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

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: 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,
$\ln \left( {\dfrac{{{P_2}}}{{{P_1}}}} \right) = - \dfrac{{\Delta {H_{vap}}}}{R} \times \left( {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right)$
Where,
$\Delta {H_{vap}}$ is the enthalpy of vaporization of the given liquid
R is the real gas constant
$T_1$ is the temperature at which the vapor pressure is known that is the starting temperature.
$T_2$ is the temperature at which the vapor pressure is to be found or the final temperature.
$P_1$ and $P_2$ are the vapor pressures at the temperatures $T_1$ and $T_2$, respectively.
3) The value of boiling point gives us the values of initial temperature and the final temperature which can be written as $T_1$ and $T_2$ 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 ${\text{1}}$ atm pressure. A liquid phase material boils when its vapor pressure is exactly equal to the exterior pressure which is ${\text{1}}$ atm and that is the value of the vapor pressure for a substance.
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