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There is some super-heated $NH_3$ in an isolated tank with initial pressure and temperature of $1600 \ kPa$ and $50^{\circ} C$ respectively. Gas leaks out from a valve on tank very slowly such that some of gas exits reversibly. In the final state, pressure is $1400 \ kPa$ and $NH_3$ is still super-heated.

The problem says that using mass and entropy balance show that $s_1=s_2$, where $s$ is the molar entropy of gas in tank.

Using the total entropy balance for inside of tank as control volume, we get

$$(m_2 s_2 -m_1 s_1)_{C.V.}=-\dot{m_e} s_e \Delta t$$

What is $s_e$ now? It's not constant! How can I reduce this equation to $s_2=s_1$.

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closed as off-topic by sammy gerbil, Kyle Kanos, Jon Custer, Yashas, Rory Alsop Dec 10 '17 at 23:09

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    $\begingroup$ As @Rishabh Jain points out, the equation should read $$\left(\frac{d(ms)}{dt}\right)_{C.V.}=-\dot{m}_es$$where $$\dot{m}_e=-\frac{dm}{dt}$$ $\endgroup$ – Chet Miller Dec 8 '17 at 13:10
  • $\begingroup$ @Chester Miller : For the first equation, why is the right side specific entropy the same as that on left side? $\endgroup$ – user115350 Dec 8 '17 at 18:01
  • $\begingroup$ That is the current value of the specific entropy in the tank (value leaving the tank, assuming tank is well-mixed). $\endgroup$ – Chet Miller Dec 8 '17 at 18:08
  • $\begingroup$ Thanks for the information :) Sir, take a look at my new question? physics.stackexchange.com/questions/373318/… $\endgroup$ – Ghartal Dec 8 '17 at 18:40
  • $\begingroup$ @ChesterMiller : Thanks. Now why, in the problem, it said "Gas leaks out from a valve on tank very slowly such that some of gas exits reversibly"? Seems that this wouldn't change the first equation. $\endgroup$ – user115350 Dec 8 '17 at 19:40
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Entropy is an extensive quantity that depends on mass. As some mass has been lost, entropy has been decreased but the molar entropy still remains the same.

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