The rigid tank shown below contains 5 kg of a noncondensable gas with a molecular weight of 44.1 g/mol. The tank also contains water vapor.
The gas in the tank is at 140 kPa and 80^{o}C and the relative
humidity is 72%. 









Assuming the gas in the tank behaves as an ideal gas, calculate the mass of water vapor in
the tank. 


Read : 
The keys here are the
definition of relative humidity and the relationship between mole fraction
and partial pressure.


We need to assume the humid air behaves as an ideal gas in order to determine the
partial pressure of water from the given h_{r}. We can use the mole
fraction of water in
the gas to determine the mass of water in the gas. 










Given: 
m_{NCG}
= 
5 
kg NCG 



P_{tot}
= 
140 
kPa 

MW_{NCG} = 
44.1 
g NCG / mole NCG 


h_{r}
= 
72% 


T = 
80 
^{o}C 
















Find: 
m_{H2O}
= 
??? 
kg 
















Assumptions: 
1 The gas in the tank
behaves as an ideal gas. This must be verified. 












Equations
/ Data / Solve: 


















Let's begin with the
definition of relative humidity: 



Eqn 1 











The vapor pressure is
equal to the saturation pressure at the system temperature. 











We can find this in
the saturation temperature section of the steam tables: 
P*_{H2O} (70^{o}C) 
47.41 
kPa 











Plug P*_{H2O} and h_{r} into Eqn 1 to get the partial pressure of water from the definition of
relative humidity. 

















P_{H2O} 
34.138 
kPa 











For an ideal gas: 

P_{H2O} = y_{H2O} P_{tot} 



Eqn 2 











or: 






Eqn 3 











Plugging values into Eqn 3 yields: 


y_{H2O}
= 
0.244 
mol H_{2}O / mol wet gas 











For all gases, mole
fraction is defined as: 


Eqn 4 











Where : 



n_{i} = m_{i} / MW_{i} 


Eqn 5 











Now, we solve Eqn 4 for n_{H2O} : 



Eqn 6 

















Eqn 7 











Now, we can plug the
numbers into equations… 



















Eqn
5 yields : 
n_{NCG}
= 
113.38 
moles NCG 





Eqn
7 yields : 
n_{H2O}
= 
36.56 
moles H_{2}O 











Finally, Eqn 5 can be rewritten as : 
m_{i} = n_{i} MW_{i} 


Eqn 8 











We can answer the
question posed by plugging numbers into Eqn 8 : 














Data: 
MW_{H20 }= 
18.016 
g H_{2}O / mol H_{2}O 

m_{H2O}
= 
658.7 
g H_{2}O 










Verify: 
Test if ideal: 
Ideal
Gas EOS : 



Eqn 3 











Solve for the molar
volume : 




Eqn 4 

















V 
20.97 
L/mole 











Therefore, since V > 20 L/mole, we can treat the
wet gas as an ideal gas. 












Answers : 
m_{H2O} = 
659 
g H_{2}O 















