This is fun "math" compared to the CO2 enhanced oil recovery liquid rich nat gas estimations in the permian basin I'm working on (I know, TMI).

Here is a simplified but actual example of the "math" -

Problem using .850 ATM, 1 mole of O2 in 1 liter of water, 25 degree Celsius which is about 77 degrees F (the 298 in the equation is 273+25 since the formula uses degrees kelvin not celsius, much less Fahrenheit)

PV=nRT

.850(v) = 1.00(.0821)298

v = 28.8 L O2

Moles × molecular mass = grams

1.00 moles O2 × 32.0 = 32.0 grams O2

Finally, we take the mass and divide it by volume to find density:

32.0 grams ÷ 28.8 L = 1.11 g/L

Now substituting 0 degrees Celsius (freezing point of water, 32 F) -

.850(v) = 1.00(.0821)273

v = 26.4 L O2

32.0 grams ÷ 26.4 L = 1.21 g/L

Everything else held constant, the density of oxygen in the one liter of water keeping everything else constant was 9.2% higher when the temperature was 32 degrees F versus 77 degrees F.

*Disclaimer - I don't do these sorts of problems on a regular basis.

I will throw another problem at you that is just as interesting. How many calories are required to run x miles at your age/weight/pace in 77 degree ambient temps vs 32 degrees? :smile:

I'll give you a clue, very high temps and very low temps (especially) require your body to consume massive quantities of energy!

*Edit - Vegas needs to hire me after I graduate.