Diethyl ether has an enthalpy of vaporization of 29.0 kJ/mol. The compound has a
ID: 749149 • Letter: D
Question
Diethyl ether has an enthalpy of vaporization of 29.0 kJ/mol. The compound has a vapor pressure of 1.00 atm at 35.0Explanation / Answer
from clasius claperyon equation we have............................................................. ln(p1/p2)=(E/R)*(1/t2-1/t1) where p1,p2 are vapour pressures at temperatures t1,t2 and E is enthalpy of vapourisation and R is gas constant....................................... given p1=1atm t1=35+273=308kelvin E=29kj/mol. R=8.3145 J mol-1 K-1..... and p2=225mm of hg converting it into atmospheres we get p2=0.296atm (since 1atm =760 mm of hg) substituting in the equation we get........... ln(1/0.296)=(29000/8.3145)*(1/308-1/t2).................................................................................. =>1.2174=2525.71*(1/t2-1/308)............................................................................................... =>1/t2-1/308=4.82*10^-4............................................................................................................... =>1/t2=3.2467*10^-3+4.82*10^-4................................................................................................=>1/t2=3.7288*10^-3.....=>t2=268.19kelvin=268.19-273=-4.81'C......................................... ............................................................................................................................................................ PLEASE RATE THE ANSWER..:)
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