A gas tank is divided into two parts by a stationary barrier. The tank is held at constant temperature and constant pressure. A small hole is pierced in the barrier, so that particles can move. Show that μ 1 = μ 2

Question

A gas tank is divided into two parts by a stationary barrier. The tank is held at constant temperature and constant pressure. A small hole is pierced in the barrier, so that particles can move. Show that       μ    1    =      μ    2

Phillip Fletcher · Accepted Answer

F  =      F    1    (      T    1    ,      V    1    ,      N    1    )  +      F    2    (      T    2    ,      V    2    ,      N    2    ). For constant temperature process,  d  F  =  d      F    1    +  d      F    2    =  −      p    1    d      V    1    +      μ    1    d      N    1    −      p    2    d      V    2    +      μ    2    d      N    2    =  0, at equilibrium.Now   d      N    1    =  −  d      N    2    ,      p    1    =      p    2  , hence  d  F  =  −      p    1    (  d      V    1    +  d      V    2    )  +  (      μ    1    −      μ    2    )  d      N    1    =  0. Since       p    1    (      T    1    ,      V    1    ,      N    1    )  = constant, in which       T    1   is also constant, then       V    1   must be a function of       N    1  . Therefore you cannot straight away equate the coefficient of   d      N    1   to zero to obtain       μ    1    =      μ    2  , because it is not an independent variation, i.e. variations   d      V    1   and   d      V    2  . This dependence is a result of pressure being constant.This problem is eliminated if you use G instead, as you may verify yourself.

A gas tank is divided into two parts by a stationary barrier. The tank is held at constant temperature and constant pressure. A small hole is pierced in the barrier, so that particles can move. Show that mu_1=mu_2

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