For an ideal gas, we have

$$C_p - C_V = nR$$

$C_p$ is heat capacity at constant pressure,
$C_V$ is heat capacity at constant volume,
$n$ is amount of substance, and
$R=N_\mathrm A\cdot k_\mathrm B=8.314\,462\,618\,153\,24\ \mathrm{J\ mol^{-1}\ K^{-1}}$[source] is the molar gas constant.

How can I prove this?


3 Answers 3



Consider $U = U(V,T, p)$. However, assuming that it is possible to write an equation of state of the form $p = f(V,T)$, I don't have to explicitly address the $p$ dependence of $U$, and I can write the following differential:

$$\mathrm{d}U = \underbrace{\left ( \frac{\partial U}{\partial V} \right)_T}_{\pi_T} \mathrm{d}V + \underbrace{\left ( \frac{\partial U}{\partial T} \right)_V}_{C_v} \mathrm{d}T \tag{1}$$

so one writes,

$$ \mathrm{d}U = \pi_T \mathrm{d}V + C_v\mathrm{d}T \tag{2}$$

also, for an ideal gas, the internal pressure ($\pi_T$) $= 0$

Additionally for changes in internal energy at constant pressure,

$$ \left (\frac{\partial U}{\partial T}\right)_p = \pi_T \left (\frac{\partial V}{\partial T}\right)_p +C_v \tag{3}$$

Here, I take the opportunity to define two new quantities, namely $\alpha$ and $\kappa_T$ (the expansion coefficient and the isothermal compressibility respectively)

$$\alpha = \frac{1}{V}\left (\frac{\partial V}{\partial T}\right)_p$$

and, $$\kappa_T = \frac{-1}{V}\left (\frac{\partial V}{\partial p}\right)_T$$

Rewriting (3) using $\alpha$

$$ \left (\frac{\partial U}{\partial T}\right)_p = \alpha \pi_T V +C_v \tag{4} $$

Now, although the constant volume heat-capacity is defined as $C_v = \left (\frac{\partial U}{\partial T} \right)_v$, if we let $\pi_T = 0$ in equation (4), we get $C_v = \left (\frac{\partial U}{\partial T} \right)_p$

This holds true for a perfect gas, and one can quickly obtain the desired relation at this stage.

Derivation: Difference between constant volume and constant pressure heat capacities for a perfect gas

Consider, $$C_p - C_v = \overbrace{\left (\frac{\partial H}{\partial T} \right)_p}^{\text{definition of} \ C_p} - \overbrace{\left (\frac{\partial U}{\partial T} \right)_v}^{\text{definition of} \ C_v} \tag{5}$$

Introducing, $H = U+ pV = U+nRT$, and exploiting $C_v = \left (\frac{\partial U}{\partial T} \right)_v = \left (\frac{\partial U}{\partial T} \right)_p$, equation (5) yields:

$$ C_p - C_v = \left (\frac{\partial (U+nRT)}{\partial T} \right )_p - \left (\frac{\partial U}{\partial T} \right )_p = nR$$

Derivation: Difference between constant volume and constant pressure heat capacities (general case)

However, as my contribution to this discussion I would like to derive a relation between heat capacities that is universally true for any substance, not just a perfect gas. So let's return to equation (5):

$$C_p - C_v = \overbrace{\left (\frac{\partial H}{\partial T} \right)_p}^{\text{definition of} \ C_p} - \overbrace{\left (\frac{\partial U}{\partial T} \right)_v}^{\text{definition of} \ C_v} \tag{5}$$

Here, we substitute $H = U + pV $ and obtain,

$$ C_p -C_v = \overbrace{\left( \frac{\partial U}{\partial T} \right)_p}^{\text{evaluated in} \ (4)}+ \left( \frac{\partial (pV)}{\partial T} \right)_p - C_v \tag{6}$$

The first partial derivative was already taken care of in equation (4). For the second one, since the derivative is to be evaluated at constant pressure, we can do the following

$$\left( \frac{\partial (pV)}{\partial T} \right)_p = p \overbrace{\left( \frac{\partial V}{\partial T} \right)_p}^{\alpha V}$$

Putting all of this together, one obtains

$$C_p -C_v = \alpha \pi_T V + \alpha pV = \alpha(p+ \pi_T)V \tag{7}$$

At this stage, I will make use of the following relation (derived in additional comments) $$\pi_T = T \left (\frac{\partial p}{\partial T}\right )_v - p$$ After substituting this into (7) we get:

$$ C_p -C_v = \alpha T V \left( \frac{\partial p}{\partial T} \right)_V \tag{8}$$

I wish to transform the last remaining partial derivative, and to do so I consider $ V = V(T,p) $ which yields the following differential

$$ \mathrm{d}V = \left( \frac{\partial V}{\partial T} \right)_p \mathrm{d}T + \left( \frac{\partial V}{\partial p} \right)_T \mathrm{d}p $$

At constant volume, $\mathrm{d}V = 0$ so one gets, $$\left( \frac{\partial V}{\partial T} \right)_p \mathrm{d}T = - \left( \frac{\partial V}{\partial p} \right)_T \mathrm{d}p$$

$$\overbrace{ \left( \frac{\partial V}{\partial T} \right)_p}^{\alpha V} = \overbrace{-\left( \frac{\partial V}{\partial p} \right)_T}^{\kappa_T V} \left( \frac{\partial p}{\partial T} \right)_V$$

Note: One can avoid all of this work, and simply invoke the Euler Chain Rule

Rearranging, $$\left( \frac{\partial p}{\partial T} \right)_V = \frac{\alpha}{\kappa_T}$$

We can finally substitute this into (8) to get

$$C_p -C_v = \frac{\alpha^2 TV}{\kappa_T} \tag{9}$$

This is true for any substance, not just a perfect gas. Now, for a perfect gas $ pV = nRT$ holds true, and thus $\alpha = \frac{1}{T}$ and $\kappa_T = \frac{1}{p}$. Making these substitutions into (9), we get our desired result

$$C_p -C_v = nR $$

Additional Comments

  • This might seem like an unnecessarily complex, and not to mention convoluted way to get to the desired result (especially, in light of a much simpler method presented by @orthocresol), however, I think the deriving the expression for a general case first, and then reducing it to the special case is illuminating. Moreover, in spirit and approach, it is not that far from what @orthocresol did.

  • Physical Significance of certain terms/quantities

    • $\pi_T$ is called the internal pressure (it has the dimensions of pressure) and is a consequence of the interactions between molecules. For an ideal gas it is necessarily zero.
    • $\alpha$ i.e the expansion coefficient is the fractional change in volume that accompanies a rise in temperature. A large volume of $\alpha$ implies that the sample responds very strongly to changes in temperature.

    • Similarly, $\kappa_T$ is a measures of the response to a change in pressure. The negative sign insures that $\kappa_T$ is a positive quantity, because a pressure increase causes a decrease in volume ($\mathrm{d}V$ is negative)

    • Since equation (9) holds true for any substance, for solids and liquids one might be tempted to say $C_p \approxeq C_V$ because $\alpha$ is small for solids and liquids. However, one must be careful because $\kappa_T$ can be small as well, which makes the fraction $\frac{\alpha^2}{\kappa_T}$ large. In other words, even though a little work has to be done to push back the atmosphere when a solid expands, a great deal of work will go into pulling the atoms apart.
  • Supplementary Derivation: For a system where, $N$ doesn't change the fundamental equation of thermodynamics is:

$$\mathrm{d}U = T\mathrm{d}S -p\mathrm{d}V$$

This seems to suggest that, $U = U(S,V) $. Thus, one can write the following differential and after comparing to the one above can equate $T$ and $-p$ (as indicated by annotations) with the partial derivatives given below:

$$\mathrm{d}U = \underbrace{\left ( \frac{\partial U}{\partial S}\right)_V}_{T} \mathrm{d}S + \underbrace{\left ( \frac{\partial U}{\partial V}\right)_S}_{-p}\mathrm{d}V$$

Moreover, dividing both sides of the fundamental equation by $\mathrm{d}V$ (yeah, I know), and imposing constraint of constant temperature) we can manipulate it into the following form:

$$\overbrace{\left( \frac{\partial U}{\partial V} \right)_T}^{\pi_T} = \overbrace{\left ( \frac{\partial U}{\partial S}\right)_V}^{T}\left ( \frac{\partial S}{\partial V}\right)_T - \overbrace{\left ( \frac{\partial U}{\partial V}\right)_S}^{-p}$$

Thus, we have $$ \pi_T = T\left ( \frac{\partial S}{\partial V}\right)_T - p$$

Invoking, the Maxwell Relation

$$ \left ( \frac{\partial S}{\partial V}\right)_T = \left ( \frac{\partial p}{\partial T}\right)_V$$

one finally gets, $$\pi_T = T \left (\frac{\partial p}{\partial T}\right )_v - p$$


The heat capacities are defined as

$$C_p = \left(\frac{\partial H}{\partial T}\right)_{\!p} \qquad \qquad C_V = \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{1}$$

and since $H = U + pV$, we have

$$\begin{align} C_p - C_V &= \left(\frac{\partial H}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{2} \\ &= \left(\frac{\partial U}{\partial T}\right)_{\!p} + \left(\frac{\partial (pV)}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{3} \end{align}$$

Ordinarily, we have $\mathrm{d}(pV) = p\,\mathrm{d}V + V\,\mathrm{d}p$, but under conditions of constant pressure $\mathrm{d}p = 0$ and so we can write

$$C_p - C_V = \left(\frac{\partial U}{\partial T}\right)_{\!p} + p\left(\frac{\partial V}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{4}$$

For an ideal gas, $V = nRT/p$ ($n$ being constant throughout this whole discussion - or else $nR$ in the original equation makes no sense!) and so

$$\left(\frac{\partial V}{\partial T}\right)_{\!p} = \frac{nR}{p} \tag{5}$$

Substituting $(5)$ into $(4)$ gives

$$C_p - C_V = \left(\frac{\partial U}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} + nR \tag{6}$$

At this stage, you have two choices.

The easy way

From the equipartition theorem, the internal energy of an ideal gas is given by

$$U = \left(\frac{\text{Degrees of freedom}}{2}\right)nRT \tag{7}$$

The number of degrees of freedom doesn't matter, because it is a constant for any given gas. The important point is that $U = U(T)$, i.e. internal energy is a function of only temperature. Therefore,

$$\left(\frac{\partial U}{\partial T}\right)_{\!p} = \left(\frac{\partial U}{\partial T}\right)_{\!V} = \frac{\mathrm{d}U}{\mathrm{d}T} \tag{8}$$

and the desired result, $C_p - C_V = nR$, immediately follows.

The "easy way" is described in a large number of textbooks, or web pages. I was even told to use it in one of my tutorials, at Oxford no less. It is not incorrect. However, I don't like the easy way, because it invokes the equipartition theorem, which (in my opinion) makes it only half a proof, unless one proves the equipartition theorem along with it.

In fact, you don't need to use equipartition to derive this result. The only thing that we need to use is $pV = nRT$. If you are not convinced: read on!

The hard way

Without invoking the equipartition theorem, we don't know that $U = U(T)$. In general, we would expect $U$ to be a function of all three variables: $U = U(V, T, p)$. However, for an ideal gas, $p$ itself is a function of $(V,T)$: $p(V,T) = nRT/V$; so, we can eliminate the $p$-dependence of $U$, since the $p$-dependence is adequately described by the $(V,T)$ dependence. The total differential for $U = U(V,T)$ is

$$\mathrm{d}U = \left(\frac{\partial U}{\partial V}\right)_{\!T}\,\mathrm{d}V + \left(\frac{\partial U}{\partial T}\right)_{\!V}\,\mathrm{d}T \tag{9}$$

Using a similar argument, we could also treat $U$ as a function of $p$ and $T$ instead. The total differential for $U = U(p,T)$ is

$$\mathrm{d}U = \left(\frac{\partial U}{\partial p}\right)_{\!T}\,\mathrm{d}p + \left(\frac{\partial U}{\partial T}\right)_{\!p}\,\mathrm{d}T \tag{10}$$

And the third total differential we need is that of $V = V(p,T)$:

$$\mathrm{d}V = \left(\frac{\partial V}{\partial p}\right)_{\!T}\,\mathrm{d}p + \left(\frac{\partial V}{\partial T}\right)_{\!p}\,\mathrm{d}T \tag{11}$$

Substituting $(11)$ into $(9)$, we have

$$\begin{align} \mathrm{d}U &= \left(\frac{\partial U}{\partial V}\right)_{\!T} \left[\left(\frac{\partial V}{\partial p}\right)_{\!T}\,\mathrm{d}p + \left(\frac{\partial V}{\partial T}\right)_{\!p}\,\mathrm{d}T \right] + \left(\frac{\partial U}{\partial T}\right)_{\!V}\,\mathrm{d}T \tag{12} \\ &= \left[\left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial p}\right)_{\!T}\right]\mathrm{d}p + \left[\left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial T}\right)_{\!p} + \left(\frac{\partial U}{\partial T}\right)_{\!V}\right]\mathrm{d}T \tag{13} \end{align}$$

Equations $(10)$ and $(13)$ have very similar forms: they both look like $\mathrm{d}U = a\,\mathrm{d}p + b\,\mathrm{d}T$. We can equate the coefficients of $\mathrm{d}T$ in both equations to get

$$\left(\frac{\partial U}{\partial T}\right)_{\!p} = \left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial T}\right)_{\!p} + \left(\frac{\partial U}{\partial T}\right)_{\!V} \tag{14}$$

and therefore, looking back to equation $(6)$ and substituting equation $(14)$ in, we see that

$$\begin{align} C_p - C_V &= \left(\frac{\partial U}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} + nR \tag{6} \\ &= \left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial T}\right)_{\!p} + nR \tag{15} \end{align}$$

The final step is that, for an ideal gas,

$$\left(\frac{\partial U}{\partial V}\right)_{\!T} = 0 \tag{16}$$

and the proof of this can be found in this question: Internal pressure of ideal gas. This yields

$$C_p - C_V = nR \tag{17}$$

as desired.

  • $\begingroup$ I'd just add that this only works for a monoatomic gas. The vibrations of the atoms adds more degrees of freedom for a polyatomic gas. This leads down a rabbit hole to a discussion about perfect gases and ideal gases. ;-) $\endgroup$
    – MaxW
    Commented Sep 24, 2016 at 6:13
  • 6
    $\begingroup$ No, it applies for all ideal gases. The exact value of $C_p$ and $C_V$ differs from monoatomic to diatomic to polyatomic, but the relation $C_p - C_V = nR$ holds true for all. $\endgroup$ Commented Sep 24, 2016 at 6:27
  • 1
    $\begingroup$ Strictly speaking the perfect (ideal) gas does not have any internal molecular nature, but is just a mass point that does not interact with others. The total energy is just its kinetic energy and there is no potential energy arising from interaction between them. What ever the gas the difference between heat capacities is caused by expansion work done at constant pressure :) $\endgroup$
    – porphyrin
    Commented Nov 29, 2016 at 9:12
  • $\begingroup$ @orthocresol Does it apply to all ideal gases or perfect gas? With ideal gas the Cp and Cv are sensitive to temperature. And I have read that Cp and Cv changes at a different rate with respect to temperature. If that is true, then the Meyer's relation will not be valid at higher temperatures. With perfect gas since there is no temperature dependence of Cp and Cv the Meyer's relation will hold at all temperature. $\endgroup$ Commented Sep 17, 2022 at 6:27
  • $\begingroup$ @AbhishekPG sorry, please ask a new question. $\endgroup$ Commented Sep 17, 2022 at 7:47

Some elegant descriptions have already been given but it is useful to consider in words what is actually happening. When a system absorbs heat q its temperature is raised and the average heat capacity between two temperature is $q/\Delta T$. As the temperature difference becomes infinitesimally small the actual heat capacity at that temperature is produced.

However, the heat needed to raise a substance by a given temperature depends on the circumstances, namely if this is done at constant volume or pressure, for if it is done at the latter, then some work has to be done against the prevailing pressure and the heat absorbed differs from the increase in internal energy by the amount of work done or $q=dU+pdV$.

If the heat absorbed is equal to the increase in internal energy then this has been done at constant volume and no work is involved, thus $$ C_V=\left ( \frac{\partial U}{\partial T}\right )_V$$ This equation is quite general and applies to any system.

At constant pressure work is done and as $q=dU+pdV$ $$ C_p = \left ( \frac{\partial U}{\partial T}\right )_p + p\left ( \frac{\partial V}{\partial T}\right )_P$$ Next we can study the effect of heating from one temperature to another at the same pressure in two parts. The first part is that the increase in energy is the same as it would be if the substance was heated from the first temperature to the second at constant volume, and the second the change as the system is brought to the original pressure at constant temperature. This is

$$ \left ( \frac{\partial U}{\partial T}\right )_p= \left ( \frac{\partial U}{\partial T}\right )_V + \left ( \frac{\partial U}{\partial V}\right )_T \left ( \frac{\partial V}{\partial T}\right )_p$$

(This is obtained by letting $dU= \left ( \frac{\partial U}{\partial T}\right )_VdT + \left ( \frac{\partial U}{\partial V}\right )_TdV$ and then differentiating by T.

The last two equations may now be combined by substituting $\left ( \partial U/\partial T\right )_p$ into $C_p$; $$C_p=C_V + \left( p+ \left ( \frac{\partial U}{\partial V}\right )_T \right) \left ( \frac{\partial V}{\partial T}\right )_p $$ and for a perfect gas this has a simple form as $(\partial U/\partial V)_T = 0$ and $(\partial V/\partial T)_p=R/p$ which produces $$ C_p = C_V+R$$


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