Question

At the normal melting point of NaCl, 801 degrees C, its enthalpy of fusion is 28.8 kJ / mol. The density of the solid is 2.165 g/cm^3 , and the density of the liquid is 1.733 g/ cm^3. What pressure increase is needed to raise the melting point by 1.00 degrees C?

Answer 1

Answer: The increase in pressure is 0.003 atm

Explanation:

To calculate the final pressure, we use the Clausius-Clayperon equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = initial pressure which is the pressure at normal boiling point = 1 atm

$P_2$ = final pressure = ?

$\Delta H$ = Enthalpy change of the reaction = 28.8 kJ/mol = 28800 J/mol     (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $801^oC=[801+273]K=1074K$

$T_2$ = final temperature = $(801+1.00)^oC=802.00=[802+273]K=1075K$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1})=\frac{28800J/mol}{8.314J/mol.K}[\frac{1}{1074}-\frac{1}{1075}]\\\\\ln P_2=3\times 10^{-3}atm\\\\P_2=e^{3\times 10^{-3}}=1.003atm$

Change in pressure = $P_2-P_1=1.003-1.00=0.003atm$

Hence, the increase in pressure is 0.003 atm