Question

The vapor pressure of liquid iodomethane, CH3I, is 40.0 mm Hg at 249 K.A sample of CH3I is placed in a closed, evacuated container of constant volume at a temperature of 404 K. It is found that all of the CH3I is in the vapor phase and that the pressure is 72.0 mm Hg. If the temperature in the container is reduced to 249 K, which of the following statements are correct
Choose all that apply
A. The pressure in the container will be 104 mm Hg.
B. No condensation will occur.
C.Some of the vapor initially present will condense.
D. Only pentane vapor will be present.
E. Liquid pentane will be present.

Answer 1

Explanation:

It is given that vapor pressure of liquid iodomethane is 40.0 mm Hg. So, if we calculate the vapor pressure according to the given values and if its value will be greater than the the given vapor pressure of iodomethane then it means that some of the vapors has converted into liquid state.

As the given values are as follows.

      $P_{1}$ = 72.0 mm Hg,       $T_{1}$ = 404 K

      $P_{2}$ = ? ,               $T_{2}$ = 249 K

As volume is constant so, according to Gay-Lussac's law pressure is directly proportional to temperature.

                      $P \propto T$         (at constant volume)

or,                    $\frac{P}{T}$ = k

Therefore, the formula to calculate the value of $P_{2}$ is as follows.

              $\frac{P_{1}}{T_{1}}$ = $\frac{P_{2}}{T_{2}}$

            $\frac{72.0 mm Hg}{404 K}$ = $\frac{P_{2}}{249 K}$

                 $P_{2}$ = 44.37 mm Hg

As calculated vapor pressure is more than the given vapor pressure. Hence, the liquid will convert into gas.

As a result, no condensation will occur and only vapors of iodomethane will be present.