Question

When 1 mole of ice melts at 0 °C and a constant pressure of 1 atm, 1440 cal of heat are absorbed by the system. The molar volume of ice and water are 0.0196 and 0.0180 L, respectively calculate $\Delta E$

• A. $1430.03cal$
• B. $1500.0cal$
• C. $1450.0cal$
• D. $1440.03cal$

Answer 1

Answer: (D)

$1440.03cal$

Answer 2

$q = 1440cal$

$\mathbf{\therefore\Delta}\mathbf{H = 1440cal}$ (absorbed)

given $H_{2}O(s)$ ⇌ $H_{2}O(l)$

$P\Delta V = 1 \times$ (0.0180–0.0196) =– 0.0016 L atm

$\because$ (1L atm = 24.206 cal)

$\therefore - 0.0016Latm = - 0.039cal$

$\Delta H = \Delta E + P\Delta V$ ⇒$\Delta E = \Delta H - P\Delta V$

$= 1440 - ( - 0.039) = 1440.039cal$