ALEKS: Understanding the connection between vapor pressure, boiling point, enthalpy of vaporization

[@arixlxra]

Thank you for this video! I was stuck on this problem on ALEK for some time. I am onto the last week of my six-week Chemistry summer course and videos like this that can quickly, but thoroughly explain topics help tremendously. Thank you!

[@nathanwhyte7568]

I love you and your videos. You explain everything so well

[@selenawellenstein4170]

Thank you for making this make sense!

[@KazmaAttack]

Perfect explanation, thank you! Please keep making videos!

[@VEAR31]

I have a Liquid Oxygen container at -196C, as per heat transfer I calculated a vaporization daily rate of 0.1036% at ATM pressure. if I increase the container pressure to 0.1519 kg/cm2 how much will the vaporization rate decrease?? How can I calculate this? I'm kind of lost, I was trying to use this ecuation but then I think probably is not the right one for my problem. Thank you for the video!!

[@RoxiHulet]

Which equation are you referring to? I don’t do any calculations in this video, and I don’t reference any equations.

[@VEAR31]

@@RoxiHulet sorry, not equation, but specific subject. I've been looking whole internet but I can not find how does pressure affect vaporization rate

[@RoxiHulet]

@@VEAR31 I don’t know if an equation for this exists. How did you calculate the 0.1036% and why are you unable to use the same equation with your new pressure?

[@VEAR31]

@@RoxiHulet Heat Flux per sec (w)= ((surface area * temperature difference) / (insulation Thickness / Insulation Therm. Conductivity)) Daily Vaporization Rate = (Heat Flux * 86400sec) / (Latent Heat of Vaporization / Total Liquid mass) Daily Vaporization Rate = 0.1036% At my understending that is considering ATM pressure, and if I increase the vessel pressure then the Vaporization rate must decrease (At least it is what I understend) and that is what I want to add to the thermal calculation. Thank you very much for your respond

[@RoxiHulet]

@@VEAR31 I’m sorry, I’m not familiar with that equation at all. I think you’ll have better luck reaching out to an engineer or physicist.

[@Obamnaz]

Made good sense of it THANK YOU

[@orlandomartinez4767]

I need more videos