Calculating Pressure - to Integrate or Not?

[erok81]

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:

\frac{dP}{dT}=\frac{L}{T\Delta V}

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...

\frac{\Delta P}{\Delta T}=\frac{L}{T \Delta V}

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.

\frac{dP}{dT}=\frac{L}{T\Delta V}

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

Which, after adding P_i to both sides, the result is:

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

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?

 

[Redbelly98]

The second method is integration.

\frac{dP}{dT}=\frac{L}{T\Delta V}

\int ^{P_{f}}_{P{_i}} dP= \frac{L}{\Delta V} \int ^{T_{f}}_{T{_i}}\frac{dT}{T}
.
.
.
So my question is twofold, is the integration method correct?

Since the Clausius-Clapeyron relation involves dP/dT and not ΔP/ΔT, then integration is the correct method, strictly speaking.

And this problem isn't the best example, but when is it needed to integrate vs. only use the Δ values?

If you examine the integral you did, the question to ask is, "does the integrand change by very much in between the two limits of integration?"

The integrand is 1/T in the integral \int ^{T_{f}}_{T{_i}}\frac{dT}{T}. What is the percentage change in 1/T, in going from 0°C to -1°C? Is that a large or a small change?