The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.500 atm? (∆Hvap = 28.5

Question

2 years ago

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kJ/mol)

1 answer:

NNADVOKAT [17] · Answer 1 · 2023-07-02T11:17:29+03:00

For a normal boiling point of a liquid is 282 °C, the temperature is mathematically given as

T2=181.55°C\

<h3> What temperature (in °C) would the vapor pressure be 0.500 atm? </h3>

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