Calculating Oxygen Cylinder Duration

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To calculate the duration of an oxygen cylinder with a gauge pressure of 2,000 psi and a volume of 16 L, it's essential to account for the gas being under pressure. The relationship between the internal and external pressures can be expressed using the equation (PV)inside = (PV)outside. By determining the volume of gas at atmospheric pressure, the flow rate of 1.8 L/min can then be used to find the total time the cylinder will last. The discussion emphasizes that while the calculations may seem straightforward, understanding the pressure dynamics is crucial. Ultimately, the solution involves calculating the effective volume of gas available at atmospheric conditions.
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Homework Statement



A standard cylinder of oxygen used in a hospital has gauge pressure = 2.0E3 psi (1.38E4 kPa) and
volume = 16 L (0.016 m3) at T = 295 K. How long will the cylinder last if the flow rate, measured at atmospheric pressure, is
constant at 1.8 L/min?


Homework Equations



PV=nRT
n=m/M
flowrate=volume/time

The Attempt at a Solution



I assume it's not as simple as 16/1.8?

I used PV=(m/M)RT to solve for mass, but that got me no where.
 
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The gas in the cylinder is under pressure. So it is equivalent to many more liters at atomospheric pressure. Find out how many and then it is as simple as you think.
 
Dick said:
The gas in the cylinder is under pressure. So it is equivalent to many more liters at atomospheric pressure. Find out how many and then it is as simple as you think.

So nRT is constant correct?

So (PV)_inside=(PV)_outside?

(1.38*10^4kPa)(.016m^3)=(P_atmosphere)(V)

Solve for V, then use that in t=V/flowrate?

Thanks,
Brandon
 
you may have to use calculus to solve this
 
proton said:
you may have to use calculus to solve this

No, it's the lowest level physics class.
 
bpw91284 said:
No, it's the lowest level physics class.

Right. Just assume the gas is delivered at atmospheric pressure and outside temperature.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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