# Example Problem with Complete Solution

8A-6 : Entropy Change as Compressed Liquid Ammonia Expands 5 pts
Consider the rigid tank shown below. It is divided into two equal volumes by a barrier. The left-hand side (LHS) is a perfect vacuum and the right-hand side (RHS) contains 5 kg of ammonia at 300 kPa and -10oC. 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 T1 and P1.  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, P2 and specific volume, and allows us to use the Ammonia Tables to determine the specific entropy and the total entropy at state 2.  DS = S2 - S1 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 P2 200 kPa
T1 -10 oC Eqn 1
P1 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 T1 and P1, so we can look up S1 in the Subcooled Liquid Table of the Ammonia Tables or in the NIST Webbook.
S1 0.54252 kJ/kg-K
At state 2, we only know the value of one intensive variable, P2.  So, we need to determine the value of another intensive variable before we can use the Ammonia Tables to determine S2.
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 :
V1 0.0015336 m3/kg V1 0.0076680 m3
Then, we can use the given relationship in Eqn 1 to determine V2: V2 0.015336 m3
Then, determine the specific volume at state 2 using: Eqn 4
V2 0.0030672 m3/kg
Now, we know the values of two intensive variables, so we can go back to the Ammonia Tables or NIST Webbook and determine S2 by interpolation.
At P2 = 200 kPa : Vsat liq 0.0015068 m3/kg Since Vsat liq < V2 < Vsat vap, state 2 is a saturated mixture.
Vsat vap 0.5946 m3/kg
We can determine x2 from the specific volume, using: Eqn 5
x2 0.002631 kg vap/kg
Then, we can use the quality to determine S2, using: Eqn 6
At P2 = 200 kPa : Ssat liq 0.387505 kJ/kg-K
Ssat vap 5.5998 kJ/kg-K S2 0.40122 kJ/kg-K
Finally, we can plug values into Eqn 2 : DS -0.70651 kJ/K
The 2nd Law tells us that: Eqn 7
where Sgen > 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 ! 