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24 July, 01:08

The air in a hot-air balloon at 763 torr is heated from 14.0°C to 31.0°C. Assuming that the moles of air and the pressure remain constant, what is the density of the air at each temperature? (The average molar mass of air is 29.0 g/mol.)

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  1. 24 July, 03:13
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    1.237g/L = density at 14°C

    1.167g/L = density at 31°C

    Explanation:

    Density is the ratio between mass and volume of a determined substance

    From ideal gas law:

    PV = nRT:

    Where P is pressure, V volume, n moles, R constant gas law, and T absolute temperature

    P/RT = n/V

    You can obtain moles / L. now:

    moles / L * (molar mass of the gas (g/mol)

    g/L

    You can obtain the density of the gas in g/L

    763torr are:

    763torr * (1atm / 760torr) = 1.004 atm

    14°C and 31.0°C in absolute temperature are:

    14°C + 273.15 = 287.15K

    31°C + 273.15 = 304.15K

    Replacing, density at 14°C:

    1.004atm / (0.082atmL/molK*287.15K) * 29g/mol = density

    1.237g/L = density at 14°C

    And at 31.0°C:

    1.004atm / (0.082atmL/molK*304.15K) * 29g/mol = density

    1.167g/L = density at 31°C
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