Answer to Question #87585 in Molecular Physics | Thermodynamics for Sridhar

Question

calculate the change in melting point of ice when it is subjected to a pressure of 100 atm. Density of ice is 0.917g/c.c and latent heat of ice is 336J/g
Answer :72.26k

Answer 1

Clausius–Clapeyron relation:

$$\frac{dP}{dT}=\frac{Q}{T \Delta V}$$

where$Q=336\, J/g$ - latent heat of ice, $\Delta V = \left(\frac{1}{0.917}-1\right) \, cm^3 /g$ is the specific volume change of unit mass of the phase transition.As we know, melting point of ice at pressure 1atm at temperature

$$T_0=0^\circ C = 273.15K$$

Assume no (or tiny) density of ice and water dependence on temperature and pressure.

Therefore,

$$\int dP=\int \frac{QdT}{T \Delta V} \Rightarrow \frac{T}{T_0}=e^{-\frac{\Delta P \Delta V}{Q}} = e^{-\frac{99\,atm \, 0.0905 \, cm^3 /g}{336\,J/g}}=0.997337$$

Hence

$$\Delta T=T-T_0 = 272.423 - 273.15=-0.72749 K$$