Heat of Vaporization from Vapor Pressure (Example) 

No, this equation is correct. Keep in mind that there is a minus sign in front of delta H in the equation

l think he multiplies 3613 k x 8.314 = 30.038. i dunno how he gets 30.043.

+Matthew Johnson III So long as the temperatures are matched with their corresponding pressures, it doesn't matter which set is listed as P1, T1 or P2, T2. You can do the math if you would like to prove it to yourself. Choose 53.1 kPa for P1 and 24.1 deg C for T1 and see if it works.

HI, Matthew, If the other answer isn't working for you: I find that if t1 or p1 is a higher number compared t2 or p2, it indicates that either pressure or temperature is lowering (less heat will be needed to vaporize) , so heat of vaporization should be positive, therefor it should be p1 over p2. Remember, positive enthalpy means, change of vaporization, where heat is being absorbed by the substance. and because the pressure is low, heat is much easier to absorb) HOWEVER, if p1 or t1 is a lower number compared to p2 or t2, that indicates the temp or pressure is on the rise (more heat will be needed to vaporize), and heat of vaporization should be negative, making it p2 over p1 Keep in mind, Negative enthalpy means, it is change of condensation, heat is being released from the substance

ln (3.99) = 1.38

@@LearnChemE thanks

The equation is correct. Note that there is a negative sign in front of the delH_vap.

Oi

Chup

Noob

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you are stupid and moron