Thank you so much man. May I ask why the partial pressure of vapor (or water cause it is the same) is independent of how much liquid that has evaporated to vapor (like in this problem there are only a few grams of vapor when heating to 100 C). Cause a few grams of vapor do not occupy that much...
Wow finally, thank you so much! Last questions: Do I just clarify my answer that because at 100 degrees celsius by Pv-diagram we see that the state of water at this temperature is saturated liquid/vapor and this only occurs at the saturation pressure, therefore the partial pressure of vapor...
Yeah my bad, it was around 133 kPa for pressure of air at 100 C. When it comes to the amount of water vapor, M=PV/RT right? Where P equals saturated pressure and V = 2L then? But the most important question: Is the pressure going to be saturated pressure (1 atm ) at 100 C + 133 kPa which is the...
I got around 150 kPa for pressure of air at 100 C. So the total pressure inside the box is 150 + 100 kPa, aka the saturated pressure + pressure of air at 100 C?
But what is the total pressure then? Pressure at 100 C of air + the saturation/vapor pressure at 100 C? Can’t the amount of liquid in vapor be calculated by ideal gas eq?
So what I got for air pressure is the relation P1/T1=P2/T2. Do I just add this with the saturated pressure of water vapor? And what about the liquid that is left? Is it independent of how much water liquid that has vaporized? I am so confused
I might sound stupid here but will the partial pressure of water at 100C be the saturated pressure? And according to my calculations, the mixture between air+ vapor is 3.029 liters I think. Pressure of air at 100 C = RT/V where V is still 5 L and T=100 C right?
Homework Statement
A five-liter pot is filled with 2 l of liquid water and 3 l of dry air at the ambient pressure of 1 atm and temperature 10◦C. Subsequently, the pot is closed and heated up. What is the pressure inside the pot going to be when the temperature reaches 100◦C?
Homework...