For a normal boiling point of a liquid is 282 °C, the temperature is mathematically given as
T2=181.55°C\
What temperature (in °C) would the vapor pressure be 0.500 atm?
Generally, the equation for the gas is mathematically given as
ln(p1/p2)=dHvap/R(1/T2-1/T1)
Therefore
ln(1/0.26)=23500/8.214(1/T2-1/555)
T2=181.55^C
In conclusion
T2=181.55°C
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