Calculate the mass in grams of CO2 gas which will occupy 100cm3 at 250 degrees C and 5 atmospheres pressure?

2 Answers

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  • 6 years ago
    Favorite Answer

    Let’s convert the volume to liters by dividing by 1000. V = 0.1 L

    Convert the temperature to ˚K. T = 250 + 273 = 523

    P * V = n * R * T

    5 * 0.1 = n * 0.08205 * 523

    n = 0.5 ÷ 42.91215 = 0.011651712

    This is the number of moles of CO2.

    Mass = Number of moles * Mass of 1 mole

    Mass of 1 mole of CO2 = 12 + 32 = 44 grams

    Mass = (0.5 ÷ 42.91215) * 44 = 0.512675315 grams

    The mass is approximately 0.513 grams

    Let’s use the following equation to determine the volume at STP.

    P1 * V1/T1 = P2 * V2/T2

    5 * 100/ 523 = 1 * V2/273

    V2 = 273 * 5 * 100/ 523 = 136,500/523 cm^3

    At STP, the volume of 1 mole of a gas is 22.4 liters. Let’s convert the volume to liters by dividing by 1000.

    V = 136,500/ 523 ÷ 1000 = 136.5/523 = 0.260994264 L

    Use the following equation to determine the number of moles of CO2.

    136.5/523 ÷ x = 22.4/1

    x = 136.5/523 ÷ 22.4 = 0.01165153 moles of CO2

    Mass = Number of moles * Mass of 1 mole

    Mass of 1 mole of CO2 = 12 + 32 = 44 grams

    Mass = (136.5/523 ÷ 22.4) * 44 = 0.512667304 grams

    Since both of these answers are same, I believe the answer is correct.

  • donpat
    Lv 7
    6 years ago

    Hello Confused?! ....

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    Apply the Ideal Gas Law :

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    PV = nRT = ( m /M ) ( R ) ( T )

    m = ( P ) (V ) ( M ) / (R ) ( T )

    V = ( 100 cu cm ) ( 1 cu m / 10^6 cu cm ) ( 1000 L / 1 cu m ) = 0.1000 L

    T = 250.0 + 273.2 = 523.2 K

    m = ( 5 atm ) ( 0.1000 L ) ( 44.01 g/gmol ) / ( 0.08205 atm - L/gmol - K ) ( 523.2 K )

    m = 0.5126 g <------------------------------------------------------

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