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7 February, 04:20

A sample of gas contains 6.25 * 10-3 mol in a 500.0 mL flask at 265°C. What is the pressure of the gas in kilopascals?

Which variables are given?

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  1. 7 February, 04:33
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    Answer: 55.7 kPa

    Explanation:

    Given variables are:

    Volume of gas (V) = 500.0mL

    convert volume in milliliters to liters

    (since 1000mL = 1L

    500.0mL = 500.0/1000 = 0.5L)

    Temperature (T) = 265°C

    Convert temperature in celsius to Kelvin

    (265°C + 273 = 538K)

    Pressure (P) = ?

    Number of mole (n) = 6.25 x 10^-3 moles

    - Molar gas constant (R) is a constant with a value of 0.0821 atm dm3 K-1 mol-1

    Then, apply ideal gas equation

    pV = nRT

    p x 0.5L = 6.25 x 10^-3 mole x (0.0821 atm L K-1 mol-1 x 538K)

    0.5L•p = 0.276 atm L

    Divide both sides by 0.5L

    0.5L•p/0.5L = 0.276 atm L/0.5L

    p = 0.55 atm

    The pressure in atmosphere is 0.55. So, convert it to kilopascal (kPa)

    Since 1 atm = 101.325 kPa

    0.55 atm = Z

    cross multiply

    Z x 1 atm = 101.325 kPa x 0.55 atm

    Z = (101.325 kPa x 0.55 atm) / 1 atm

    Z = 55.7 kpa

    Thus, the pressure of the gas is 55.7 kilopascals
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