YACHT: i need help with the ideal gas laws.

[MikeMike]

ok, we need to find experimental, and theoretical molar volume for CO2

what i have is.

that we collected .2482 grams of CO2.

now if the ideal gas law, turned into solving for V =

V = (nRTk)/P

so n = m/M
or
n = .2482(g)/40(g/mol)
n = .006205(mol)

so now V = (.006205(mol) * 8.31 (KPa(l)/mol(k)) * 294k)/ P

we had a pressure of 718 mm of Hg, thus we take P = 718(mm hg)/760 (mm Hg/Kpa) * 101.325KPa = 95.725 KPa

now we have

V = (.006205 (mol) * 8.31 KPa l / mol k * 294K)/ 95.725 KPa

after doing the math, we have .158 l which makes no sense wat so ever, as it should be more or less around 24L.

where did i go wrong?

MIKE

 

[Howard]

Can you show me on paper? Reading text is hard...

 

[MikeMike]

bahhhh
give me a second to draw it up in paint.

MIKE

 

[Howard]

Are you sure it's 24L?

 

[Howard]

1 mole of any gas at SATP has a volume of 24.8 L... 95kPa is pretty close to 100kPa and temperature is standard...

The amount of CO2 you have is a lot less than 1 mol. I think your original answer was correct.

 

[Legendary]

Can I ask why you multiplied by 101.325 KPa? that sorta FUBARs your units.

 

[MikeMike]

because it is explained in the pic i am about to show.....

http://www.brownwestern.com/hmwk.gif

MIKE

 

[Fenixgoon]

edit: note how few moles of carbon dioxide you have

STP = 22.4 L... molar mass of CO2 = 44g/mol

those little things will change your values and therefore, answers as well.

and 95kPa is still 6,325 pascals off of 1 atm

294 K is above standard temp (273K)

did you include the ideal gas constant? 0.08206 L atm / mol K

when using the ideal gas law, your pressure value must be in atmospheres to divide correctly

without using sig figs until the end, i got 0.144 liters

by the way, i;m curerntly in AP chemistry.. just in case you were wondering if i know what im doin lol

 

[MikeMike]

I AM TEH RETRADED

i should use one mol of it instead of what our masurement was.

who wants the honors of placing the dunce cap, and beating me with a long hard rod?
sorry michael jackson, i dont meet the requirement to be under 15, and you dont meet the long requirement.

with one mol, the answer becombes 25.52L

MIKE

 

[Eli]

Originally posted by: nourdmrolNMT1
I AM TEH RETRADED

i should use one mol of it instead of what our masurement was.

who wants the honors of placing the dunce cap, and beating me with a long hard rod?
sorry michael jackson, i dont meet the requirement to be under 15, and you dont meet the long requirement.

MIKE

HA!

 

[Howard]

Yeah, that's what I figured. And a 6kPa difference is pretty damn small, whoever pointed that out!