Article: The formula used for calculating vapor pressure given a change in the vapor pressure over time is known as the Clausius-Clapeyron equation (named for physicists Rudolf Clausius and Benoît Paul Émile Clapeyron). This is the formula you'll use to solve the most common sorts of vapor pressure problems you'll find in physics and chemistry classes. The formula looks like this: ln(P1/P2) = (ΔHvap/R)((1/T2) - (1/T1)). In this formula, the variables refer to:   ΔHvap: The enthalpy of vaporization of the liquid. This can usually be found in a table at the back of chemistry textbooks.  R: The real gas constant, or 8.314 J/(K × Mol).  T1: The temperature at which the vapor pressure is known (or the starting temperature.)  T2: The temperature at which the vapor pressure is to be found (or the final temperature.)  P1 and P2: The vapor pressures at the temperatures T1 and T2, respectively. The Clausius-Clapeyron equation looks tricky because it has so many different variables, but it's actually not very difficult when you have the right information. The most basic vapor pressure problems will give you two temperature values and a pressure value or two pressure values and a temperature value — once you have these, solving is a piece of cake.  For example, let's say that we're told that we have a container full of liquid at 295 K whose vapor pressure is 1 atmosphere (atm). Our question is: What is the vapor pressure at 393 K? We have two temperature values and a pressure, so we can solve for the other pressure value with the Clausius-Clapeyron equation. Plugging in our variables, we get ln(1/P2) = (ΔHvap/R)((1/393) - (1/295)). Note that, for Clausius-Clapeyron equations, you must always use Kelvin temperature values. You can use any pressure values as long as they are the same for both P1 and P2. The Clausius-Clapeyron equation contains two constants: R and ΔHvap. R is always equal to 8.314 J/(K × Mol). ΔHvap (the enthalpy of vaporization), however, depends on the substance whose vapor pressure you are examining. As noted above, you can usually find the ΔHvap values for a huge variety of substances in the back of chemistry or physics textbooks, or else online (like, for instance, here.)  In our example, let's say that our liquid is pure liquid water. If we look in a table of ΔHvap values, we can find that the ΔHvap is roughly 40.65 kJ/mol. Since our H value uses joules, rather than kilojoules, we can convert this to 40,650 J/mol.  Plugging our constants in to our equation, we get ln(1/P2) = (40,650/8.314)((1/393) - (1/295)). Once you have all of your variables in the equation plugged in except for the one you are solving for, proceed to solve the equation according to the rules of ordinary algebra.  The only difficult part of solving our equation (ln(1/P2) = (40,650/8.314)((1/393) - (1/295))) is dealing with the natural log (ln). To cancel out a natural log, simply use both sides of the equation as the exponent for the mathematical constant e. In other words, ln(x) = 2 → eln(x) = e2 → x = e2.  Now, let's solve our equation: ln(1/P2) = (40,650/8.314)((1/393) - (1/295)) ln(1/P2) = (4,889.34)(-0.00084) (1/P2) = e(-4.107)  1/P2 = 0.0165 P2 = 0.0165-1 = 60.76 atm. This makes sense — in a sealed container, increasing the temperature by almost 100 degrees (to almost 20 degrees over the boiling point of water) will create lots of vapor, increasing the pressure greatly
What is a summary of what this article is about?
Write the Clausius-Clapeyron equation. Plug in the variables you know. Plug in your constants. Solve the equation.