1.

Calculate W and DeltaU for the conversion of 1 mole of water into 1 mole of steam at a temperature of 100^(@)C and at a pressure of 1 atmosphere. Latent heat of vaporisation of water is 9720 cal/"mole".

Answer»

SOLUTION :p=1 atm
=76 cm
=76xx13.6xx981 "dynes//cm"^(2)`
=1.013xx10^(6) "dymes//cm"^(2).
`V_(1)`= volume of 1 "mole" of water at `100^(@)C=18` mL.
`V_(2)` = volume of 1 "mole"of steam at `100^(@)C`
`=(373)/(273)xx22400=30605 mL`. (Charles.s law)
Now we have,
`W=-p(V_(2)-V_(1))` ...(Eqn.2)
`=-1.013xx10^(6)XX(30605-18) ergs
`=(-1.013xx10^(6)xx30587)/(4.18xx10^(7))"calories"`
=-741 "calories".
Again we have,
`DeltaU=q+W`
`=9720+(-741)`
`=8979 "calories"`.


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