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20 April, 20:50

What is the temperature of 0.5 moles of water vapor that occupies 120 dm3 and applies a pressure of 15,000 Pa to its container?

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  1. 20 April, 20:52
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    Use the ideal gas law:

    PV = nRT

    so, T = PV / nR

    n=0.5

    V = 120 dm^3 = 120 L (1 dm^3 = 1 L)

    R = 1/12

    P = 15,000 Pa = 0.147 atm (1 pa = 9.86 10^{-6})

    Put the values:

    T = PV / nR

    T = (0.147) (120) / (0.5) (1/12)

    T = 426 K
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