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

Step-by-step explanation:

As the given data is as follows.

`\mu_(Ar)` =
`kTln(n)/(n_(Q))`

and,
`\mu_{H_(2)O}` =
`-\Delta`

T = 80.5 K

According to the given condition,

`kTln(n)/(n_(Q))` =
`kTln(P)/(P_(Q))` =
`-\Delta`

`(n)/(n_(Q))` =
`e^{(-\Delta)/(kT)}`

p = nkT =
`n_(Q)kTe^{(-\Delta)/(kT)}`

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

`n_(Q)` =
`(10^(30)) * (4)^{(3)/(2)} * ((80.5)/(300))^{(3)/(2)}`

=
`3.52 * 10^(31) m^(-3)`

Therefore, the required pressure is
`2.14 * 10^(4) Pa`.