$1\ cm^{3}$ of water at its boiling point absorbs 540 cal of heat of become steam with a volume of $1671cm^{3}.$ If the atmospheric pressure=$1.013 imes 10^{5}\ Nm^{- 2}$and the mechanical equivalent of heat =$419\ J\ cal^{- 1},$ the energy spent in this process in overcoming intermolecular forces is

Question

$1\ cm^{3}$ of water at its boiling point absorbs 540 cal of heat of become steam with a volume of $1671cm^{3}.$ If the atmospheric pressure=$1.013 	imes 10^{5}\ Nm^{- 2}$and the mechanical equivalent of heat =$419\ J\ cal^{- 1},$ the energy spent in this process in overcoming intermolecular forces is

Dr. Abubacker · Accepted Answer

(c) According to first law of thermodynamics, $\mathrm{\Delta}Q = \mathrm{\Delta}U + \mathrm{\Delta}W$
$	herefore\ \mathrm{\Delta}U = \mathrm{\Delta}Q - \mathrm{\Delta}W$
$= 540 - \frac{p\left( V_{2} - V_{1} ight)}{J}$
$= 540 - \frac{1.013 	imes 10^{5} 	imes \left\lbrack (1671 - 1) 	imes 10^{- 6} ightbrack}{4.2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 540 - 40 = 500\ cal$

Explanation

Reviewed by Dr. Abubacker

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