While teaching us thermodynamics our chemistry teacher told us that enthalpy change of a real gas is a function of temperature, pressure etc. but the enthalpy change of an ideal gas is only a function of temperature. He didn't give us any explanation for this fact.

But I am curious to know why it is so ?

I have tried to answer it myself but cannot verify it.

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    $\begingroup$ Let assume hypothesis it depends on other factors as well. How would be this energy stored within an ideal gas? $\endgroup$
    – Poutnik
    Commented Aug 15, 2021 at 6:36
  • $\begingroup$ @Poutnik Sorry I didn't understand what you meant, please elaborate your comment. Assume what ? $\endgroup$
    – Tushar
    Commented Aug 15, 2021 at 8:10
  • $\begingroup$ The alternative hypothesis to "ideal gas enthalpy depends only on T" is "ideal gas enthalpy depends on T and other factors(e.g. p)". If so, then how?(rhetorical question) $\endgroup$
    – Poutnik
    Commented Aug 15, 2021 at 8:35
  • $\begingroup$ Based on the first and 2nd laws of thermodynamics, it is possible to prove mathematically that, for a gas that obeys the equation of state PV=nRT, the the internal energy U and enthalpy H are functions only of T. Have you learned about the 2nd law of thermodynamics yet, and are you familiar with the equation dH=TdS+VdP? $\endgroup$ Commented Aug 15, 2021 at 12:35
  • 1
    $\begingroup$ @ChetMiller I have studied the 2nd low of thermodynamics today ( and obviously I am very cofused for now regarding order, disorder and entropy ) but I haven't encountered the equation mentioned by you, maybe I get to learn about in upcoming days. $\endgroup$
    – Tushar
    Commented Aug 15, 2021 at 15:26

2 Answers 2


Let's start from the definition of enthalpy change (not going to formal definition because it's very non-intuitive) here's how I like to define it:

Enthalpy change is the amount of heat absorbed ($\mathrm dH\gt0$) or given ($\mathrm dH\lt0$) by the system at constant pressure of system.

$$ \mathrm dH=Q_p $$ $$ \begin{array}{l} \therefore \mathrm d U=Q_p+W \\ \Rightarrow Q_p=\mathrm dU-W=\mathrm dU+p_{\text{ext}}\, \mathrm dV \end{array} $$

Now as the pressure of the ideal gas system is constant so it must be equal to the external pressure. $$ \begin{array}{l} p_{\text{ext}}=p_\mathrm g \\ \Rightarrow p_\mathrm g=\frac{nRT}{V_\mathrm g} \end{array} $$ $$ \begin{array}{l} \Rightarrow p_\mathrm g V_\mathrm g=nRT\\ \Rightarrow p_\mathrm g\mathrm dV=nR\,\mathrm dT=p_{\text{ext}}\,\mathrm dV \end{array} $$ The above equation is valid only if the composition of the gas does not change. Now using this equation $$ \begin{aligned} \mathrm dH &=\mathrm dU+p_{\text{ext}}\,\mathrm dV \\ &=\mathrm dU+nR\,\mathrm dT \end{aligned} $$

Now as we know that $\mathrm dU$ is a function of temperature only for ideal gas (as according to the kinetic theory of gases there is no intermolecular interaction between molecules of ideal gases so they possess kinetic energy only which itself is a function of temperature only) so we can say that

$$\mathrm dH=f(T)$$

That is, enthalpy change of an ideal gas is only a function of temperature given that the composition of ideal gas system do not change.

Please do correct me if I am getting something wrong.

  • $\begingroup$ See my comment above. $\endgroup$ Commented Aug 15, 2021 at 12:35

There's a simpler way to see this.

By definition, $H = U + pV$. We would like to show that, for an ideal gas, $H(n, p, T) = H(n, T)$, or equivalently that $$\left(\frac{\partial H}{\partial p}\right)_{n, T} = 0.$$

Clearly, $$\left(\frac{\partial H}{\partial p}\right)_{n, T} = \left(\frac{\partial (U + pV)}{\partial p}\right)_{n, T} = \left(\frac{\partial (U + nRT)}{\partial p}\right)_{n, T} = \left(\frac{\partial U}{\partial p}\right)_{n, T} = 0,$$ because an ideal gas satisfies $pV = nRT$ and $U = U(n, T)$.


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