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I'm pretty confused. So I've been told that the equation of state for $n$ moles of some type of gas is $P(V-b) = nRT$. That's not quite like an ideal gas. But then the relations $C_p - C_v = R$, and $\gamma = \frac{C_p}{C_v} = 5/3$ hold for this gas, and at the very least the last equation is definitely for ideal gases.

So if a gas has these properties, and only one mol is considered, is it fair to assume it is an ideal gas? Using the above relations gave me $C_v = \frac{3}{2} R$, which again is for ideal gases.

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The gas you are describing is not precisely an ideal gas, but is pretty close. In an ideal gas, the molecules are dots, they don't have volume ; moreover, threr are no interactions except for the elastic collisions that allows the gas to thermalize.

The gas you describe is is a gas with no interactions, but with molecules of finite volume. Having molecules of finite volume reduces the space avaliable for molecules to move around, hence the $V-b$ factor instead of just $V$ for an ideal gas.

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  • $\begingroup$ It would be fair to approximate it as an ideal gas then! Thanks. $\endgroup$
    – user13948
    Jan 20, 2016 at 16:47
  • $\begingroup$ In a way yes, as you pointed out many formulas are similar. $\endgroup$
    – Dimitri
    Jan 20, 2016 at 16:50
  • $\begingroup$ I'll double check exactly how different any formulae I use would be if volume were taken into account, before I use them. $\endgroup$
    – user13948
    Jan 20, 2016 at 16:55
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The heat capacities of this gas behave the same as for an ideal gas. The PVT behavior is not. So, what is the problem? Note that, for this gas, even though $C_v$, $C_p$, and U are functions only of temperature, like an ideal gas, H is a function not only of temperature but also of pressure p: $dH=C_pdT+bdP$.

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  • $\begingroup$ You might want to explain what $b$ is (compressibility? bulk modulus? the correction factor from the question?). $\endgroup$
    – wyphan
    Dec 4, 2020 at 23:58
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    $\begingroup$ b is the volume parameter in the OP equation of state. $\endgroup$
    – Chet Miller
    Dec 5, 2020 at 0:50
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    $\begingroup$ Related: chemistry.stackexchange.com/questions/71728/… Set $a = 0$, and you recover the exact same answer. $\endgroup$
    – wyphan
    Dec 5, 2020 at 0:52

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