Calculate the amount of work done, in joules, when 2.5 mole of H₂O vaporizes at 1.0 atm and 25^$\circ$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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The work done during vaporization is calculated using the formula W = -PΔV. For 2.5 moles of water vaporizing at 1 atm, the volume change ΔV can be calculated using the ideal gas law: ΔV = nRT/P. Substituting the values (n = 2.5 moles, R = 0.0821 L·atm/mol·K, T = 298 K, P = 1 atm), we find ΔV = 61.1 L. The work done is then W = -1 atm * 61.1 L = -61.1 L·atm. Converting to joules, W = -61.1 L·atm * 101.3 J/L·atm = -6,190 J or -6.19 kJ.

Calculate the Amount of Work Done, in Joules, When 2 ∘\circ∘

Calculate the Amount of Work Done, in Joules, When 2 $\circ$