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11 June, 05:43

A sample of fluorine gas occupies 810 milliliters at 270 K and 1.00 atm. What volume does the gas occupy when the pressure is doubled, and the temperature increases to 400 K

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Answers (2)
  1. 11 June, 05:54
    0
    The answer to your question is Volume = 600 ml

    Explanation:

    Data

    Volume 1 = 810 ml

    Temperature 1 = 270°K

    Pressure 1 = 1 atm

    Volume 2 = ?

    Pressure = 2 atm

    Temperature 2 = 400°K

    Process

    1. - To solve this problem, use the Combine gas law.

    P1V1 / T1 = P2V2 / T2

    - Solve for V2

    V2 = P1V1T2 / T1P2

    2. - Substitution

    V2 = (1) (810) (400) / (270) (2)

    3. - Simplification

    V2 = 324000 / 540

    4. - Result

    V2 = 600 ml
  2. 11 June, 06:10
    0
    600mL

    Explanation:

    The following were obtained from the question:

    V1 = 810mL

    P1 = 1atm

    T1 = 270K

    V2 = ?

    P2 = 2atm (since the pressure is doubled)

    T2 = 400K

    The new volume can be obtained by doing the following:

    P1V1/T1 = P2V2/T2

    1 x 810/270 = 2 x V2 / 400

    Cross multiply to express in linear form

    270 x 2 x V2 = 810 x 400

    540 x V2 = 324000

    Divide both side by 540

    V2 = 324000/540

    V2 = 600mL

    The new volume of the gas is 600mL
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