Ok_Construction5119
Ok_Construction5119 t1_ixlaldr wrote
Reply to comment by montbarron in How does Jupiter hold all the gasses in a spherical shape? by flatulent_tarantula
Yes but there is a temperature differential. Gases expand and contract with temperature, so the densities are constantly changing. Even if there would be a difference at constant temperature, the heating and cooling dominates the equation.
Ok_Construction5119 t1_ix1i7gs wrote
Reply to comment by Still_Not-Sure in If you froze soda for a year, would be still fizz when thawed out? by The_Grand_Canyon
The systems before the phase change from liquid to solid and after the phase change back from solid to liquid will be identical and so will their equilibrium states. It will just take time for the CO2 to dissolve back. It does not matter how quick or slow the phase change itself is.
Ok_Construction5119 t1_ix1dwgu wrote
Reply to comment by urzu_seven in If you froze soda for a year, would be still fizz when thawed out? by The_Grand_Canyon
This implies a rigid container that performs no pressure-volume work, but a bottle or can will change things!
Ok_Construction5119 t1_iww47r2 wrote
Reply to comment by foundmyreddit in If you froze soda for a year, would be still fizz when thawed out? by The_Grand_Canyon
This is not true. Solubility of solids decreases as temperature decreases, but solubility of gases increases as temperature decreases. See this image for reference:
https://chem.libretexts.org/@api/deki/files/125741/CNX_Chem_11_03_gasdissolv_2.jpg?revision=1
To answer OP's question, if, in an isolated system, you returned the system to its original state, the CO2 would redissolve into the solution.
Soda as you buy it is at equilibrium because of the pressure, around 2 psi. The CO2 spontaneously dissolves into the solution, and will again as long as you leave the receptacle unopened. For further information, you can read this:
Ok_Construction5119 t1_iwriglk wrote
Reply to comment by OneTreePhil in What would the pressure be like in a body of water in free fall? by OneTreePhil
Yes, your density to mass conversion is correct.
V,h2o [m^3 ] * rho,h2o [kg/m^3 ] = V,earth [m^3 ] * rho,earth [kg/m^3 ]
With the density of water, volume of earth, and density of earth constant, you can use this equality to find the required volume, which will be proportional to the ratio of densities.
As you have already done, multiplying volume by density yields mass, and with constant density, volume is directly proportional to mass.
Lots of words to say yes, you are correct in terms of volume required to give the same gravity given the above assumptions.
This next paragraph relies on many invalid assumptions: However, due to the different volume and identical gravity, the pressure would increase more slowly with depth than on earth. This is because water is in fact slightly compressible, and under such enormous forces it would in fact be more dense as you approached the core. If you took the density as an average density, which is inaccurate, you would see a pressure/depth change similar to how it is on earth, with a similar pressure to the earth's core at the core of your big water droplet, and a similar pressure at the depth of the challenger deep that you would see on earth.
Again, p = rho g h, so if g = G,earth, and rho = rho,h20, then p will increase linearly with the 'height' of the water above you
Ok_Construction5119 t1_iwrf69o wrote
Reply to comment by Chemomechanics in What happens to air when it escapes into outer space? by CentralOregonKush
The 100km we chose as the boundary for outer space is the arbitrary thing.
Ok_Construction5119 t1_iwrf0ai wrote
It just becomes more disordered due to the second law of thermodynamics. It will reach its lowest energy state and highest entropy value based on the temperature and pressure (a few degrees K and 0, respectively) as time approaches infinity.
Ok_Construction5119 t1_iwre4op wrote
Reply to comment by OneTreePhil in What would the pressure be like in a body of water in free fall? by OneTreePhil
Yeah, the pressure of your 14 mile droplet is noticeably less than atmospheric pressure
Ok_Construction5119 t1_iwrdsz1 wrote
The surface tension of water is negligible compared to the other forces at work here. The "pressure" from water is due to the weight, not the mass, making it entirely dependent on gravity (P = rho * g * h). Thus, if g = 0, as in the case of freefall, P is also 0.
If you are asking about the pressure changing due to the water itself, it would have to be much larger than 14 miles to have enough mass to exert gravity that would be detectable by human senses.
As stated earlier, surface tension is negligible compared to these forces. According to wikipedia, the surface tension of water is 72.8 mN/m, which is again so small you could barely feel it. This is what is known as an intensive property, meaning it is not dependent on the amount of a substance.
As gravity is directly proportional to mass, it would need to have the same mass as the earth to have the same pressurization capabilities due to gravity. See newton's law of universal gravitation if you are still curious.
Ok_Construction5119 t1_ivn8viq wrote
Reply to comment by Tehnizzim in How does extracting venom from animals help us create antidotes? by asafen
Horses are bigger, can tolerate a higher dose (due to greater muscle mass), and produce more antibodies (in terms of volume, not concentration)
Ok_Construction5119 t1_jdsirpc wrote
Reply to comment by huckerbjk in Humans experienced a massive population expansion in a very small amount of time. What are the evolutionary consequences and benefits of such an event, massive popular of a species in a small amount of time? by bent_over_life
Heart disease rarely kills you before you have the chance to reproduce. That's why our teeth are only built to last 40 years or so