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

Step-by-step explanation:

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

`\ln((P_2)/(P_1))=(\Delta H)/(R)[(1)/(T_1)-(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((P_2)/(1))=(28800J/mol)/(8.314J/mol.K)[(1)/(1074)-(1)/(1075)]\\\\\ln P_2=3* 10^(-3)atm\\\\P_2=e^{3* 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