Question

A liquid has a vapor pressure of 0.309 atm at 76.5 °C and an enthalpy of vaporization of 34.2 kJ/mol. At 45.1 °C, the vapor pressure is Blank 1. Round your answer to three decimal places and do NOT include units.

Answer 1

Using the Clausius-Clapeyron equation, we calculate the vapor pressure at 45.1 °C based on known vapor pressure at 76.5 °C and the enthalpy of vaporization. Temperatures are converted to Kelvin before applying to the formula, leading to a solution for P2 after logarithmic and exponential calculations.

Step-by-step explanation:

To calculate the vapor pressure at 45.1 °C, we use the Clausius-Clapeyron equation, which is a way to quantify the change in vapor pressure with temperature. The equation is:

ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)

In this equation, P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively, ΔHvap is the enthalpy of vaporization, R is the universal gas constant which is approximately 8.314 J/mol·K, and T1 and T2 are the temperatures in Kelvin. First, we convert temperatures from Celsius to Kelvin by adding 273.15.

T1 = 76.5 °C = 349.65 K

T2 = 45.1 °C = 318.25 K

Next, we apply the values into the Clausius-Clapeyron equation:

ln(P2/0.309 atm) = (-34200 J/mol)/(8.314 J/mol·K) * (1/318.25 K - 1/349.65 K)

Solving for P2 gives us the vapor pressure at 45.1 °C. Using a calculator for the exponential and logarithm functions, we find P2.