The specific volume of liquid water is $1.001 \mathrm{mL} \mathrm{g}^{-1}$ and that of ice is $1.0907 \mathrm{mL} \mathrm{g}^{-1}$ at ${ }^{\circ} \mathrm{C}$. If the heat of fusion of ice at this temperature is $333.88 \mathrm{J} \mathrm{g}^{-1}$, the rate of change of melting point of ice with pressure in $\mathrm{deg} \mathrm{atm}^{-1}$ will be
(a)-0.0075
(b)0.0075
(c)0.075
(d)-0.075
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
dT/dP = TΔV/ΔH = 273.15 K × (−0.0906×10⁻⁶ m³ g⁻¹)/333.88 J g⁻¹ = −7.4×10⁻⁸ K Pa⁻¹ ≈ −0.0075 K atm⁻¹ (a).