CH. 96% aromero356@cnm.edu 06/24/18 The vaporization of 1 mole of liquid water (

Question

CH. 96% aromero356@cnm.edu 06/24/18 The vaporization of 1 mole of liquid water (the system) at 100.9°C, 1.00 atm, is endothermic. H,O(1) + 40.7kJ-? H2O(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 30.62 L,respectively. 1st attempt Part 1 (2 pts) i See Periodic Table See Hint Calculate the work done on or by the system when 2.05 mol of liquid H20 vaporizes. Part 2 (2 pts) See Hint Calculate the water's change in internal energy kJ 05/15> SUBMIT

Explanation / Answer

Part 1

Work done by the system = - P (V2 - V1)

= - 1 atm x [(30.62 L) - (18.80 mL x 1L/1000 mL)]

= - 30.6012 atm-L x 101.3 J/L-atm

?W = - 3099.90 J

Part 2

Change in internal energy

?E = ?Q + ?W

For 1 mol liquid to vaporize ?Q = 40.7 kJ

For 2.05 mol liquid to vaporize ?Q = 2.05 x 40.7 = 83.435 kJ

?W = - 3099.90 J x 1kJ/1000 J = - 3.0999 kJ

?E = 83.435 - 3.0999 = 80.3351 kJ

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