A 0.0372-m3 container is initially evacuated. Then, 4.65 g of water is placed in the container, and, after some time, all of the water evaporates. If the temperature of the water vapor is 368 K, what is its pressure

18.3 kilopascals

Explanation:

We are given that the volume of this container is 0.0372 meters^3, that the mass of water is 4.65 grams, and that the temperature of this water vapor ( over time ) is 368 degrees Kelvins. This is a problem where the ideal gas law is an " ideal " application.

First calculate the number of moles present in the water ( H2O ). Water has a mass of 18, so it should be that n, in the ideal gas law - PV = nRT, is equal to 4 / 18. It is the amount of the substance.

We now have enough information to solve for P in PV = nRT,

P( 0.0372 ) = 4 / 18( 8.314 )( 368 ),

P ≈ 18,276.9

Pressure ≈ 18.3 kilopascals

Hope that helps!

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