Ammonia
is contained in sealed test tube at 25^{o}C. The test tube is
slowly cooled until liquid ammonia droplets condense on the inside of the test tube. At this point, the temperature of the ammonia is 20^{o}C. 
Determine the initial pressure
in the test tube, before
the cooling process
began. 


Read : 
We know the initial
temperature of the ammonia. If we knew
the intial temperature, we could look up the specific volume in the Superheated Vapor Tables for ammonia. There is only one pressure that yields this
value of the specific volume when the system is at 25^{o}C. Therefore, we could
also look at the problem in the following way. We know the intitial temperature of the ammonia and IF we also knew the
specific volume of the ammonia, we could use the Superheated
Vapor Tables to work backwards and determine the
initial pressure! That is what we are
going to need to do in this problem. 











When the 1st droplet
of liquid appears on the wall of the glass vessel, the vapor inside the
vessel is a satuated vapor. We can
look up the properties of this saturated vapor in the Ammonia
Tables.
Since the vapor is saturated at 20^{o}C, it must be superheated at 25^{o}C. But at both the
initial and final state the specific volume must be the same because neither
the mass nor the volume of the system changed! This is the key to the
problem. Because we know the values of
2 intensive variables at the initial state, specific volume and temperature,
and the initial state is a pure substance in a single phase, we can determine
the values of ALL other properties! In
this case we need to determine the pressure. 










Given: 
T_{1} 
25 
^{o}C 


Find: 
P_{1} 
??? 
kPA 

T_{2} 
20 
^{o}C 







x_{2} 
1 
kg vap/kg 
















Assumptions: 
1 The system contains
saturated vapor in the final state. 













Equations
/ Data / Solve: 


















The specific volume of
the system is equal to the specific volume of saturated ammonia vapor at T_{2}. 











We can look up this
value in the Saturated Temperature table of the Ammonia Tables at 20^{o}C : 

















V_{sat vap} 
0.62373 
m^{3}/kg 











Next, we scan the Superheated R134a Tables to
determine the 2 pressures between which this value of specific volume falls,
at the given temperature of 25^{o}C. 











Here are the data at 25^{o}C: 

P (kPa) 
V (m^{3}/kg) 








100 
1.1381 








200 
0.59465 








400 
0.30941 








500 
0.25032 








600 
0.21035 








700 
0.18145 














We need to interpolate
between 100 kPa and 200 kPa to determine the system
pressure that corresponds to our value of specific volume at a temperature of
25^{o}C. 















Eqn 1 

















Eqn 2 











slope 
184.0134 
kPa/(m^{3}/kg) 


P 
194.6 
kPa 










Verify: 
The assumption cannot be
verified. 
















Answers : 
P_{1} 
195 
kPa 















