Question

Calculate the vapor pressure of a liquid in torr at 90.70celsius if it has a vapor pressure of43.5 torr at 28.2 Celsius and a enthalpy of vaporization 21.4kj/mol r =8.314 Jk

the answer is 188.66 solve off that I need the steps

Answer 1

After converting all necessary units, inserting the known values, and solving, the vapor pressure is found to be approximately 188.66 torr.

Step-by-step explanation:

To calculate the vapor pressure of a liquid at 90.7°C, given its vapor pressure of 43.5 torr at 28.2°C and an enthalpy of vaporization (∆Hvap) of 21.4 kJ/mol, we can use the Clausius-Clapeyron equation:

ln(P2/P1) = (∆Hvap/R) * (1/T1 - 1/T2)

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• Let P1 = 43.5 torr (the vapor pressure at 28.2°C)
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• Convert the temperatures from Celsius to Kelvin:
  
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  • T1 = 28.2 + 273.15 = 301.35 K
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  • T2 = 90.7 + 273.15 = 363.85 K
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• ∆Hvap = 21.4 kJ/mol = 21400 J/mol (since 1 kJ = 1000 J)
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• R = 8.314 J/mol·K (the universal gas constant)

First, we rearrange the equation to solve for P2:

P2 = P1 * exp[(∆Hvap/R) * (1/T1 - 1/T2)]

Inserting the values, we get:

P2 = 43.5 torr * exp[(21400 J/mol) / (8.314 J/mol·K) * (1/301.35 K - 1/363.85 K)]

After calculating, we find that P2 ≈ 188.66 torr, which is the vapor pressure at 90.7°C.