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    (question posted below) The question states this is math based, so you dont need to know what the equations are. I found the answer, but I don't understand the steps, so I was wondering why they are as they are.

    To get A in terms of a, you should add the log of the factor between both units to the A parameter. So:

    AkPa = ATorr + log10 \frac {101.325}{760} (because 1 Torr = 0.133 kPa or 760 Torr = 101.325 kPa)

    I dont understand why you do this? (why log and why not add it to B and C as well, why just A)

    To change from log to ln, you should multiply A and B by ln(10). I dont understand why you do this, and why not multiply C as well.

    • Study Helper

    Study Helper
    (Original post by G A B R I E L)
    If you're still stuck, drop me a PM, and I'll add some details to the thread sometime later today, otherwise not, as the Latex is going to be a pain in the ....
    • Study Helper

    Study Helper
    (Original post by G A B R I E L)
    As you're not a maths student, I'll give you a full description. Lets start from:

    \displaystyle\log_{10}P^\text{sa  t}_\text{Torr}= a-\frac{b}{T_C+c}

    I've changed the format slightly, but hopefully the meaning is clear.

    Now we know that 760 Torr = 101.325 kPa

    Or 1 Torr = 101.325/760 kPa.

    So inverting that we have

    1 kPa = 760/101.325 Torr.

    So if P^\text{sat}_\text{kPa} is the pressure in kPa, then the pressure in Torr is P^\text{sat}_\text{kPa}\times\fr  ac{760}{101.325}

    So substituting this into our equation we get:

    \displaystyle\log_{10}\left(P^\t  ext{sat}_\text{kPa}\times\frac{7  60}{101.325}\right)= a-\frac{b}{T_C+c}

    And so, splitting the log:

    \displaystyle\log_{10}P^\text{sa  t}_\text{kPa}+\log_{10}\frac{760  }{101.325}= a-\frac{b}{T_C+c}


    \displaystyle\log_{10}P^\text{sa  t}_\text{kPa}= -\log_{10}\frac{760}{101.325}+a-\frac{b}{T_C+c}

    And using the fact that -log x = log(1/x) we have:

    \displaystyle\log_{10}P^\text{sa  t}_\text{kPa}= \log_{10}\frac{101.325}{760}+a-\frac{b}{T_C+c}

    Now lets do the temperature conversion. I think it's clear that T_C = T_K-273

    So we get:

    \displaystyle\log_{10}P^\text{sa  t}_\text{kPa}= \log_{10}\frac{101.325}{760}+a-\frac{b}{T_K-273+c}

    All that remains is the log conversion. Now,

    \log_{10}x=\frac{\ln x}{\ln 10}

    So substituting this into the left hand side of our equation:

    \displaystyle\frac{\ln P^\text{sat}_\text{kPa}}{\ln 10}= \log_{10}\frac{101.325}{760}+a-\frac{b}{T_K-273+c}

    Then multiplying through by ln 10 and grouping terms together we have:

    \displaystyle\ln P^\text{sat}_\text{kPa}= \ln 10 \times \left(\log_{10}\frac{101.325}{76  0}+a\right)-\frac{\ln 10\times b}{T_K+(c-273)}

    So now just compare terms with the modern formula. I've had a bottle of wine, but hopefully that's correct. Do check it through, and any questions, get back to me.
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Updated: February 12, 2010
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