Why is the cyclic volume integral of P / T larger than the one of P e x t / T in an irreversible process?

Question

Why is the cyclic volume integral of   P      /    T larger than the one of       P    e    x  t      /    T in an irreversible process?

Miah Carlson · Accepted Answer

Probably this question as stated doesn&#039;t make sense. However the derivation as set makes perfect sense with a slight difference. After getting rid of the cyclic integral corresponding to entropy we&#039;re left with:  ∮                              đ                    q        T    =  ∮                    P                  e          x          t                    −      P        T    d  VFor a given infinitesimal contribution to the integral at the rhs, if       P          e      x      t        −  P  &amp;gt;  0 the system will be spontaneously compressed and   d  V  &amp;lt;  0, while if       P          e      x      t        −  P  &amp;lt;  0 the system will spontaneously expand and   d  V  &amp;gt;  0, therefore      (          P              e        x        t              −    P    )    d  V  ≤  0will be true throughout the whole cycle. Then:  ∮                              đ                    q        T    ≤  0

Why is the cyclic volume integral of P/T larger than the one of P_ext/T in an irreversible process?

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