Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 327.0 Torr at 45.0 °C. Calculate the value of ΔH°vap for this liquid. Calculate the normal boiling point for this liquid.

Answer #1

By Clausius-Clapeyron equation

logP2/P1 = delta H/(2.303*R) * (1/T1-1/T2)

P1=92 torr P2=372, T1=296k, T2=318

deltaH = heat of vaporisation in cal/mole

T2 and T1 are temp in kelvin

R=1.99 cal/mole-k

log372/92=0.6067

2.303*R=2.303*1.99=4.5829cal/mole-k ...

1/T1-1/T2=0.0002337K

Therefore, deltaH=0.6067*4.5829/0.0002337 =11897 cal or
11897/239kj=49.78kj

Part2

Let normal boiling point is T2

Corresponding vap pressure P2 would be atm pr =760 torr

log760/92=0.9170 , P1=92 torr

2.303*R=2.303*1.99=4.5829cal/mole-k ...

1/T1-1/T2= (1/296-1/T2) , T1=273+23=296k

1/T1-1/T2=( 0.003378 - 1/T2)

Heat of vaporisation = 47.9kj given=47.9*239cal = 11448cal 1 kJ =
239.0057

Therefore, 0.9170*4.5829/11448=0.003378-1/T2 ...

0.0003671=0.003378-1/T2

1/T2=0.003378-0.0003871=0.0030109 ...

T2=332K or 59 ^{o}C

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