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Is that enough information?
pV=nRT applies only to gases.
The bottom It should say you filled the bottom of the can with 3ml of water.. So Can I ask do you know how you would solve for it?
The lab above just states the info: Soda can filled with 3ml of water empty one, and this is a question not sure if they go together or not but here is a question. 1.)How many moles of water vapor were in the can? Assume pressure is 1atm and temperature is 100C Hint* Use PV=nRT as you said. Just lost to how to work this?
@373K the water would boil, start to evaporate, but that's undergoing a phase change (g) to (l) so those phases would be in equilibrium. I guess my issue is that you could say PV/RT = n but that's only if the volume V is the volume occupied by the gas, not liquid, that's issue of contention.
So if you were to go to work this problem how do you think I should do it? Having the info I know?
they didn't give you the total volume of the can?
Well its just a soda can? Do you need to know the ml? or what exactly
they tell me its 680ml
ok that changes it
@373K the water would boil, start to evaporate, but that's undergoing a phase change (g) to (l) so those phases would be in equilibrium. so when the vapor pressure equals the external pressure the substance would start to boil. so you're given 1 atm. so that's the pressure you use bc that's equal the vapor pressure. (so you know 1atm, R, and 373K) I guess my issue is that you could say PV/RT = n but that's only if the volume V is the volume occupied by the gas, not liquid, that's issue of contention. so you would need to find the number of molecules of water in the gaseous phase. but I think you need the total volume of the container. (1atm)(677/1000)/(0.08x373) = 0.0227 mol H2O
I don't think that it matters whether you have filled the bottom with 3mL of water b/c for the ideal gas law you can only use with gases not liquids. I asked for total volume b/c you need that to figure out what the number of moles of gas are in the container. you know that once it boils there's a certain fraction of h2o gas molecules, which they want you to find.
Strange thing is they don't tell me it? It is for summer school I found that strange too?
So you would say try (1atm)(677/1000)/(0.08x373) = 0.0227 mol H2O as the answer and work?
the key is that when a substance boils the vapor pressure = external pressure and they gave you 1 atm,
sorry use 680mL sorry that's a typo not 677.
the key is that they're asking for the number of moles of vapor, it's undergoing a boiling so not all 3mL of that liquid would have evaporated right away
Yes that is true, so you did (1atm)(677/1000)/(0.08x373) = 0.0227 mol H2O and what is 0.08 and the 373? where are you getting those from and then 0.0227 being your answer is it?
oh so you had 100 degrees celsius, so it has to be in kelvin (C+273) = k so 100C+273 = 373. 0.08 is the Gas constant units mol L atm K and I converted 680mL to liters so 0.68L
and your pressure is in atmospheres atm which is just 1 atm
so all together it's 0.68L(atm)/(0.08L mol atm K^-1)(373K) = 0.0227mol notice that the units cancel out
Okay got it, and there is a few other questions you may be able to help me a bit. If you want?
I can spare a few
2.) How many grams of water vapor were in the can? Hint convert moles of water vapor into grams of water vapor (1mole=18.02grams) @Photon336
Okay so the question says, (convert moles of water vapor into grams of water vapor) so whenever you want to convert something from moles to grams you do the following: moles x ( grams/mol) (notice when you multiply them that because moles x (1/moles) cancels out. so for our purpose you do the following 0.0227 mol H2O x (18.02g/mol) moles cancel out and you have 0.0227x18.02 = 0.408g of H20 are in the gaseous state.
Okay thanks and there is just one more I don't understand, its How much volume did the water vapor take up when it condensed into a liquid? Hint (H2O= 1mL H2O)@Photon336
Okay so we have 3.0ml of water. we know that we used 0.4 grams of water roughly was evaporated.
so since the density of water is about 0.9718g/ml and we have 0.4grams in of water in the vapor phase that's 0.9718 (g/ml) x (0.4)g = 0.4 mL of water so 0.4/1000 = 4x10^-4 Liters
So 4x10^-4 Litrs would be your final answer?