O See page 3 Question (2 points) The vaporization of 1 mole of liquid water (the

Question

O See page 3 Question (2 points) The vaporization of 1 mole of liquid water (the system) at 100.9°C, 1.00 atm, is endothermic. H,0(1) + 40.7kJ H,Olg) 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 dit See Periodic Table O See H Part 1 (1 point) Calculate the work done on or by the system when 1.25 mol of liquid H2O vaporizes. SUBMIT AN 9 OF 13 QUESTIONS COMPLETED arch o o D e 9 a 1 e a XPS

Explanation / Answer

Temperature = 100 0C

Pressure = 1 atm

volume at 100 0C = 18.8 mL for liquid water

Volume at 100 0C = 30.62 mL for water vapour

Workdone w = -2.303nRT log (V2/V1)

W = -2.303 * 1.25 * 8.314 * (100+273) log(30.62/18.8)

w = -1891.23 J

here temperature is constant so it is isothermal process then dE = 0

change in internal energy dE = 0 for an isothermal process.

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