Solve Clausius Clapeyron Eqn for Vapor Pressure @95C

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SUMMARY

The discussion focuses on solving the Clausius-Clapeyron equation to determine the vapor pressure of diethyl ether at 95.0°C. The user correctly identifies the enthalpy of vaporization (∆Hvap) as 29.1 kJ/mol and the initial vapor pressure (P1) at 25.0°C as 0.703 atm. The equation is rearranged to P2 = P1 exp(∆Hvap/R (1/T1 - 1/T2), leading to a calculated vapor pressure (P2) of 2.19 atm at 95.0°C. The user confirms the methodology but has not yet verified the final result.

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  • Understanding of the Clausius-Clapeyron equation
  • Knowledge of thermodynamic concepts such as enthalpy of vaporization
  • Familiarity with temperature conversion from Celsius to Kelvin
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devon10
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It has been years scene I have taken a math class and I need help to see if I am setting the equation up correctly and if I have solved the problem correctly.

Diethyl ether has ∆Hvap of 29.1kJ/mol and a vapor pressure of 0.703 atm at 25.0 C. What is its vapor pressure at 95.0 C?

I converted the temp to K and kJ to J

In P2/P1=delta H vaporization/R (1/T2- 1/T1)

rearranged to

P2/P1= antilog (delta H vap/R (1/T2- 1/T1))

Then muliply by P1
P2= P1 antilog (delta H vap/R (1/T2- 1/T1))

antilog is also exp so...

P2= P1 exp(delta H vap/R (1/T2- 1/T1))

P2= .703 atm exp (29100/8.3145 (1/368-1/298)

P2= 2.19 atm
 
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Looks OK. I have not checked the final result.

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