How do you solve this problem? The ambient temperature is 85.0°F and the humidity of the surrounding air is reported to be 54.0%

Question

The ambient temperature is 85.0°F and the humidity of the surrounding air is reported to be 54.0%. Using the Clausius-Clapeyron equation and the boiling point of water as 100.0°C at 760 torr, calculate the vapor pressure (in torr) of water in the air. Use 40.7 kJ/mol as the ∆Hvap of water.

Answer 1

First, let's convert the given temperature from Fahrenheit to Kelvin, which is the unit of temperature used in the Clausius-Clapeyron equation.

T(K) = (T(°F) - 32) * 5/9 + 273.15 T(K) = (85.0°F - 32) * 5/9 + 273.15 = 302.0389 K

Next, let's convert the given ΔHvap from kJ/mol to J/mol, which is the unit of energy used in the Clausius-Clapeyron equation.

ΔHvap(J/mol) = ΔHvap(kJ/mol) * 10^3 ΔHvap(J/mol) = 40.7 kJ/mol * 10^3 = 40700 J/mol

The Clausius-Clapeyron equation is as follows:

ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)

Here, P1 is the vapor pressure at the boiling point of water (760 torr), T1 is the boiling point of water in Kelvin (373.15 K), and T2 is the ambient temperature in Kelvin (302.0389 K). We're solving for P2, the vapor pressure of water at the ambient temperature.

Rearranging the Clausius-Clapeyron equation to solve for P2 gives us:

P2 = P1 * e^(-ΔHvap/R * (1/T2 - 1/T1))

The value of R, the ideal gas constant, should be 8.314 J/(mol*K) to match the units of ΔHvap.

Substituting the given values into this equation gives us:

P2 = 760 torr * e^(-40700 J/mol / 8.314 J/(molK) * (1/302.0389 K - 1/373.15 K)) P2 = 760 torr * e^(-16.7086) ≈ 760 torr * 5.610^-8 ≈ 4.26*10^-5 torr

This is the saturation vapor pressure, the maximum amount of water vapor that could be present at this temperature. However, the humidity is only 54%, so the actual vapor pressure will be 54% of this value:

P = 0.54 * P2 P = 0.54 * 4.2610^-5 torr ≈ 2.310^-5 torr

Please note that the vapor pressure is extremely low, and it is due to the low temperature compared to the boiling point of water.