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Homework Help: Ideal Gas Laws equation help

  1. Feb 29, 2004 #1
    I have a few homework problems on the Ideal Gas Laws, was wondering if someone could help me out with a few of the problems.

    1) Hemoglobin has a molecular mass of 64 500 u. Find the mass (in kg) of 873 molecules of hemoglobin.

    Here's what I did: (64500 g/mol/6.022x10^23 mol^-1)(873 kg/1000 g)=9.35x10^26 kg but the computer program I use tells me it's wrong...can u find the mistake cause I don't know where it is... :frown:

    2) A cylindrical glass of water (H2O) has a radius of 6.01 cm and a height of 13.9 cm. The density of water is 1.00 g/cm3. How many moles of water molecules are contained in the glass?

    Not sure how to work this one out, I was thinking of using V=pi*r^2 and n=N/NA

    3) Suppose that a blimp contains 5020 m3 of helium (He) at an absolute pressure of 1.10 x 105 Pa. The temperature of the helium is 276 K. What is the mass (in kg) of the helium in the blimp?

    Should I use pV=nRT for this problem?

    4) A bicycle tire whose volume is 4.1 x 10-4 m3 has a temperature of 283 K and an absolute pressure of 3.76 x 105 Pa. A cyclist brings the pressure up to 7.15 x 105 Pa without changing the temperature or volume. How many moles of air must be pumped into the tire?

    Does the equation P1V1/T1=P2V2/T2 sound right?
  2. jcsd
  3. Feb 29, 2004 #2


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    Think about #1. Obviously a few hundred molecules of hemoglobin are not going to weigh as much as the Sun!

    - Warren
  4. Feb 29, 2004 #3


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    873 molecules (1 mol/ 6.20*10^23 molecules) (64500 g /mol) (1 kg / 1000 g)

    Is what I get. It's hard to follow your grouping because you stacked the fractions, I think you should be getting 10^-22 instead of 10^26.
  5. Feb 29, 2004 #4
    For #1, even if I type in 9.3504x10^-22 it still says it's wrong....grrrrrrrrr!! I thought I had the right answer... :frown:
  6. Feb 29, 2004 #5


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    Try using fewer significant digits.

    - Warren
  7. Feb 29, 2004 #6


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    "22" IS incorrect. Take five, watch a little TV, then do the calculation one step at a time rather than all at once.

    2)? Yeah --- pay attention to the details as in 1).

    3)? Can you think of an approach that doesn't use the ideal gas?

    4)? Look at the definitions for the terms in the ideal gas expression; ask yourself what is happening when you pump up a bicycle tire --- i.e., what remains constant and what changes.
  8. Mar 1, 2004 #7
    Ok, I'm just about to give up here. For the first problem I've tried the answer 9.3505x10^-22, 9.4x10^-22, 9.35x10^-22 AND it's still saying that the answer is wrong. Do I have the right numbers or is the computer program going nuts?!? Also, for the third problem I used the equation n=PV/RT and I got the answer 1.834x10^10. Is this right? Ahhhhhh please help!!!!
  9. Mar 1, 2004 #8
    Actually for the third problem I figured out the answer to equal 7.3407684x10^13 but the program is saying the answer is wrong! Ahhhhhhhhhhh, I'm going to go insane!!!

  10. Mar 1, 2004 #9
    1. 1 u = 1.66 x 10^-27 kg
    therefore 64 500 u = 1.02707 x 10^-22 kg

    2. mass = volume x density
    = pi(r^2)h x 1 kg/m^3
    = 1.58 x 10^-3 kg

    no. of moles of H20 = mass/molar mass = 8.76 x 10^-5
    => no. of molecules = 8.76 x 10^-5 x 6.023 x 10^23
    = 5.3 x 10^19 molecules

    3. n = pv/RT = (1.1x10^5 * 5020)/(8.31 * 276) = 240 761 moles
    mass = molar mass x n
    mass = 963 kg

    4. n(1) = [P(1)V]/[RT] and n(2) = [P(2)V]/[RT]
    so delta n = n(2) - n(1) = [V/RT][P(2) - P(1)]
    delta n = [(4.1x10^-3)/(8.31*283)]*[(7.15 - 3.76)*10^5]
    delta n = 0.59 moles
  11. Mar 1, 2004 #10
    Ok, I found out the answer to the first and last problems but I'm not sure how they got the answers. Can anyone explain it to me.

    For the first question the answer is 9.4x10^20 kg

    For the last question the answer is 0.06 mol

    Thanks for all the help! :smile:
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