# Volume of CO2 from a Tank at 30psi

1. Jul 12, 2018

### sean882

First off, this is not a homework question nor assignment. I'm working on building a solar shower to mount on my car out of PVC, and don't feel like pumping large volumes of air with a bike pump. I'm looking at a portable air pump, or compressed gas in a tank with a regulator (and blow off valve).

I'm posting here rather than the projects section as my question is a direct use of the ideal gas law; I'm not looking for critiques on the design itself.

Goal: How many gallons, at 30psi and 70°F, would a 20oz CO2 tank (assume properly filled with 20oz of gas) expand to fill? How many times could a 5 gallon tank be filled at 30psi from the 20oz tank?

Question: I believe I figured out the volume at STP correctly (1ATM of pressure, 0°C). When I relate that to my desired conditions, I used PV/T = P2V2/T2. When relating atmospheric pressure to gage pressure (30psi, 2.04 ATM), can I simply add 1 ATM to the desired gage pressure, making my P2 value 3.04 ATM? See starred (in margin) line in my work.

Work and attempt is attached. Thanks for your help and review! It's been about 10 years since I did this in school, and haven't really had to apply it between then and now.

-Sean

#### Attached Files:

• ###### CO2 vs Compressor.pdf
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380.2 KB
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2. Jul 12, 2018

### Staff: Mentor

As the CO2 leaves the tank, what happens to the pressure in the tank?

3. Jul 12, 2018

### sean882

I'd expect that, up until most of the CO2 has left the tank, the pressure in the tank would remain constant at roughly 860 psi (with minor fluctuations with temperature). The CO2 should primarily be in the liquid form in the tank if filled properly, converting to gas phase as gas escapes (slightly lower pressure, some liquid turns to gas, 'rinse & repeat').

After all the liquid had turned to gas, the pressure would drop, and this pressure should equalize to the pressure of the vessel being filled.

4. Jul 12, 2018

### Staff: Mentor

The universal gas constant in convenient units is 10.731 (psi-ft^3)/(degR-lb_mole). You have $20 oz =1.25 lb=0.159 lb-mole$ So, $$V=\frac{nRT}{P}=\frac{(0.159)(10.731)(562)}{(44.7)}=21.4\ ft^3$$

Last edited: Jul 12, 2018
5. Jul 12, 2018

### sean882

My gut feeling says that is a bit much - how did you get to 562 for temperature?

Also, for my knowledge - it is acceptable and proper to simply add standard pressure of 14.7psi to the desired gage pressure of 30psi?

Thanks for your help,
-Sean

6. Jul 12, 2018

### Staff: Mentor

Oops. My mistake. R = F + 460. So it should only be 530 R, and V = 20.3 ft^3.

As far as the pressure part is concerned, that is done correctly. Psia = Psig + 14.7

7. Jul 14, 2018

### sysprog

I think you should go with the air pump. Something like this (about \$100 @ Walmart) is a good thing to have in your car or truck anyway. This one can deliver 100 psi (I don't know the cfm), and will automatically limit to a preset pressure:

8. Jul 17, 2018

### gmax137

@Chestermiller - shouldn't that be $$1.25~lb~*~ \frac {lb-mole} {44~lb}~=~0.028~lb-mole$$

That would give 3.6 ft^3 or 27 gallons, just exactly what the OP calculated in his attached pdf.

9. Jul 17, 2018

### Staff: Mentor

Yes. I have no idea how I ended up with that 0.159 lb-moles. Should have double checked.

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