Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Astrid on February 25, 2013, 09:10:49 PM
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Hi everyone, I just want to see if I did this problems correctly.
This is the question:
A mass of helium is needed to fill a 5.5 L balloon. The temperature is 25 degrees C. The pressure is 751 torr. What is the mass?
- I assume you use the equation PV = nRT
I solved for N, since everything else is known.
I came up with:
n = PV/RT
= (751 torr) (5.5 L) / (0.08206 L-atm/mol ) (25 degrees C)
= 2013.4 g
I feel like I did something wrong... Do I need to find the molar mass of helium first or something?
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Just at first look, your units are inconsistent. Units of R should be L atm K−1 mol−1, and the pressure and temperature also have wrong units. Once all the units cancels out, the answer should have units of mole and not gram.
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Oops, I did type that one part wrong. (With the units for R). But the question itself just asks for pressure in torr. I think that's the way I'm supposed to do it.
I'm confused about where I get the moles from. No molar mass is given, and there's no grams to multiply by to get the molar mass. The most important thing is .. Did I rearrange the equation right to solve this problem?
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Yes you have rearranged the equation correctly, don't need to worry about that. You need to convert 'torr' into 'atm' (because the units of R have 'atm' in it and not 'torr'), similarly R has units of K and not oC if you know what I mean. You just need to plug these figures into the equation. Note than R also has a 'mol' unit in it and not 'g'
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Alright, thanks a lot. I think I know what you mean. I'm going to convert C to K and work on this a bit more. I think I understand now though!
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Take molar mass of He from the periodic table.
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First you must rember that n is equal to mass over molar mass (µ), mu
n = m/µ
PV=nRT
Substituting the equation we get PV=m/µ RT
So to get the mass we derive m as follows: m=PVµ/RT
Solving for the mass of helium:
Step 1: Molar mass of He is 4.00 g/mol
Step 2: Convert 751 torr to atm
751 torr(mmHg/torr)(1atm/760mmHg) = 0.988 atm
Srep 3: Convert Celcius to Kelvin
K = 25+273.15 = 298.15K
Step 4: Substitute and solve for the mass of Helium
m=PVµ/RT
m = (0.988 atm)(5.5L)(4.00g/mol) / (0.0821 L-atm/K-mol)(298.15K)
m = 0.89 g Helium
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You can self check your own work by doing a ball park check or reality check.
Once you have a ball park number in your mind you can compute the precise value and see how close.
You know that
At 1 Atmosphere of pressure (1 ATM or 760 Torr) and 0 degrees Celsius a mole of Ideal Gas occupies about 22.4 Liters.
You also know that
One mole of Helium is about 4 grams.
We can loosely estimate
relatively close to 1 ATM and 0 C
about one quarter (.25) the volume of a mole at 1 ATM and 0 C
that we likely have around 1 gram of Helium (.25 of 4 grams)
So we would expect something near 1 gram
2013.4 g does not look promising
0.89 g looks promising
Side note
For myself, seeing that the temperature is warmer than 0 C and the pressure is less than 760 Torr
I would expect something less than 1 gram since the gas is "thinner"
but that is just me thinking n=PV/RT algebraically
holding everything constant except pressure and moles -- pressure decrease moles decrease
holding everything constant except temperature and moles -- temperature increase moles decrease