[legge]

The heat of melting of ice at 1 atm and 0 degree c is +5.9176kj/mol. The density of ice under these conditions is 0.917g/cm^3 and the density fo water is 0.9998g/cm^3



What will the melting point be at 10 atm? Derive necessary equations and make necessary asumptions.



DeltaH and deltaS can be easily calculated, but i was thinking that dp/dt=(s2-s1)/(v2-v1). however i don't know if the formula is correct or how to apply to this problem. Thank you.

[Hunt]

You're on the right track. For phase transition , use delta S = delta H / T. My guess is you have to assume enthalpy and volume remain constant.
dP/dT=(S2-S1)/(V2-V1) = (H2 - H1)/ T(V2 - V1)
dP = [delta H_trs / delta V] dT /T
P = [delta H_trs / delta V] Ln T + C
C is a const of integration , can be determined by plugging the values of the variables from the given. Remember that density = rho = m / V.