estimate the heat of vaporization of water if the vapor pressure is 17.54 torr at 20C and 149.4 torr at 60C. R=8.314 J/mole K. I can't remember how to set up the problem. Would someone be able to send me the formula? Thank you

Hvap = Rln(P2/P1)

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(1/T1 - 1/T2)

Rln(P2/P1)

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(1/T1 - 1/T2)

To estimate the heat of vaporization of water, you can use the Clausius-Clapeyron equation. The equation relates the vapor pressure of a substance at one temperature to its vapor pressure at another temperature, and the heat of vaporization. The formula for the Clausius-Clapeyron equation is as follows:

ln(P₂/P₁) = (-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.

To solve for the heat of vaporization, we need to rearrange the equation:

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

Now let's plug in the given values:

P₁ = 17.54 torr (converted to atm: 17.54/760 = 0.0230 atm)
T₁ = 20°C = 293.15 K
P₂ = 149.4 torr (converted to atm: 149.4/760 = 0.1968 atm)
T₂ = 60°C = 333.15 K

Now substitute these values into the equation:

ΔHvap = -8.314 J/mol K * (1/333.15 K - 1/293.15 K) * ln(0.1968 atm/0.0230 atm)

Calculating this expression will give you the estimation of the heat of vaporization of water.