When 1g of water at 100⁰C is converted iinintintointo steam aatat 100⁰C,it occupies a volume of 1671cc at normal atmospheric pressure. Find the increase in internal energy of the molecules of steam

here we shall use the first law of thermodynamics

ΔQ = ΔU + ΔW

so, change in internal energy will be

ΔU = ΔQ - ΔW 

here,

ΔQ = mL =1g x 540Cal/g = 540 Cal

ΔU  = change in internal energy (to be calculated)

ΔW = change in work done

so,

ΔW = pΔV 

where

ΔV = 1671cm3 - 1cm3 = 1670cm3 = 1670 x 10-6 m3

p = hρg = 0.76 x 13600 x 9.8 ~ 101293 N/m2

so,

ΔW = 101293 x 1670x10-6 = 169.16J

or

ΔW = 169.16 / 4.2 = 40.3 Cal

thus, we have

 ΔU = 540 Cal - 40.3 Cal

so,

 ΔU = 499.7 Cal = 2098.74 J