Answer:
100.8 °C
Explanation:
The Clausius-clapeyron equation is:
[tex]ln\frac{P_{1} }{P_{2}} =[/tex]-Δ[tex]\frac{H_{vap}}{r} (\frac{1}{T_{2}}-\frac{1}{T_{1}} )[/tex]
Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'
Isolating for T2 gives:
[tex]T_{2}=(\frac{1}{T_{1}} -\frac{Rln\frac{P_{2}}{P_{1}} }{Delta H_{vap}}[/tex]
(sorry for 'deltaHvap' I can not input symbols into equations)
thus T2=100.8 °C
