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)

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

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.