Is there any way to theoretically obtain values of specific volume, internal energy, enthalpy and entropy of air, assuming it's an ideal gas?

Can it be done using ideal gas laws?

  • $\begingroup$ Do you mean changes ($\Delta$) in these properties? $\endgroup$
    – Bob D
    Commented Aug 13, 2020 at 8:37
  • $\begingroup$ Well, really, I want an explanation. How does one calculate these. To answer your question, sir––yes. $\endgroup$
    – Qwin
    Commented Aug 13, 2020 at 8:50
  • $\begingroup$ @bobd Sir, do you think there is a way? $\endgroup$
    – Qwin
    Commented Aug 13, 2020 at 10:11
  • $\begingroup$ Sure. See my answer. $\endgroup$
    – Bob D
    Commented Aug 13, 2020 at 12:35

1 Answer 1


You can determine such properties by combining the ideal gas law with the first and/or second laws.

For example, by combining the ideal gas law and the first law of thermodynamics, you can show that, for any process, the change in specific internal energy is a function of temperature only according to

$$\Delta u=c_{V}\Delta T$$

For the derivation see here: $\Delta U$, $C_p$, $C_v$ for an ideal gas process

In a similar way you can show, for air as an ideal gas and a constant pressure process,

$$\Delta h=c_{P}\Delta T$$

Combining the ideal gas equation, first and second laws we can show:

$$\Delta s=R ln\frac{v_2}{v_1}$$

as follows:

From the second law, for a reversible isothermal process we have

$$q=T\Delta s$$

From the first law we have

$$\Delta u=q-w$$

Since, for an ideal gas isothermal process, $\Delta T=0$, then $\Delta u=0$, and $q=w$, therefore

$$T\Delta s=w=\int Pdv$$

From ideal gas law, one mole of gas


$$P=\frac {RT}{v}$$

$$T\Delta s=\int \frac {RT}{v}dv$$

and finally

$$\Delta s=R ln\frac{v_2}{v_1}$$

Hope this helps.


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