I have to calculate the vapor pressure of water at 25.0 degrees celicus. The heat of vaporization of water is 40.65 kj/mol and the vapor pressure of water is 92.7 Torr at 50 degrees celcius. I know I should use log10 P=A- B/C+T but I don't know how to work it out.

I am not familiar with your equation. I would use the Clausius-Clapeyron equation (and please note the correct spelling of celsius).

ln(p2/p1) = dHvap/R(1/T1-1/T2)
If you let p1 = 92.7 and T1 = 50C+273; then solve for p2 = ? when T2 = 25+273

To calculate the vapor pressure of water at a specific temperature using the equation you mentioned, you need to use the given values and ascertain the necessary constants. Let's break down the equation and go through the steps:

1. The formula you mentioned is the Antoine equation, which relates the vapor pressure of a substance to its temperature. The general form of the equation is: log10 P = A - (B / (C + T)), where P represents the vapor pressure and T is the temperature in degrees Celsius.

2. In order to use this equation, you need to determine the values of the constants A, B, and C specific to water. These constants vary for different substances. For water, the following values are commonly used:
- A = 8.07131
- B = 1730.63
- C = 233.426

3. Substitute the given temperature (25.0 degrees Celsius) into the equation.

log10 P = A - (B / (C + T))
log10 P = 8.07131 - (1730.63 / (233.426 + 25.0))

4. Calculate the expression within the brackets on the right side of the equation.

log10 P = 8.07131 - (1730.63 / 258.426)

5. Evaluate the bracketed expression, dividing the value of B by the sum of C and T.

log10 P = 8.07131 - 6.695

6. Subtract the value within the brackets from A.

log10 P = 1.37631

7. Raise both sides of the equation to the power of 10 to eliminate the logarithm.

P = 10^(1.37631)

8. Calculate the result.

P ≈ 27.27 Torr

So, the vapor pressure of water at 25.0 degrees Celsius would be approximately 27.27 Torr, using the given equation and constants.