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25 July, 16:08

Many laboratory gases are sold in steel cylinders with a volume of 43.8 L. What mass (in grams) of argon is inside a cylinder whose pressure is 17615kPa at 23∘C?

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  1. 25 July, 16:29
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    It's very simple ... if we remember value of Universal Gas Constant R and Ideal Gas Law, so ...

    Ideal Gas Law

    pV = nRT, where:

    p - pressure (in kPa),

    V - volume (in L),

    n - number of moles (in mol),

    R - universal cas constant (in kPa * L / mo l * K),

    T - temperature (in K)

    n = m/M, where:

    n - number of moles,

    m - mass (in grams),

    M - molar mass of ingredient (in g/mol) - you find this at Periodic Table.

    pV = nRT - - - > pV = mRT/M - - - > pVM = mRT - - - > pVM/RT = m

    p = 17615 kPa

    T = 273.15 + 23 = 296.15 K

    V = 43.8 L

    R = 8.314 kPa * L / mol * K

    M (for argon) = 39.948 g/mol

    and

    m = (17615 kPa * 48.3 L * 39.948 g/mol) / (296.15 K * 8.314 kPa * L / mol * K)

    m = 13803.93 grams of Argon
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