- #1

Parzeevahl

- 6

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- Homework Statement
- Show that a substance with a -ve slope for its fusion curve, such as ordinary ice or bismuth, contracts upon melting.

- Relevant Equations
- Clausius-Clapeyron equation:

Delta T (change in temp.) = (T * (v(f) - v(i)) * Delta P) / (h(f) - h(i))

where 'Delta P' is the change in pressure.

The question says that the process is melting, so temperature must increase.

Hence, Delta T > 0.

Also, it is given that the slope for its fusion curve is -ve, which means that as we increase temperature, the pressure will decrease.

So, Delta P < 0.

The question asks to prove that the substance contracts upon melting, i.e., Delta v (change in vol.) < 0.

So, in the right-hand side of the Clausius-Clapeyron equation, for both Delta P and Delta v = (v(f) - v(i)) to be -ve, (T / (h(f) - h(i))) must be +ve, as in the left-hand side, Delta T > 0. So essentially the problem comes down to proving that (T / (h(f) - h(i))) > 0.

Now, we are using absolute temperature here, T is always greater than or equal to zero.

So, my question is,

Hence, Delta T > 0.

Also, it is given that the slope for its fusion curve is -ve, which means that as we increase temperature, the pressure will decrease.

So, Delta P < 0.

The question asks to prove that the substance contracts upon melting, i.e., Delta v (change in vol.) < 0.

So, in the right-hand side of the Clausius-Clapeyron equation, for both Delta P and Delta v = (v(f) - v(i)) to be -ve, (T / (h(f) - h(i))) must be +ve, as in the left-hand side, Delta T > 0. So essentially the problem comes down to proving that (T / (h(f) - h(i))) > 0.

Now, we are using absolute temperature here, T is always greater than or equal to zero.

So, my question is,

**how do I prove that h(f) - h(i), i.e., the change in enthalpy, is +ve in this melting process?**