Question

When $1$ mole of ice melts at $0^{\circ }C$ and at a constant pressure of $1atm,1440cal$ of heat is absorbed by the system. If the molar volume of ice and water are $0.0196$ and $0.0180L$, then the value of $\Delta H$ and $\Delta U$ will be

• A. $1440cal,1440.039cal$
• B. $1440cal,1438cal$
• C. $1440.039cal,1440cal$
• D. $1438cal,1440cal$

Answer 1

Answer: (A)

Since, heat absorbed at constant pressure, is $q=1440cal$
$\therefore \Delta H=1440cal$
Given: $H_{2}O(s)\rightleftharpoons H_{2}O(l)$
$\Delta V=(0.0180-0.0196)$
$=-0.0016L$
$\therefore p\Delta V=-1atm\times 0.0016L$
$=-0.0016L$ atm
$=-0.039cal$
(\because 1{\,} L \text { atm }=24.20{\,} cal )
Using, $\Delta H=\Delta U+p\Delta V$
$\Delta U=\Delta H-P\Delta V$
$=1440-(-0.039)$
$=1440.039cal$