The heat of vaporization of a liquid may be obtained from the approximate integr

Question

The heat of vaporization of a liquid may be obtained from the approximate integrated form of the Clausius-Clapeyron equation,

For water, the vapor pressure is measured to be 9.2 Torr at 283.15K and 55.3 Torr at 313.15K. Taking R to be 8.3145 J/K*mol, calculate delta H. TAKING THE LIMIT OF ERROR (95 PERCENT CONFIDENCE) IN A TEMPERATURE MEASUREMENT TO BE 0.1K AND THAT IN A PRESSURE MEASUREMENT TO BE 0.1 TORR, CALCULATE THE UNCERTAINTY IN DELTA H. A handbook gives 43,893 for delta H at 25 degrees Celsius.

I have already calculated delta H, I need help with the part that is in ALL CAPS. Please show all your work! Thank you!

Explanation / Answer

so given formula for DH = R(T1T2/(T2 - T1))ln(p2/p1)
consider
so DH is a product of T1T2/(T2 - T1) and ln(p2/p1)
error in T2 - T1 = 2dT [ where dT is error in temperature]
let T2 - T1 = T'
and error in t' = dT' = 2dT

also relative error in f = ln(p2/p1) = ln(p2) - ln(p1)
relative error in ln(p2) is dP/P2

so, absolute error in ln(p2) is dP*ln(p2)/P2
so, absolute error in ln(p1) is dP*ln(p1)/P1

so df = dP[ln(p2)/P2 + ln(p1)/P1]
where f = ln(p2/p1)

hence total error in DH is
d(DH) = DH*sqroot((dT/T1)^2 + (dT/T2)^2 + (dT'/T')^2 + (df/f)^2)
here DH is change in enthalapy
dT/T1 is relative error in T1 measurement = 0.1/283.15 = 3.531*10^-4
dT/T2 is relative error in T2 measurement = 0.1/313.15 = 3.193*10^-4
dT'/T' = 2dT/(T2 - T1) = 2*0.1/(30) = 0.006667
also, dT is error in temperature measurement
and df/f = dP[ln(p2)/P2 + ln(p1)/P1]/(ln(p2/p1)) = 0.1[ln(55.3)/55.3 + ln(9.2)/9.2]/ln(55.3/9.2) = 0.0174

so, d(DH) = 43893*0.0186 = 818.1489 J

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