The formula for the work done by adiabatic expansion of gas is

$$W = C_V\,\mathrm dT.$$

Why the volumetric heat capacity $C_V$ is constant and not the heat capacity at constant pressure $C_p?$ If the system is at constant volume, it is a isovolumetric process and hence no work can be done in that case.

  • $\begingroup$ Probably better to write $dW = C_V dT$, since it is a differential amount of work. $\endgroup$
    – Buck Thorn
    Commented Feb 21, 2021 at 10:37

3 Answers 3


Pretty simple, $\Delta U = nC_V\Delta T$ no matter what the process is. This is called a state function, i.e. it is independent of the path taken and only depends on the final and initial conditions here initial and final temperature.

However, work done is path dependent, so it depends on how you change pressure and volume. Let's say you take a gas from $p_1,$ $V_1,$ $T_1$ to $p_2,$ $V_2,$ $T_2.$ You can say $\Delta U = nC_V(T_2 - T_1),$ however we need to know the process (isothermal, adoabetic, polytropic) for finding work done.

  • 2
    $\begingroup$ One caveat... This only applies to an ideal gas. In a real gas, the internal energy is also a function of volume. $\endgroup$ Commented Oct 15, 2017 at 12:08

Part 1. Let's consider first law of thermodynamics.

$\delta Q = dU + \delta W$ (classical sign convention)


$\delta Q = dU - \delta W$ (IUPAC sign convention)

In the case of adiabatic process, $\delta Q = 0$.

$0 = dU - \delta W$

$dU = \delta W$

$dU = C_VdT$ (by definition)

$C_VdT = \delta W$

After integration, assuming CV is constant.

$C_V\Delta T = W$

Part 2. Why not Cp?

By definition, $\delta Q_p = dH = C_pdT$. Obviously, we don't have an isobaric process here.

P. S. Yes, those sign convention differences also make me feel bad.


The total differential of the internal energy (at constant composition and allowing only pV work) is usually written as

$$dU = -pdV + TdS$$

However the differential $dU$ can be written in terms of other variables, for instance p and T, in which case:

$$\begin{align} dU &= \left(\frac{\partial U}{\partial V} \right)_T dV + \left(\frac{\partial U}{\partial T} \right)_V dT \\ &= \left[-p + T \left( \frac{\partial S}{\partial V} \right)_T \right] dV + C_V dT \\ &= \left[-p + T \left( \frac{\partial P}{\partial T} \right)_V \right] dV + C_V dT \end{align}$$

The last equality follows from one of Maxwell's relations. This expression btw is known as the thermodynamic equation of state. For an ideal gas it follows that

$$dU = C_V dT $$

Note that the change in energy is independent of changes in pressure and volume, it depends only on changes in T.

If the heat capacity is constant over the temperature change of interest then you may write

$$ \Delta U = C_V \Delta T $$


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