what is the heat of vaporization if the vapor pressure of liquid water at 25 °C is 23.8 Torr and at a boiling point of 100 °C is 760 Torr.

so technically (P1, T1) is 25 °C 23.8 Torr and (P1, T2) is 100 °C is 760 Torr? Or is it the other way around? so what is R?

Use th Clausius-Clapeyron equation and solve for delta H vap.

is this set up correct?


ln(760torr/23.8torr) = -delta Hvap/ 8.31447 (1/373K - 1/298K)

To find the heat of vaporization, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance at two different temperatures to its heat of vaporization.

The Clausius-Clapeyron equation is given by:

ln(P₁/P₂) = -ΔHvap/R * (1/T₂ - 1/T₁)

Where:
P₁ and P₂ are the vapor pressures at temperatures T₁ and T₂ respectively,
ΔHvap is the heat of vaporization,
R is the ideal gas constant (8.314 J/(mol·K)),
T₁ and T₂ are the temperatures in Kelvin.

In this case, we are given the vapor pressures at two temperatures: 23.8 Torr at 25 °C (298 K) and 760 Torr at 100 °C (373 K).

Using the equation, we can rearrange it to solve for ΔHvap:

ΔHvap = -R * (T₂ - T₁) / (ln(P₂/P₁))

Substituting the values:

ΔHvap = -8.314 J/(mol·K) * (373 K - 298 K) / ln(760 Torr / 23.8 Torr)

Calculating this expression will give us the heat of vaporization.

Please note that the units of pressure, temperature, and the gas constant need to be consistent for accurate results in this equation.

Do you need any further assistance with the calculation?