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ira [324]
3 years ago
12

Predict the boiling point of water at a pressure of 1.5 atm.

Chemistry
1 answer:
Lina20 [59]3 years ago
7 0

Answer:

100.8 °C

Explanation:

The Clausius-clapeyron equation is:

ln\frac{P_{1} }{P_{2}} =-Δ\frac{H_{vap}}{r} (\frac{1}{T_{2}}-\frac{1}{T_{1}}  )

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:

T_{2}=(\frac{1}{T_{1}} -\frac{Rln\frac{P_{2}}{P_{1}} }{Delta H_{vap}}

(sorry for 'deltaHvap' I can not input symbols into equations)

thus T2=100.8 °C

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