Calculating Oxygen Volume in Hospital Tanks Using Ideal Gas Law

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SUMMARY

The discussion focuses on calculating the volume of oxygen in hospital tanks using the Ideal Gas Law under different conditions. The initial conditions are 65.0 atm and 288 K, while the conditions in the patient's room are 1.00 atm and 297 K. The correct approach involves using the relationship (PV/RT)1 = (PV/RT)2, which allows for the cancellation of the gas constant R and the number of moles n, simplifying the calculation without needing to solve for n explicitly. The initial calculation of 67 m^3 was incorrect due to misunderstanding the application of the Ideal Gas Law.

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colton4286
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Oxygen for hospital patients is kept in special tanks, where the oxgen has a pressure of 65.0 atm and a temp. of 288 K. The tanks are stored in a separate room, and the oxygen is pumped to the patient's room, where it is administered at a pressure of 1.00 atm and a temp. of 297 K. What volume does 1.00 m^3 of oxygen in the tanks occupy at the conditions in the patient's room?

What I did was solve for the # of mols (n) given the tank information: n= PV/RT and used this n value to solve for volume: V= nRT/P. But I got 67 m^3 which is a much larger number than I expected. Am I heading in the right direction on this problem? Thanks!
 
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That sounds exactly right.
Note that you don't actually have to solve for the number of moles, all you need to know is that it stays constant.

n = PV/RT and n is constant, therefore:
(PV/RT)1 = (PV/RT)2 where 1 and 2 correspond to before and after the pumping, i.e. 1 is at 65 atm 288K and 2 is at 1atm 297K.
Also R is constant, so you can cancel that out.

Cheers
 

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