Utilizing the Clausius-Clapeyron Equation to Determine Propane Pressure

What is the pressure (in psi) of propane in a tank of liquid propane at 24.0 °C?

How can we calculate the pressure of propane in the tank at 24.0 °C using the Clausius-Clapeyron equation?

Answer:

To determine the pressure of propane in a tank at 24.0 °C, we can utilize the Clausius-Clapeyron equation:

The Clausius-Clapeyron equation relates the vapor pressure of a substance at two different temperatures to its enthalpy of vaporization (∆Hvap) and the gas constant (R).

The equation is written as: ln(P2/P1) = (∆Hvap/R) * (1/T1 - 1/T2)

First, we need to convert the given temperatures to Kelvin:

T1 = -42.0 °C + 273.15 = 231.15 K

T2 = 24.0 °C + 273.15 = 297.15 K

Substitute the values into the rearranged equation to solve for P2:

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

Given P1 as 1 atm (equivalent to 14.7 psi) and ∆Hvap as 18.8 kJ/mol, plug in the values and calculate the expression to find the pressure of propane in the tank at 24.0 °C.

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