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Question: If given a boiling point, how do you find vapor pressure?...

If given a boiling point, how do you find vapor pressure?

Explanation

Solution

The vapor pressure is the pressure that is exerted by gas-phase molecules on the liquid. The boiling point can be elaborated as the temperature at which a liquid turns into a gas. One can find the relation between the two terms via the Clausius-Clapeyron equation.

Complete step by step answer:

  1. First of all we will learn about the vapor pressure concept wherein in chemistry the term vapor pressure is the pressure which is exerted on the walls of a container that is sealed when a substance in it evaporates that is converted to a gas phase.
  2. Now to find out the vapor pressure at a given temperature let us use the Clausius-Clapeyron equation as below,
    ln(P2P1)=ΔHvapR×(1T21T1)\ln \left( {\dfrac{{{P_2}}}{{{P_1}}}} \right) = - \dfrac{{\Delta {H_{vap}}}}{R} \times \left( {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right)
    Where,
    ΔHvap\Delta {H_{vap}} is the enthalpy of vaporization of the given liquid
    R is the real gas constant
    T1T_1 is the temperature at which the vapor pressure is known that is the starting temperature.
    T2T_2 is the temperature at which the vapor pressure is to be found or the final temperature.
    P1P_1 and P2P_2 are the vapor pressures at the temperatures T1T_1 and T2T_2, respectively.
  3. The value of boiling point gives us the values of initial temperature and the final temperature which can be written as T1T_1 and T2T_2 and can be put in the Clausius-Clapeyron equation.
  4. Now in this equation, we can put the values of boiling point and initial temperature and find out the values of partial pressure.

Note:
While solving the question it is important to note that under normal conditions everything boils at 1{\text{1}} atm pressure. A liquid phase material boils when its vapor pressure is exactly equal to the exterior pressure which is 1{\text{1}} atm and that is the value of the vapor pressure for a substance.