Question

Calculate the change in pressure required to change the freezing point of water at 1°C and 0°C if heat of fusion of ice is 333.5 j/cm^3 and density of water is 0.998g/cm^3 and of ice is 0.9168g/cm^3?

Answer 1

I'm getting `-"7.53 atm"` to cause `DeltaT_"fus" = +"1 K"` from `0^@ "C"` to `1^@ "C"`, indicating the slope is NEGATIVE.

---

I assume you mean, calculate the change in pressure associated with a change in the freezing point of water from `0^@ "C"` to `1^@ "C"`.

I will forgo the derivation and present the Clapeyron Equation:

`(DeltaP)/(DeltaT) = (DeltabarS_(tr))/(DeltabarV_(tr)) = (DeltabarH_(tr))/(T_(tr)DeltabarV_"tr")`

where:

• `P` is pressure in `"atm"`
• `T_"tr"` is the normal phase transition temperature in `"K"`.
• `barS` is molar entropy in `"J/mol"cdot"K"`.
• `barV` is molar volume in `"L/mol"`.
• `barH` is molar enthalpy in `"J/mol"`.

First, let's evaluate the right-hand side. The molar volumes are:

`barV_"ice" = [(0.9168 cancel"g")/cancel("cm"^3) xx (1000 cancel("cm"^3))/("1 L") xx ("1 mol")/(18.015 cancel"g water")]^(-1) = "0.01965 L/mol"`

`barV_"water" = [(0.998 cancel"g")/cancel("cm"^3) xx (1000 cancel("cm"^3))/("1 L") xx ("1 mol")/(18.015 cancel"g water")]^(-1) = "0.01805 L/mol"`

Therefore,

`DeltabarV_("ice"->"water") = "0.01805 L/mol" - "0.01965 L/mol" = -"0.0016 L/mol"`

i.e. ice contracts when it melts. Hence, the right-hand side is:

`(DeltabarH_(tr))/(T_(tr)DeltabarV_"tr") = (333.5 cancel"J""/"cancel"g" xx (18.015 cancel"g")/(cancel"1 mol"))/("273.15 K" cdot (-0.0016 cancel"L""/"cancel"mol")) xx (cancel"1 L"cdot"atm")/(101.3cancel"J")`

`= -"7.53 atm/K"`

So, given the change in temperature is `"1 K"`,

`color(blue)(DeltaP) = DeltaT(-"7.53 atm/K") = color(blue)(-"7.53 atm")`