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30 December, 12:32

Temperature of 1.4 moles of gas's with pressure of 3.25 atmosphere in a 4.738-liter container

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  1. 30 December, 15:07
    0
    -139 °C

    Step-by-step explanation:

    We can use the Ideal Gas Law and solve for T.

    pV = nRT Divide each side by nR

    T = (pV) / (nR)

    dа ta:

    p = 3.25 atm

    V = 4.738 L

    n = 1.4 mol

    R = 0.082 06 L·atm·K⁻¹mol⁻¹

    Calculation:

    T = (3.25 * 4.738) / (1.4 * 0.082 06)

    T = 134 K

    T = (134 - 273.15) °C

    T = - 139 °C
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