Two rigid tanks are connected by a large duct, as shown below. Tank A initially
contains 3 kg of hydrogen gas at 125^{o}C and 175
KPa. Tank B initially
contains 5 kg of methane gas at 85^{o}C and 65
KPa. 







a.) Can either gas be treated
as an ideal gas?


b.) The valve in the duct is opened and the gases in the
tanks mix. If both
tanks eventually cool
to room temperature (25^{o}C) after mixing, what will the final pressure be at equilibrium? 





Read : 
The key here is to
assume that both the pure gases and the final mixture of gases behave as
ideal gases. We can immediately verify
that the pure gases in their initial state are ideal gases, but we cannot
verify that the final mixture of gases is ideal until we solve the problem
and determine the final pressure. 













Choose the contents of
both tanks as the system. The fact
that links the initial and final states of this system is that the total
number of moles in the system does not change. This is a closed system. 












Given: 
m_{H2 }= 
3 
kg 



T_{B}
= 
85 
^{o}C 


T_{A}
= 
125 
^{o}C 




358.15 
K 



398.15 
K 



P_{B}
= 
65 
KPa 


P_{A}
= 
175 
KPa 



T_{equ }= 
25 
^{o}C 


m_{CH4}
= 
5 
kg 




298.15 
K 












Find: 
P_{equ}
= 
??? 
kPa 


















Assumptions: 
 Both pure gases, as
well as the final mixture, behave as ideal gases. 














Equations
/ Data / Solve: 



















Part a.) 
A diatomic gas can be
considered ideal when the following criterion is satisfied: 


Eqn 1 













V_{N2}
= 
18.92 
L/mol 


Where: 
R 
8.314 
J/mol K 


V_{O2}
= 
45.81 
L/mol 



















Since both molar
volumes are much greater than 5 L/mole, it is safe to consider both gases to be ideal gases. 












Part b.) 
The key to solving the
problem is to ASSUME that the equilibrium mixture will be an ideal gas: 


Eqn 2 













Let's begin by
determining how many
moles of gas are initially in each tank. 


Eqn 3 













Then we can determine the total moles of gas in the
system : 

Eqn 4 













MW_{N2 }= 
28.01 
g/mol 



n_{N2}
= 
107.1 
mole N_{2} 


MW_{O2 }= 
32.00 
g/mol 



n_{O2}
= 
156.3 
mole O_{2} 








n_{total} = 
263.4 
mole total 












The total number of
moles in the system does not changes as the gases mix! 


The system, consisting
of both tanks, is closed. 













Next, we can use the IG EOS to determine the volume of
each tank and then the total volume of the system. 














Eqn 5 
Where (for ideal gases) : 

Eqn 6 








and 




V_{A}
= 
2.03 
m^{3} 




Eqn 7 


V_{B}
= 
7.16 
m^{3} 








V_{tot}
= 
9.18 
m^{3} 



















From the IG EOS we derive the following
equation for the equilibrium pressure
: 


Eqn 8 



















P_{eq} 
71.08 
kPa 












Verify: 
Now,
calculate the molar volume at the equilibrium state, just to be sure it is
still safe to treat the gas as an ideal gas ! 


















V_{eq}
= 
34.87 
L/mol 
















Since V_{eq} is greater than 5 L/mole, we were justified in using the Ideal
Gas EOS for the equilibrium state as well. 












Answers : 
a.) 
Since both molar
volumes are much greater than 5 L/mole, it is safe to consider both gases to be ideal gases. 













b.) 
P_{eq}
= 
71.1 
kPa 
















