A pressure gauge on
a rigid steel tank reads 50 kPa. The tank holds 2.1 kg of air
and 0.250 kg of water vapor at 70^{o}C. Calculate the relative humidity of the air in the tank. 









Read : 
The keys here are the
definition of relative humidity and the relationships between mass, moles,
molecular weight, mole fraction and partial pressure.


We need to assume the humid air behaves as an ideal gas in order to determine the
partial pressures from the given m_{water} and m_{BDA}. 










Given: 
T 
70 
^{o}C 



m_{tot} 
2.100 
kg wet air 

P_{tot} 
50 
kPa 



m_{H2O} 
0.250 
kg H_{2}O 










Find: 
h_{r}
= 
??? 
% 
















Assumptions: 
1 Air is a
noncondensable gas. 







2 Humid air behaves as
an ideal gas. 















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* (80^{o}C) = 
47.39 
kPa 











For an ideal gas: 


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


Eqn 2 











Test if ideal: 
Ideal
Gas EOS : 



Eqn 3 











Solve for the molar
volume : 




Eqn 4 

















V 
57.06 
L/mole 











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













For all gases, mole
fraction is defined as : 











Eqn 5 











Where : 



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


Eqn 6 











Data: 

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






MW_{BDA} = 
28.96 
g bonedry air / mol
bonedry air 












Since we know the
mass of water and bonedry air in the tank, as well
as their molecular weights, we can calculate the number of moles of water and BDA in the tank using Eqn 6. 











Then, we can calculate
the mole fraction of water in
the gas in the tank, using Eqn 5. Next, we can use the
given total pressure to calculate the partial pressure of water in the gas using Eqn 2. 











Here are the numerical
results: 

n_{H2O} 
13.88 
mol H_{2}O 







m_{BDA} 
1.850 
kg BDA 







n_{BDA} 
63.88 
mol BDA 







y_{H2O} 
0.178 
mol H_{2}O / mol wet gas 





P_{H2O} 
8.92 
kPa 













Finally, we
can calculate the relative humidity using Eqn 1: 










h_{r}
= 
18.8% 
% 










Verify: 
1  The assumption is
sound since the normal boiling points of oxygen and nitrogen are 68 K and 70
K and our system is at 343
K. 

2  The ideal gas
assumption was verified above because V = 57 L/mol which is greater than 20 L/mole. The limit for
nondiatomic gases applies because of the water in the air. 










Answers : 
h_{r}
= 
19% 
% 















