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Calculate the amount of work done, in joules, when 2.5 mole of H2O vaporizes at 1.0 atm and 25°C. Assume the volume of liquid H2O is negligible compared to that of vapor. (1 L·atm = 101.3 J)

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Answer:

6.195 x 10 3 J oule

Explanation:

all given values

Temperature =25+ 273= 298K

Pressure= 1 atm

ideal gas constant = R=0.0821 L.atm/K/mol

number of moles n= 2.5

volume =?

using ideal gas equation

PV=nRT

for volume V=nRT/P

=2.5×.0821×298/1

=61.16 L

As the direction of work done is opposite to the direction of motion so work done will be negative  

W=-PV

=-1atm× 61.16 L

=101.3× 61.61

 = - 6195.96 J

= 6.195 x 10 3 J

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