Question

Calculate the vapor pressure for a mist of spherical water droplets of radius 3.70×10−8m surrounded by water vapor at 298 K. The vapor pressure of water at this temperature is 25.2 Torr, the density of water is 998 kg⋅m−3, and the surface tension of water is 71.99 mN⋅m−1.

Answer 1

Answer:

The vapor pressure for a mist is $P= 25.92\ Torr$

Explanation:

From the question we are given that

        The radius is  $r = 3.70 *10^{-8} m$

        The temperature is $T = 298K$

       The vapor pressure of water $P_o = 25.2\ Torr$

      The density of water is  $\rho = 998 kg.m^{-3}$

      The surface tension of water is $\sigma = 71.99 m N \cdot m^{-1}$

Generally the equation of that is mathematically represented as

                            $ln (\frac{P}{P_0} ) = \frac{2 \sigma M}{r \rho RT}$

 Where  P is the vapor pressure for mist

               R is the  ideal gas constant = 8.31

      making P the subject in the formula

    $P = e^ {\frac{2 \sigma M}{r \rho RT}} * P_0$

        $= e^{\frac{2 *(0.07199)(0.018)}{(3.70*10^{-8})(998)(8.31)(298)} } * 25.2$

        $P= 25.92Torr$