xam 2 A Fall 2014 KEY-105TSCHE 002 Hald open suny CH5HW 65% facey106@canton.e 10

Question

xam 2 A Fall 2014 KEY-105TSCHE 002 Hald open suny CH5HW 65% facey106@canton.e 10/30/17 H20() + 40.7kJ-->M20(g) Assume at exactly 100.0°C and 1,00 atm total pressure, 1.00 mole of liquid water and 1.00 mole of water vapor occupy 18.80 mL and 30.62L respectively 2nd attempt Part 1 (1 point) l See Periodic Table See Hint Calculate the work done on or by the system when 4.85 mol of liquid H2O vaporizes. x 104 15 Part 2 (1 point) OSee Hint Calculate the water's change in internal energy 05,14 > VIEW SOLUTION 9 OF 14 QUESTIONS COMPLETED

Explanation / Answer

Work.done W = P delta V

= 1 (0.03062- 0.01880L)

= 0.01182 atm.L

= 0.01182*101.33 J

= 1.20 J

For 4.85 mole = 4.85* 1.2 = 5.82 J

As work done by the system so.value is negative.

Now , change in internal.energy= W + q

= [-5.82 + 4.85(40700) ]

= 197389.2 J

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