $\implies dU=dQ+nR.dT+V.dP$. how we can prove the equation above from these assumptions or am I missing any other equations here? $\endgroup$
    – Foad
    Feb 5, 2018 at 12:14
  • $\begingroup$ Your original question doesn't need entropy or the Second Law. dU=dQ+dW (i.e. the First Law) is all that's needed. $\endgroup$ Feb 5, 2018 at 12:24
  • $\begingroup$ I think there must be another assumption place. for example the assumption that in $dQ=C.dT$ the heat capacity is constant when P or v are constant. $\endgroup$
    – Foad
    Feb 5, 2018 at 12:43
  • $\begingroup$ I don't think we're assuming constancy of $c_v$. It is, in general, a function of temperature. $\endgroup$ Feb 5, 2018 at 12:52
  • $\begingroup$ I think that's the case. check this $\endgroup$
    – Foad
    Feb 5, 2018 at 12:55

The answer is simpler than implied by the discussion. It is just that the pressure of an ideal gas is proportional to T at fixed V, N. So T dP/dT = P. Now (dU/dV)_T = [T (dS/dV)_T - P] dV and (dS/dV)_T = (dP/dT)_V is a Maxwell relation. Put it together and U(T,V) for an ideal gas (or any system where P = NRT*f(V)) is independent of V.

One other point. P, V, N do not uniquely define the thermodynamic state. A counter example is water near freezing. At 1 bar pressure (for example) there is a minimum in the volume around 4C. That means V(T) is uniquely specified whereas T(V) has two solutions. So T,P,N is unique but P,V,N is not. Think of y=x^2 if its clearer: x specifies y, but y does not specify x (can be + or -)


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