Question

For mercury, ∆H_{fus}^{o} = 2.292 kJ
mol^{-1}, and its normal freezing point is 234.3 K. The
change in molar volume on melting ∆V_{fus} = 0.517
cm^{3} mol^{-1}. Calculate the melting point of
mercury at 20.0 bar.

Answer #1

we know that

dP/dT= deltaH fusion/ deltaVT

dP= (deltaHfusion/deltaV)(dT/T)

when integrated, the equation becomes

P2-P1= (deltaHfusion/deltaV)*dT/T,

deltaHfuson= 2.292Kj/mole= 2.292*1000J/mole. 101.3 L.atm= 1 J

hence deltaH fusion= (2.292*1000/101.3)L.atm/mole= 22.63 L.atm/mole

delaV fusion = 0.517 cm3/mole, 1000CC= 1L

deltaV fusion = 0.517/1000 L/mole = 0.517*10^{-3} L/mole
=

hence dP= (22.63/0.517*10^{-3})*dT/T=43764
dT/T

when integrated P2- P1= 43764* ln (T2/T1) give normal freezing point is 234.3 at 1 atm

given P1= 1 atm and T1= 234.3 K, P2= 20 bar= 20*0.9869 atm =19.74 atm

19.74-1= 43764*ln(T2/ 234.3)

ln(T2/234.3)= 0.000428

T2/ 234.3= 1.000428

T2= 234.3*1.000428=234.4 K

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