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31 December, 16:23

During the time 0.325 mol of an ideal gas undergoes an isothermal compression at 22.0∘c, 352 j of work is done on it by the surroundings. part a if the final pressure is 1.76 atm, what was the initial pressure?

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  1. 31 December, 17:16
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    According to the equation of the work done W:

    When W = nRT㏑ (Pi/Pf) and we have:

    W = 352 joule & n = 0.325 & T by kelvin = 22+273 = 295 K & Pf = 1.76

    by the substitute:

    352 = (0.325 x 8.314 x 295) ㏑ (Pi/Pf)

    ∴㏑ (Pi/Pf) = - 352 / (0.325 x 8.314 x 295)

    ㏑ (Pi/Pf) = - 0.44

    Pi/Pf = 0.644

    ∴ Pi = 1.76 x 0.644 = 1.13 atm
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