Question

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

Answer 1

Answer:

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.

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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!