Question

The volume of gas at 20°C and 1.00 bar that adsorbed on a cold surface was 1.52 cm3 at 56.4 kPa and 2.77 cm3 at 108 kPa. Find the equilibrium constant, and volume of a monolayer.

Answer 1

Step-by-step explanation:

According to Langmuir isotherm,

`\theta = (k * P)/(1 + k * P)`

where,
`\theta` = fraction coverage by gas molecules

k = rate constant

P = partial pressure of gas

Also,
`(V)/(V_(monolayer)) = (k * P)/(1 + k * P)`

or
`(V_(monolayer))/(V) = (1 + k * P)/(kP)`

`(V_(monolayer))/(V) = (1)/(kP) + 1`

k =
`((V)/(V_(monolayer)))/(P + (1 - (V)/(V_(monolayer))))`

Now, equation
`P_(1)` and
`P_(2)` for
`V_(monolayer)` as follows.

`((V_(1))/(V_(monolayer)))/(P + (1 - (V_(1))/(V_(monolayer))))` =
`((V_(2))/(V_(monolayer)))/(P + (1 - (V_(2))/(V_(monolayer))))`

`(P_(1)(V_(monolayer) - V_(1))/(V_(1))` =
`(P_(2)(V_(monolayer) - V_(2))/(V_(2))`

`V_(monolayer) = (P_(1) - P_(2))/((P_(1))/(V_(1))) - (P_(2))/(V_(2))` .......... (1)

As it is given that
`P_(1)` is 56.4 kPa,
`P_(2)` is 108 kPa,
`V_(1)` is
`1.52 cm^(3)` and
`V_(2)` is
`2.77 cm^(3)`.

Therefore, putting these values into equation (1) as follows.

`V_(monolayer) = (P_(1) - P_(2))/((P_(1))/(V_(1))) - (P_(2))/(V_(2))`

=
`(56.4 kPa - 108 kPa)/((56.4 kPa)/(1.52 cm^(3))) - (108 kPa)/(2.77 cm^(3))`

= 27.44
`cm^(3)`

Thus, we can conclude that volume of monolayer is 27.44
`cm^(3)`.