Iris, in order to isolate P1, you must get rid of the natural log (ln) on the left side of the equation. You do this by raising both sides by to the natural log exponent, e, since e^ln(x) = x. When you raise the left side of the equation, ln (P1/P2), you get exp^(ln (P1/P2)), which equals P1/P2. Multiply both sides by P2 and you then isolate P1.
Basically, we have this math equation right? And this ln, stupid ln, it’s legit a bully blocking ur way. Can’t solve. So, the only way to fix this? Eliminate it. when u have ln(P1)= [insert right hand side of equation], then u can make the equation into P1=e^([right side of equation]) because that’s just the definition of ln. the way he did it also works. E^ on both the sides. It pretty much does the same thing. Also yes me and ln have issues
the normal boiling point of water is 80 °c . if vapour pressure at 60°c is 0.3672 atmosphere . calculate the anthelpy of vapourization per mole of water is
The P1 and P2 are flipped as far as I know. They do not correlate with the order he has written T1 and T2. The answer may not be correct with the way this is set up unless he corrects it somewher later in the video
His equation is correct. Had he used ln (P2/P1) on the left side of the equation, he would have taken the negative value of the enthalpy of vaporization. The equation (other variation of the C-C equation) would become: ln (P2/P1) = - delta H-vap/R * (1/T2 - 1/T1).
