Question #b39e8

Question

Question #b39e8

1 Answer

Impact of this question

1780 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-20T05:38:35

By Clausius - Clapeyron equation weknow

$color(red)(ln(P_1/P_2) = (DeltaH_"vap")/R xx (1/T_2 - 1/T_1)$

Where

$P_1$ is the vapor pressure of the liquid at $T_1K$

$P_2$ is the vapor pressure of the liquid at $T_2K$

$R$ is the universal gas constant, equal to $"8.314J mol"^(-1)"K"^(-1)$

$DeltaH_"vap"= "enthalpy of vaporization of the liquid"$

Given

$T_1=273+5=278K$

$P_1=8.54mm$ of Hg

$T_2=273+120=393K$

$DeltaH_"vap"=41.8kJmol^-1"=41.8xx10^3Jmol^-1$

$P_2=?$

Inserting these in above equation

$color(blue)(ln(P_1/P_2) = (DeltaH_"vap")/R xx (1/T_2 - 1/T_1)$

$=>color(blue)(ln(8.54/P_2) = (41.8xx10^3)/8.314xx (1/393 - 1/278)$

$=>color(blue)(ln(8.54)-lnP_2 = (41.8xx10^3)/8.314xx (1/393 - 1/278))$

$=>color(blue)(lnP_2 =ln(8.54)- (41.8xx10^3)/8.314xx (1/393 - 1/278))$

$=>color(blue)(lnP_2=7.437)$

$=>color(blue)(P_2=e^7.437~~1698mm)$

Science

Math

Humanities

... and beyond

Question #b39e8

1 Answer

Related questions

Impact of this question