Question

Argon (molecular weight 40 g/mole) is a monatomic compound. If liquid argon is confined to a container and held at a constant temperature of 80.5 K, what is the approximate vapor pressure of gaseous argon, assuming the liquid has no entropy and a binding energy of 0.1 eV? [Note: At 1 atm, the boiling point is 87.3 K.]

Answer 1

Explanation:

As the given data is as follows.

        $\mu_{Ar}$ = $kTln\frac{n}{n_{Q}}$

and,     $\mu_{H_{2}O}$ = $-\Delta$

        T = 80.5 K

According to the given condition,

               $kTln\frac{n}{n_{Q}}$ = $kTln\frac{P}{P_{Q}}$ = $-\Delta$

       $\frac{n}{n_{Q}}$ = $e^{\frac{-\Delta}{kT}}$

                  p = nkT = $n_{Q}kTe^{\frac{-\Delta}{kT}}$

Therefore, putting the given values into the above equation as follows.

               $n_{Q}$ = $(10^{30}) \times (4)^{\frac{3}{2}} \times (\frac{80.5}{300})^{\frac{3}{2}}$

                                 = $3.52 \times 10^{31} m^{-3}$

Therefore, the required pressure is $2.14 \times 10^{4} Pa$.