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A certain substance has a heat of vaporization of 34.45 kJ/mol. At what Kelvin temperature will the vapor pressure be 5.00 times higher than it was at 335 K?

I've tried this question numerous times and Sapling keeps saying its incorrect... can someone please explain what I did wrong? and how to work the problem correctly? thank you!


Use the Clausius-Clapeyron equation

ln(p1/p2)=Hvap/8.3145 (1/T1-1/T2)

where R=8.3145 J/(mol┬ĚK).

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2 Answers

By the Clausius-Clapeyron equation ln(p2/p1)=Hvap/8.3145 (1/T1-1/T2), you have

ln(5) = (34450/8.3145)(1/335 - 1/T2)

Solve for T2,

T2 = 385.11 K <==Answer

I'm not familiar with the equation, but I looked it up.  You have T1 and T2 swapped in the formula.  I got a negative result until I swapped them.  Now I get 385K.  Sounds plausible!

ln(p1/p2)=Hvap/8.3145 (1/T2-1/T1)

I assumed that you used Hvap = 3.445e4 J mol-1 K-1 since its units are KJ mol-1 K-1.


It's also possible that Kayla's book had a negative sign in the equation (some do).