Tuesday, September 27, 2011

Blog 7 Due October 3, 2011

Suppose two 200 L tanks were filled separately with helium and hydrogen gas. What mass of each gas is needed to produce a pressure of 135 atm in its respective tank at 24 degrees Celsius?
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12 comments:

  1. Varak ArmoudikianSeptember 29, 2011 at 1:03 AM

    PV=nRT
    hydrogen= (135atm)(200L)=(n)(8.31)(297K)
    n=10.9mol*2.02gH2/1mol= 22.01g H2

    helium= (135atm)(200L)=(n)(8.31)(297K)
    n=10.9mol*4.002gHe/1mol=43.62g He

    ReplyDelete
  2. NayiriSeptember 30, 2011 at 2:51 PM

    given: Volume: 200 L
    Pressure: 135 atm
    Temperature: 24 C +273= 297 K
    PV=nRT
    n= PV/RT
    n= (135)(200)/ (.08206)(297)
    n= 1107.84 mol

    Hydrogen: 1107.84 mol * (1.01g / 1 mol) = 1118.92 g Hydrogen

    Helium: 1107.84 mol * (4.003 g / 1 mol)= 4434.67 g Helium

    ReplyDelete
  3. AnonymousSeptember 30, 2011 at 4:03 PM

    PV= nRT
    for Hydrogen...
    n= (PV/RT)= (135 atm)(200 L)/(0.08206 L*atm/K*mol)(24+273.15 K)= 1107.28 mol Hydrogen (g)
    for Helium...
    n= (PV/RT)= (135 atm)(200 L)/(0.08206 L*atm/K*mol)(24+273.15 K)= 1107.28 mol Helium (g)

    1107.28 mol H x (1.01 g H/1 mol H)= 1118.35 g Hydrogen (g)

    1107.28 mol He x (4.0026 g He/1 mol He)= 4432.00 g Helium (g)

    ReplyDelete
  4. AlecOctober 1, 2011 at 8:22 AM

    PV=nRT
    (135atm)(200L)/(.082)(24+273.15K)=
    1107.52*1.01=
    1118.02gH

    1107.52*4.003=4431.97gHe

    ReplyDelete
  5. AnonymousOctober 1, 2011 at 1:27 PM

    a) PV=nRT
    n=PV/RT
    n=(135 atm*200 L)/[(0.08206 L*atm/mol*K)(297 K)]
    n=1107.8 mol H2
    1107.8 mol H2*(2.02 g H2/1 mol H2) =2237.8 g H2

    b) PV=nRT
    n=PV/RT
    n=(135 atm*200 L)/[(0.08206 L*atm/mol*K)(297 K)]
    n=1107.8 mol He
    1107.8 mol He*(4.003 g He/1 mol He) =4434.5 g He

    ReplyDelete
  6. Hasmik.October 1, 2011 at 4:23 PM

    PV = nRT

    (200 L) (135 atm) = n (.080206) (297 K)
    n = 1.10 x 10^3 moles

    Hydrogen:
    1.13 x 10^3 moles x 1.008 grams H / 1 mole H
    = 1.14 x 10^3 grams of Hydrogen

    Helium:
    1.10 x 10^3 moles x 4.003 grams He / 1 mole He
    = 4.54 x 10^3 grams of Helium

    ReplyDelete
  7. varak ghazarianOctober 2, 2011 at 7:18 PM

    We can solve the problem using PV=nRT
    (135 atm) (200 L) = n (.080206) (297 K)
    n = 1107.28 mol
    1107.28 mol H x 1.01 g H= 1118.35 g H
    If you use PV=nRT you get the same answer for helium.
    1107.28 mol He x 4.003 g He= 4432.00 g He

    ReplyDelete
  8. Alexa KovacsOctober 2, 2011 at 9:13 PM

    PV=nRT
    n=PV/RT

    n= (135 atm)(200 L)/(0.08206)(297 K)
    n=1.11*10^3 mol

    Hydrogen
    1.11*10^3 mol H* 1.01 g H/1 mol H= 1.12*10^3 g hydrogen gas

    Helium
    1.11*10^3 mol He*4.00 g He/1 mol He= 4.44*10^3 g helium gas

    ReplyDelete
  9. AnonymousOctober 2, 2011 at 10:01 PM

    PV=nRT
    PV/RT=n

    H2) n=(135atm)(200L)/(.08206)(297K)
    n=1107mol
    1107molH2 x 2.02g/mol= 2236gH2

    He) n=(135atm)(200L)/(.08206)(297K)
    n=1107mol
    1107molHe x 4.00g/mol= 4428gHe

    ReplyDelete
  10. AnnieOctober 3, 2011 at 8:05 AM

    n=(P*V)/(R*T)
    n=(135*200)/(.08206*297)
    n=1107.84 mol
    1107.84 mol * (2.02 g H2/1 mol) =2237 g H2
    1107.84 mol * (4.00 h He/1 mol)=4431 g He

    ReplyDelete
  11. Garni TerterianOctober 3, 2011 at 7:21 PM

    PV=nRT
    (200 L) (135 atm) = n (.080206) (297 K)
    n = 1107.84 moles
    Hydrogen : (1107.84 mol)*(2.02 g/1 mol) = 2237 g
    Helium: (1107.84 mol)*(4.00g/1 mol) = 4431 g

    ReplyDelete
  12. Shushan GinosyanOctober 3, 2011 at 8:07 PM

    (135atm)(200L)=(n)(8.31)(297K)
    n= 1107.84mol He and H each
    1107.84molHe*(4.00gHe/1molHe)= 4*10^3 g He
    1107.84molH*(1.008gH/1molH)= 1*10^3 g H

    ReplyDelete

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