Air consists mainly of four gases. One sample taken from Honolulu exerts a total pressure of 760.0 mmHg. Nitrogen exerts 585.2 mmHg, oxygen exerts 144.4 mmHg, and argon exerts 6.8 mmHg. What pressure in mmHg is exerted by water vapor?

23.6 mmHg.

Explanation:

From the question given above, the following data were obtained:

Total pressure (Pₜ) = 760.0 mmHg.

Pressure of nitrogen (Pₙ) = 585.2 mmHg

Pressure of oxygen (Pₒ) = 144.4 mmHg,

Pressure of argon (Pₐ) = 6.8 mmHg.

Pressure of water vapor (Pᵥ) =?

The pressure exerted by the water vapor can be obtained as follow:

Pₜ = Pₙ + Pₒ + Pₐ + Pᵥ

760 = 585.2 + 144.4 + 6.8 + Pᵥ

760 = 736.4 + Pᵥ

Collect like terms

760 – 736.4 = Pᵥ

23.6 = Pᵥ

Pᵥ = 23.6 mmHg

Thus, the pressure exerted by the water vapor is 23.6 mmHg

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