Determining entropy as function of pressure

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The discussion focuses on calculating the entropy of an ideal gas as a function of pressure, specifically addressing the challenges of integrating from zero pressure. The integral for entropy, S(p) = -∫(nR/p)dp, diverges as the lower limit approaches zero, complicating direct calculations. To determine entropy at arbitrary pressures and temperatures, one must start from a known reference state and consider a reversible path to the desired state. For instance, during an isothermal compression from pressure p1 to p2, the process can be divided into infinitesimal steps, allowing for the integration of dq/T over the entire path. The absolute entropy at a specific pressure, such as 101325 Pa at 298 K, is derived from the Third Law of Thermodynamics, which posits that the entropy of a perfect crystalline substance is zero at absolute zero. This requires heat capacity data to calculate absolute entropy at higher temperatures, enabling the determination of entropy changes during isothermal processes.
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I read this discussion but I am interested in how the entropy is obtained as a function of pressure. Namely, how can you determine a following integral for an ideal gas:

$$S(p) = -\int_{0}^{p} \frac{nR}{p}dp $$

when you need to start from 0 pressure?
 
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That integral does not converge if you make the lower integration limit approach 0. To calculate entropy as a function of pressure, you need to know the entropy at some reference state and then imagine a reversible path between that reference state and the wanted final state. Finally you integrate the quantity ##dS = \frac{dq}{T}## over that path.

For example, if you have a reversible isothermal compression of an ideal gas from initial pressure ##p_1## to final pressure ##p_2##, you can divide the process to small infinitesimal steps where the volume decreases by ##dV##, and an infinitesimal heat flow ##dq## out of the system keeps the temperature constant despite the mechanical work done on the system by compressing it. Then you just sum the ##dq/T##:s for the whole process.
 
Thank you for that nice example, however I believe that my question still persists but perhaps I put it unclearly in my first post.
So, what I basically do not understand is how I can get the entropy in arbitrary pressure and temperature? When I take my physical
chemistry book I can see quantities like ##S^{o}## at 298 K. Very nice, but how do they know that this is entropy at this specific pressure
of 101325 Pa? Obviously the change in entropy for an ideal gas due to pressure change () in isothermic process and when no reaction or mixing occurs it is ## \Delta S = -nRln(\frac{p_{2}}{p_{1}})##. But how "the absolute" entropy at arbitrary pressure is obtained?
 
Absolute entropy is calculated by assuming the 3rd Law of Thermodynamics, which says that the most stable crystalline form of any compound at absolute zero of temperature has exactly zero entropy. Then you need to have heat capacity data of the substance down to low enough temperatures to be able to calculate the absolute entropy at some final state that has a nonzero temperature.
 
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