- #1

erok81

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## Homework Statement

This was a multi-part problem using the Clausius-Clapeyron relation to calculate how much pressure needed to be put on an ice cube in order to have it melt at -1*C.

## Homework Equations

Clausius-Clapeyron relation given by:

[itex]\frac{dP}{dT}=\frac{L}{T\Delta V}[/itex]

## The Attempt at a Solution

I am confused on how exactly to solve this. I've seen it two ways. One integrates and one doesn't, leading to two different answers.

The first is use just using ΔP/ΔT as just changes. So...

[itex]\frac{\Delta P}{\Delta T}=\frac{L}{T \Delta V}[/itex]

That's really it. Then one plugs in 272k for temp, then the value for ΔV that was calculated earlier, and L. Then just solve. I don't think this is the correct way.

The second method is integration.

[itex]\frac{dP}{dT}=\frac{L}{T\Delta V}[/itex]

[itex]\int ^{P_{f}}_{P{_i}} dP= \frac{L}{\Delta V} \int ^{T_{f}}_{T{_i}}\frac{dT}{T}[/itex]

Which, after adding P

_{i}to both sides, the result is:

[itex]P_{f}=\frac{L}{\Delta V} ln(\frac{T_f}{T_i})+P_i[/itex]

And then plug in the same values, using the initial P and T as STP.

So my question is twofold, is the integration method correct? And this problem isn't the best example, but when is it needed to integrate vs. only use the Δ values?