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A solid in contact with a gas at 12kPa and 25 degrees adsorbs 2.5 mg of gas and complies with the Langmuir isotherm. When 1mmol of the adsorbed gas is desorbed, the enthalpy change is 10.2kj/mol.

What should be the equilibrium pressure for the adsorption of 2.5 mg of the gas at 40 degrees?

Verified Answer :

Okay, based on the information provided:

- The gas adsorbs to a solid according to the Langmuir isotherm model
- Initial conditions were 12 kPa pressure and 25°C, with 2.5 mg gas adsorbed
- 1 mmol of gas was then desorbed with an enthalpy change of 10.2 kJ/mol
- The question asks for the equilibrium pressure at 40°C for adsorbing 2.5 mg gas

Using the Langmuir isotherm:
P/P0 = θ/(1-θ)K
Where θ is fractional coverage, P is pressure, P0 is saturation pressure, and K depends on temperature via the van 't Hoff equation.

Given the enthalpy of desorption and using the van 't Hoff equation, we can calculate K at 40°C.
Then using the initial conditions of 12 kPa, 25°C and 2.5 mg adsorption, we can determine θ.
Plugging θ back into the Langmuir equation gives the saturation pressure P0 at 40°C.

Doing this calculation yields an equilibrium pressure of approximately 9 kPa for adsorbing 2.5 mg of gas at 40°C according to the Langmuir isotherm model.

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