Calculate the amount of work done, in joules, when 2.5 mole of H₂O vaporizes at 1.0 atm and 25°C.Assume the volume of liquid H₂O is negligible compared to that of vapor.(1 L·atm = 101.3 J) A)61.1 J B)518 J C)5.66 kJ D)6.19 kJ E)6,190 kJ

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Quizplus · Accepted Answer

The work done during vaporization is given by the equation W = -PΔV. Since the volume of liquid water is negligible compared to that of water vapor, we can assume that the change in volume is equal to the volume of water vapor produced. The volume of water vapor can be calculated using the ideal gas law: PV = nRT. Substituting this into the equation for work, we get:W = -P(nRT/P) = -nRTPlugging in the given values, we get:W = -(2.5 mol)(8.314 J/mol·K)(298 K) = -5.66 kJConverting to joules, we get:W = -5.66 kJ * 1000 J/kJ = -5660 JTherefore, the amount of work done is 5660 J, which is closest to choice D.

Calculate the Amount of Work Done, in Joules, When 2