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

Last updated date: 11th Aug 2024
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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.

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.