Question

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

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

Answer 1

Answer: (A)

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