Consider the rigid tank shown below. It is
divided into two equal volumes by a barrier. The lefthand side (LHS) is a perfect
vacuum and the righthand side (RHS) contains 5 kg of ammonia at 300 kPa and 10^{o}C. 









When the barrier is removed,
the ammonia expands and fills the entire
tank. 









When the ammonia reaches equilibrium, the pressure in the tank is 200 kPa. Calculate ΔS for the ammonia and Q
for this process. 














Read : 
The key to this
problem is that the mass of ammonia in the system does not change and the volume doubles. We can use the Ammonia Tables to determine the specific volume and specific entropy at state 1 because we know T_{1} and P_{1}. We can use the specific volume at state 1 and the known mass and volume
relationships to determine the specific volume at state 2. This gives us the
values of two intensive properties at state 2, P_{2} and specific volume, and allows us to
use the Ammonia Tables
to determine the specific entropy and the total entropy
at state 2. DS = S_{2}  S_{1} and we are done. 













The 2nd Law and the fact that entropy generation must be positive will allow us to determine
the direction of heat transfer or if the process could be adiabatic. 



Given: 
m 
5 
kg 



P_{2} 
200 
kPa 


T_{1} 
10 
^{o}C 




Eqn 1 


P_{1} 
300 
kPa 




Find: 
DS 
??? 
kJ/K 

Determine whether: 
Q = 0, Q > 0
or Q < 0 




Diagram: 
See the problem statement. 





Assumptions: 
1  
The system is the contents of the entire tank. 



2  
No work or
mass crosses the system boundary. 



3  
Changes in kinetic and potential energies are negligible. 





Equations
/ Data / Solve: 






The change in entropy can be calculated using: 

Eqn 2 




We know both T_{1} and P_{1}, so we can look up S_{1} in the Subcooled Liquid Table of the Ammonia Tables or in the NIST Webbook. 













S_{1} 
0.54252 
kJ/kgK 




At state
2, we only know the value of one intensive variable, P_{2}. So, we need to
determine the value of another intensive variable before we can use the Ammonia
Tables to determine S_{2}. 




We can determine
the
specific volume at state 2 as follows: 


Eqn 3 




We can obtain specific volume at state 1 from the Ammonia Tables or the NIST Webbook : 













V_{1} 
0.0015336 
m^{3}/kg 

V_{1} 
0.0076680 
m^{3} 




Then, we can use the
given relationship in Eqn 1
to determine V_{2}: 
V_{2} 
0.015336 
m^{3} 




Then, determine the specific volume at state 2 using: 


Eqn 4 




V_{2} 
0.0030672 
m^{3}/kg 




Now, we know the
values of two intensive variables, so we can go
back to the Ammonia Tables or NIST Webbook and determine S_{2} by interpolation. 




At P_{2} = 200 kPa : 
V_{sat liq} 
0.0015068 
m^{3}/kg 

Since V_{sat liq} < V_{2} < V_{sat vap}, state 2
is a saturated mixture. 


V_{sat vap} 
0.5946 
m^{3}/kg 








We can determine x_{2} from the specific volume, using: 

Eqn 5 




x_{2} 
0.002631 
kg vap/kg 



Then, we can use the quality to determine S_{2}, using: 


Eqn 6 




At P_{2} = 200 kPa : 
S_{sat liq} 
0.387505 
kJ/kgK 



S_{sat vap} 
5.5998 
kJ/kgK 

S_{2} 
0.40122 
kJ/kgK 




Finally, we can plug
values into Eqn 2 : 

DS 
0.70651 
kJ/K 




The 2nd Law tells us that: 


Eqn 7 




where S_{gen} > 0 and T > 0 because it is a thermodynamic temperature scale, such as the Kelvin scale. 




Therforefore, the only way for DS to be
negative is if dQ < 0. 




We conclude that heat must have been transferred out of
the system during this
process ! 













We could apply the 1st Law to evaluate Q, but it is not required. I got Q =
184.6 kJ. 



Verify: 
The assumptions made
in the solution of this problem cannot be verified with the given
information. 










Answers : 
Part a.) 
DS 
0.707 
kJ/K 















Part b.) 
Q < 0 
Heat was transferred out of the system during this process ! 












