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10 January, 19:38

What volume would 8.0g helium gas occupy at 0.0 c and 1.0 atm?

How many moles of gas molecules would you expect to find in a 2.0-L bottle of air at 0.0 C and 1.0 atm?

What would be the mass of gas inside a 3.0-L balloon at 0 C and 1.0 atm filled with

a. He (g) ? b. Co2 (g) ? c. CH4 (g) ?

How many moles of oxygen gas would be present in a 1.0-mL gas sample at 0.0 C and 1.0 atm?

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Answers (1)
  1. 10 January, 20:06
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    By using the ideal gas formula:

    1) TO get the volume: from the ideal gas formula:

    when PV = nRT

    when we have P = 1 & R = 0.0821 & T = 0°C+273 = 273 K

    and n = mass / molar mass of He

    = 8 g / 4 g/mol = 2 mol

    so by substitution:

    1 atm * V = 2 mol * 0.0821 * 273

    ∴V = 44.8 L

    2) To get moles of gas molecules we will use also the ideal gas formula:

    PV = nRT

    when P = 1 atm & V=2 L & R = 0.0821 & T = 0+273 = 273 K

    so by substitution:

    1atm * 2 L = n * 0.0821 * 273

    ∴n (the no. of moles) = 0.089 mol

    to get the no. of moles of gas molecules we will multiply by avogadro's number:

    = 0.089 * (6.02*10^23) = 5.37X10^22 molecules

    3) first we have to get n:

    PV = n RT

    when P = 1 atm & V = 3L & R = 0.0821 & T = 0+273 = 273 K

    so by substitution:

    1atm * 3L = n * 0.0821 * 273

    ∴n the no. of moles of the gas = 0.134 mol

    a) when the gas is He so the mass can be calculated from this formula:

    He Mass = no. of moles * molar mass of He

    = 0.134m * 4g/mol

    = 0.5 g

    b) when the gas is CO2 So the mass can be calculated from this formula also:

    CO2 Mass = no. of moles * molar mass of CO2

    =0.134mol * 44 g/mol

    = 5.896 g

    c) when the gas is CH4 So the mass can be calculated from this formula:

    CH4 Mass = no. of moles * molar mass of CH4

    = 0.134mol * 16 g / mol

    = 2.144 g

    4) by using also the ideal gas formula we can get the moles of O2:

    PV = nRT

    when P = 1 atm & V = 0.001 L & R = 0.0821 & T = 0+273 = 273 K

    so by substitution:

    1atm * 0.001 L = n * 0.0821 * 273

    ∴ n = 4.5X10^-5 mol
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