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How much work is required to break up 1 mole of water at 20°C into spherical droplets of radius 1 cm? Given that gamma(H_(2)O) =72.75 xx 10^(-3) Nm^(-1),density (H_(2)O)=19.015g/mole. |
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Answer» Solution :SUPPOSING 1 mole of water to be spherical, volume of 1 mole of water `=(18.015)/(0.998 xx 10^(6))` `=1.8047 xx 10^(-5) m^(3)` `therefore r= root(3)((3 xx V)/(4pi)) ((3 xx 1.8047 xx 10^(-5))/(4 xx 3.14))^(1//3) = 1.6272 xx 10^(-2) m` Surface AREA `=4pir^(2) = 4 xx 3.14 xx (1.6272 xx 10^(-2))^(2)` `=3.3273 xx 10^(-3) m^(2)` Volume of each droplet= `4/3 xx pi xx (1 xx 10^(-3))^(3) ~~ 4 xx 10^(-6) m^(3)` Number of droplets `=(1.8047 xx 10^(-5))/(4 xx 10^(-6)) =5` Surface area of 5 droplets `=5 xx 4pir^(2)` `=5 xx 4 xx 3.14 xx (1 xx 10^(2))^(2)` `=6 xx 10^(-3) m^(2)` When a DROP of 1 mole of water breaks into 5 droplets, INCREASE in surface area `=(6 xx 10^(-3) -3.3273 xx 10^(-3)) m^(2)` `=2.6727 xx 10^(-3) m^(2)` Work DONE to break up 1 mole of water=surface tension `xx` increase in area `=72.75 xx 10^(-3) xx (2.6727 xx 10^(-3)) J` `=1.94 xx 10^(-4) J (1 J = 1 Nm)` |
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