The tank, shown below, has two chambers of equal
volume. The left side holds 10 kg of air at 500 kPa and 60^{o}C. The right
side is completely
evacuated. 










When the wall that separates the two chambers within the tank is removed,
the air expands to fill the right side of the tank. Calculate the final temperature and pressure in the tank. 

Assume air behaves as an ideal gas and the process is adiabatic because the tank is wellinsulated. 














Read : 
The most important
thing to recognize in this problem is that removing the
partition is equivalent to allowing the partition to move
the to the left (in our diagram) until the ideal gas fills the entire tank. The resisting force in
this expansion process
is zero because there
is a vacuum in the left chamber of the tank. Since the resisting force is zero,
the work done by the expanding gas is also zero. If we take the ideal
gas to be our system, there is no heat transfer during the expansion either, because the tank is insulated. The 1st Law tells us that DU = 0
when no work or heat
transfer occur during a process on a closed system. Also, because U is a function of T only
for an ideal gas, T_{2} = T_{1}. Then, all we need to
do is apply the IG EOS
to determine P_{2}. 










Diagram: 
The diagram in the
problem statement is adequate. 









Given: 
m 
10 
kg 

Find: 
P_{2} 
??? 
kPa 


P_{1} 
500 
kPa 


T_{1} 
60 
^{o}C 


Assumptions: 
1  
The gas behaves as an ideal gas. 


2  
Changes in kinetic and potential energy are negligible. 


3  
The tank is perfectly insulated,
so the process is adiabatic: Q = 0. 


Equations
/ Data / Solve: 



The ideal gas is the system that we will analyze. This
is a closed system
because no mass crosses the boundary during the expansion process. 



The 1st Law for a closed system is: 


Eqn 1 



Since the restraining force overcome during the expansion is zero, the boundary work for the expansion is also zero. Combining this fact
with the 2nd and 3rd assumptions listed above,
allows us to simplify Eqn 1 to Eqn 2
: 





or: 



Eqn 2 



For an ideal gas, internal energy depends only on the temperature of the gas. If the internal energy is the same in state
2 as in state 1, then the temperature in state 2 must also be the same as the temperature in state 1 ! 















Eqn 3 



Now that we know T_{2}, we can apply the Ideal Gas EOS to both
states 1 and 2 to determine P_{2}. 




Eqn 4 


Eqn 5 



Divide Eqn 5 by Eqn
4 and cancel like terms : 


Eqn 6 



But, we know that T_{1} = T_{2} and because the system is closed,
n_{1} = n_{2}. Also, because the left and right chambers of the tank are equal
in size, V_{2} = 2 V_{1}. Therefore, Eqn 6 can be simplified as follows: 




Eqn 7 

or: 

Eqn 8 



Solving Eqn 8 for P_{2} yields : 



Eqn 9 



Putting values into Eqn 9 gives us the answer : 

P_{2} 
250.0 
kPa 


Verify: 
None of the assumptions
can be verified. 










Answers : 
P_{2} 
250 
kPa 

























