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bulgar [2K]
3 years ago
10

Water vapor is condensed steam. True False

Chemistry
1 answer:
adelina 88 [10]3 years ago
8 0

False, because water vapor, water vapour or aqueous vapor, is the gaseous phase of water. It is one state of water within the hydrosphere. Water vapor can be produced from the evaporation or boiling of liquid water or from the sublimation of ice. Unlike other forms of water, water vapor is invisible.

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If a small weather balloon with a volume of 40.0 L at a pressure of 1.00 atmosphere was deployed at the edge of Typhoon Odessa,
Feliz [49]

Answer: The volume of the balloon at the center of the typhoon is 41.7L.

Note: The complete question is given below;

If a small weather balloon with a volume of 40.0 L at a pressure of 1.00 atmosphere was deployed at the edge of Typhoon Odessa, what was the volume of the balloon when it reached the center?

The severity of a tropical storm is related to the depressed atmospheric pressure at its center. In August 1985, Typhoon Odessa in the Pacific Ocean featured maximum winds of about 90 mi/hr and pressure that was 40.0 mbar lower at the center than normal atmospheric pressure. In contrast, the central pressure of Hurricane Andrew (pictured) was 90.0 mbar lower than its surroundings when it hit south Florida with winds as high as 165 mi/hr.

Explanation:

Since no temperature changes were given, it is assumed to be constant. Therefore, Boyle's law which describes the relationship between pressure and volume is used to determine the new volume at the center of Typhoon Odessa. Mathematically, Boyle's law states that; P1V1 = P2V2

Assuming 1atm = 1 bar, 1mbar = 0.001atm, 40mbar = 0.040atm

P1 = 1.0atm, V1 = 40.0L, P2 = 1atm - 0.040atm = 0.960atm, V2 = ?

Using P1V1 = P2V2

V2 = P1V1/P2

V2 = 1.0 * 40.0 / 0.96

V2 = 41.67L

Therefore, the volume of the balloon at the center of the typhoon is 41.7L.

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