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maria [59]
3 years ago
6

Most full scuba tanks have a pressure of between 1,800 p.s.i. and 3,000 p.s.i. (pounds per square inch). Approximately 70 to 85

cubic feet of air has been squeezed into the tank. When a diver takes his air tank to be refilled, the dive shop will hook his tank to a compressor with a pressure hose and place his tank in a tub of water while it is being filled. What reasons can you give for placing the tank in water?
Chemistry
2 answers:
ArbitrLikvidat [17]3 years ago
8 0

Answer: The Diver's air tank is filled with oxygen gas.

The filling of air tank while keeping in water can be explained in reference to the Charles' Law.

Charles' Law: It states that the volume of the ideal gas is directly proportional to the absolute temperature at constant pressure.

                        V\propto T    (at constant Pressure)

The Scuba tank contains cold water and is present at high pressure. The gas which is put in the tank will be compact and would not expand. As the temperature is low inside the tank, hence by using Charles' Law, the volume will also decrease inside the tank.

When the air tank is filled, it is always kept in a tub of water because the pressure is high and hence, more oxygen gas can be filled inside the air tank.


Vlad [161]3 years ago
7 0
<span>I would think that because water has a high specific heat, the water in the tank must be cold. Because the scuba tank is at high pressure, and the water is cold, the gas put into the tank will be more compact, and will not expand. According to Charles' law, as temperature increases, volume increases and because in this situation temperature is decreasing, volume will decrease. It would be great to keep the tank into the water because under these conditions, you can store more oxygen into the tank per square inch.

I hope this helps:)</span>
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A gas with a pressure of 820.4 mmHg occupies a
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Answer:

V₂ = 1223.2 mL  

Explanation:

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Pressure of gas = 820.4 mmHg

Initial volume of gas = 900.0 mL

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Final volume = ?

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Solution:

The given problem will be solve through the Charles Law.

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V₁/T₁ = V₂/T₂

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T₁ = Initial temperature

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A volume of 5.0 L of gas at a temperature of 25°C is cooled to a temperature of -120°C at constant pressure. What is the new vol
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V1 = 5L = 5000ml V2 = ?
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CH_{3}COOH(aq)-->CH_{3}COO^{-}(aq)+H^{+}(aq)

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