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Calculating Pressure - to Integrate or Not?

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    Clausius-Clapeyron relation given by:

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

    3. 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 Pi 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?
     
  2. jcsd
  3. Mar 11, 2012 #2

    Redbelly98

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

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