- #1
Addiw777
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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?
$$S(p) = -\int_{0}^{p} \frac{nR}{p}dp $$
when you need to start from 0 pressure?