A gas adsorbs on a solid following the Langmuir isotherm. At $T$ constant, the volume adsorbed (STP) per gram is $10.0 c m^{3}$ at $8.0 kPa$ and $15.0 c m^{3}$ at $24.0 kPa$ . Predict the volume adsorbed per gram at $40.0 kPa$ in $c m^{3}$ (two decimals).

Numerical answer type, answer in cm^3, tolerance 0.05.

Answer: 16.6667 cm^3

Use

p / V = p / V_{m} + 1/ (K V_{m})

. The points give

p / V = 0.800

and

1.600

, so slope

= 0.800/16.0 = 0.0500

and

V_{m} = 20.0 c m^{3}

. Intercept

= 0.800 - 0.0500 \times 8 = 0.400 = 1/ (K V_{m})

, so

K = 0.125 kP a^{- 1}

. At

40 kPa

V = 40/ (0.0500 \times 40 + 0.400) = 16.67 c m^{3}