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?
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$\begingroup$ Do you mean changes ($\Delta$) in these properties? $\endgroup$– Bob DCommented Aug 13, 2020 at 8:37
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$\begingroup$ Well, really, I want an explanation. How does one calculate these. To answer your question, sir––yes. $\endgroup$– QwinCommented Aug 13, 2020 at 8:50
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$\begingroup$ @bobd Sir, do you think there is a way? $\endgroup$– QwinCommented Aug 13, 2020 at 10:11
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$\begingroup$ Sure. See my answer. $\endgroup$– Bob DCommented Aug 13, 2020 at 12:35
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1 Answer
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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
$$Pv=RT$$
$$P=\frac {RT}{v}$$
$$T\Delta s=\int \frac {RT}{v}dv$$
and finally
$$\Delta s=R ln\frac{v_2}{v_1}$$
Hope this helps.
