How much enthalpy of argon changes when the pressure is isothermally increased by $\pu{1 atm}$? (Gas should be considered as real gas under specified conditions) ($\pu{1 mol}$ of Argon gas is available at $\pu{25 ^\circ C}$ and $\pu{11 atm}$ conditions)

-Firstly I took the total derivative of Enthalpy function of pressure and temperature:

$$dH =\left(\frac{∂H}{∂P}\right)_T \cdot dP + \left(\frac{∂H}{∂T}\right)_P \cdot dT$$

Then divided terms with $dP$ because as the question asks I need $\left(\frac{∂H}{∂P}\right)_T$. After this I get:

$$\left(\frac{∂H}{∂P}\right)_V = \left(\frac{∂H}{∂P}\right)_T + \left(\frac{∂H}{∂T}\right)_P \cdot \left(\frac{∂T}{∂P}\right)_V$$

Third term is enthalpy change with temperature at constant pressure means $C_p$, so the equation turns into:

$$\left(\frac{∂H}{∂P}\right)_V = \left(\frac{∂H}{∂P}\right)_T + C_p \left(\frac{∂T}{∂P}\right)_V$$ so I rewrite this like: $$\left(\frac{∂H}{∂P}\right)_T =\left(\frac{∂H}{∂P}\right)_V-C_p\left(\frac{∂T}{∂P}\right)_V$$

Then I used Chain rule, which is: if $z=f(x,y)$: $$\left(\frac{∂z}{∂y}\right)_x\left(\frac{∂y}{∂x}\right)_z\left(\frac{∂x}{∂z}\right)_y=-1$$

I used this rule for writing a derivative term, $\left(\frac{∂T}{∂P}\right)_V$, with constants, and my equation turns:

$$\left(\frac{∂H}{∂P}\right)_T = \left(\frac{∂H}{∂P}\right)_V - C_p\left(\frac{\kappa T}{\alpha}\right)$$ where $\alpha$: expansion coefficient; and $\kappa T$: isothermal compressibility coefficient (kappa t).

In this way, I came up to this point but I couldn't find the exact answer. I Lost myself in thermodynamic differentials. Can anyone explain me? Where did I mistake or is my solution method incorrect?


$$\left(\frac{∂H}{∂P}\right)_T = V - T\left(\frac{∂V}{∂T}\right)_P$$

If i wasn't wrong rhs of the equation turns into = V - T ( ßV) which equals to partial derivative of enthalpy with respect to pressure at constant temperature. But i'm confused here. There are other terms that can be used for ßV. For example alphaV or Vkappa t. Which one should i use? What is partial derivative of volume with respect to temperature at constant pressure means? Question asked about isothermally increasing pressure so should i use isothermal compressibility coefficient or isobaric compression coefficient or thermal expansion coefficient or what? Really confused here..

$$\left(\frac{∂H}{∂P}\right)_T = V - TßV$$ this is the last point i came to

  • $\begingroup$ If my editing is incorrect, please feel free to correct them as you mean. $\endgroup$ Commented May 2, 2020 at 19:52
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    $\begingroup$ Thank you for the arrangement.There was a little mistake but i fixed it myself :) $\endgroup$ Commented May 2, 2020 at 20:02
  • $\begingroup$ That derivation is challenging, and requires some non-obvious manipulations. Here's my suggestion: You can probably find, in thermodynamics textsbooks, a derivation of $\left(\frac{∂U}{∂V}\right)_T = \left(\frac{\alpha T}{\kappa}\right) - p$. Take the techniques and manipulations used in that derivation, and apply them to your problem. If you are successful, you will get an answer that contains only V, $\alpha$, and T (no partial derivatives). $\endgroup$
    – theorist
    Commented May 2, 2020 at 20:48

2 Answers 2


The easiest place to begin the derivation is from the following definition for the differential form of the enthalpy:

$$dH= dU + d(PV) = VdP + TdS$$

Taking the derivative wrt P at constant T gives

$$\left(\frac{\partial H}{\partial P}\right)_T= V + T\left(\frac{\partial S}{\partial P}\right)_T$$

It remains to evaluate $\left(\frac{\partial S}{\partial P}\right)_T$. To this end you can employ one of the Maxwell relations, starting from the definition of the differential form of the Gibbs free energy

$$dG = VdP -SdT$$

Because $dG$ is an exact differential, it follows that

$$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P$$

Substituting in the original equation gives

$$\left(\frac{\partial H}{\partial P}\right)_T= V - T\left(\frac{\partial V}{\partial T}\right)_P$$

You can then proceed by making appropriate substitutions and integrating the above equation, as explained in the answer by Chet Miller.

The last equation can also be written $$\left(\frac{\partial H}{\partial P}\right)_T= V(1 - T\alpha)$$

where $\alpha$ is the thermal expansion coefficient of the substance :

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

Use of the coefficient is convenient if it does not change much over the pressure interval of interest (you can then treat it as a constant and pull it out of the integral). Sometimes other symbols such as $\beta$ are used; $\kappa$ is the isothermal compressibility and is something else - the coefficients can be related through the heat capacities but this is another topic!

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    $\begingroup$ Thanks for your answer, i appreciated! $\endgroup$ Commented May 3, 2020 at 12:42

The equation for the partial derivative of enthalpy with respect to pressure at constant temperature is given by: $$\left(\frac{\partial H}{\partial P}\right)_T=V-T\left(\frac{\partial V}{\partial T}\right)_P$$From the equation of state for your gas, you can evaluate the rhs of this equation.


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