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16 July, 14:36

Gas is confined in a tank at a pressure of 9.8 atm and a temperature of 29.0°C. If two-thirds of the gas is withdrawn and the temperature is raised to 83.0°C, what is the pressure of the gas remaining in the tank? 9.1e-21 Incorrect: Your answer is incorrect.

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  1. 16 July, 15:29
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    3.9 atm

    Explanation:

    Ideal gas law states:

    PV = nRT

    where P is absolute pressure,

    V is volume,

    n is number of moles,

    R is universal gas constant,

    and T is absolute temperature.

    The volume of the tank is constant, so we can say:

    V₁ = V₂

    Using ideal gas law to write in terms of P, n, R, and T:

    n₁ R T₁ / P₁ = n₂ R T₂ / P₂

    n₁ T₁ / P₁ = n₂ T₂ / P₂

    Initially:

    P₁ = 9.8 atm

    T₁ = 29.0°C = 302.15 K

    n₁ = n

    Afterwards:

    P₂ = P

    T₂ = 83.0°C = 356.15 K

    n₂ = n/3

    Substituting:

    n (302.15 K) / (9.8 atm) = (n/3) (356.15 K) / P

    (302.15 K) / (9.8 atm) = (1/3) (356.15 K) / P

    P = 3.85 atm

    Rounding to 2 significant figures, P = 3.9 atm.
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