Answered: If the initial pressure of the air in…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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If the initial pressure of the air in the flask was 0.985atm at 298.15k what would the corrected pressure of the air be at 327.76K. For a given liquid a plot of the lnP versus 1/T gives a straight line with a slope of -5199k.

What is the Hvap of this liquid?

Expert Solution

Step 1

We have:

Initial pressure (P1) = 0.985 atm at Temperature (T1) = 298.15 K.

Final ressure (P2) = x atm at Temperature (T2) = 327.76 K.

We know that:

The Clausius Clapeyron equation plot of ln p Vs 1/T gives a straight line, whose slope is ∆H/R.

According to the question, Slope= ∆H/R = -5199 K.

Substituting R= 8.314 J/mol K

We get: ∆H= -5199 K*(8.314 J/mol K)

∆H= - 43.224 KJ/mol.

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