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26 February, 09:11

A pocket of gas is discovered in a drilling operation. The temperature of the gas is 498º C and its pressure is 13.8 atm. At the surface, the same gas has a volume of l8.0 L at 29º C and 2.00 atm. How large was the pocket of gas originally?

and

A balloon with a pressure of 2.2 atm and a volume of 2.5 L is released in the mountains of Colorado. As the balloon ascends, the volume of the balloon gets noticeably larger and increases to a volume of 4.0 L. What is the new pressure of the balloon?

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  1. 26 February, 09:27
    0
    The answer to your question is below

    Explanation:

    1)

    Data

    Temperature 1 = 498°C

    Pressure 1 = 13.8 atm

    Volume 1 = ?

    Temperature 2 = 29°C

    Pressure 2 = 2 atm

    Volume 2 = 18 L

    Process

    1. - Convert temperature to °K

    Temperature 1 = 498 + 273 = 771 °K

    Temperature 2 = 29 + 273 = 302 °K

    2. - Use the combine law of gas

    P1V1/T1 = P2V2/T2

    Solve for V1

    V1 = T1P2V2 / P1T2

    Substitution

    V1 = (771) (2) (18) / (13.8) (302)

    Simplification

    V1 = 27756 / 4167.6

    Result

    V1 = 6.66 L

    2)

    Use Boyle's law to solve this problem

    P1V1 = P2V2

    Data

    P1 = 2 atm

    V1 = 2.5 L

    P2 = ?

    V2 = 4 L

    - Solve for P2

    P2 = P1V1/V2

    - Substitution

    P2 = (2) (2.5) / 4

    - Simplification

    P2 = 5 / 4

    - Result

    P2 = 1.25 atm
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