Question Details

The relative humidity of ambient air at 300 K is 50% with a partial pressure of water vapour equal to P𝒗. The saturation pressure of water at 300 K is P𝒔𝒂𝒕. The correct relation for the air-water mixture is

Options

A

P𝑣 = 0.5 Pπ‘ π‘Žt

B

P𝑣 = 0.5 Pπ‘ π‘Žt

C

P𝑣 = 0.622 Pπ‘ π‘Žt

D

P𝑣 = 2 Pπ‘ π‘Žt

Correct Answer :

P𝑣 = 0.622 Pπ‘ π‘Žt

Solution :

The correct option is: P𝑣 = 0.622 PοΏ½οΏ½οΏ½οΏ½π‘Žt.

Step-by-Step Explanation:

In psychrometrics and thermodynamics, an air-water vapour mixture is often analyzed using the molecular weights of the constituent gases. The ratio of the molecular mass of water vapour (Mw ≈ 18.016 g/mol) to the molecular mass of dry air (Ma ≈ 28.97 g/mol) is a fundamental constant used in calculations.

This molecular weight ratio is given by:

Mw Ma = 18.016 28.97 0.622

This constant of 0.622 is key in defining relations such as specific humidity (humidity ratio). For the given air-water mixture, this ratio establishes the correct proportional relation between the partial pressure of water vapour Pv and the saturation pressure Psat as:

Pv = 0.622 Psat

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