Question

what is the volume occupied by 21.6 g of water vapor at 98.0 kPa and 60 degrees Celsius

Answer 1

Use the ideal gas law formula: \(\sf \color{red}{PV = nRT}\) where you are solving for \(\sf \color{blue}{V}\).

Note:
P = pressure [atm]
V = volume [L]
n = mols (convert 21.6 grams of water -H\(_2\)O to mols by dividing by molar mass_
R = gas constant
T = temp [K]

Answer 2

i had done that the answer should be 33.9 L but I was not able tog get that answer

Answer 3

so i had done 21.6 divided by the molar mass which is 18.016 H20 to give me 1.20 moles of h20

Answer 4

Did you convert degrees celcius to kelvin?

Answer 5

yes i did.

Answer 6

Yes did you sub it as n in the formula?

Answer 7

Yes, make sure that you are also using the correct gas constant, R.

Answer 8

There is 8.314 and 0.08314 and both have different units.

Answer 9

should i be using 0.8314?

Answer 10

Yes

Answer 11

That's the gas constant.

Answer 12

i do not understand what i am doing wrong

Answer 13

do i need to change KpA into atm?

Answer 14

Wait @abbot isn't is 8.314?

Answer 15

That's what we use at school- I read it wrong...

Answer 16

do i need to change Kpa into atm?

Answer 17

ok

Answer 18

Does the question require atm or kPA?

Answer 19

I say use kPa that's what we use...

Answer 20

Yes, gas constant is in units of atm.

Answer 21

It's different here, maybe your right @abbot

Answer 22

We don't use atm, it's no longer required where I live.

Answer 23

1 atm = 101.325 kPa.

Answer 24

so this is what i have. 98.0 kPa times v = 1.20 moles times 0.0831 times 333

Answer 25

Therefore, your gas contant, 8.314 or 0.8314 would not be the same if you did the conversion. You need to either convert to atm, or convert 8.314 to units of kPa

Answer 26

i am still not getting the answer.

Answer 27

i ended up with 3.39 has my final answer