Solve Gas Laws Questions: Density, Mass, Volume, Pressure

AI Thread Summary
The discussion focuses on solving gas law problems related to density, mass, volume, and pressure. The first problem involves calculating the mass of 3.0 L of air at a higher pressure and temperature, with an initial attempt yielding around 21 grams. The second problem addresses the volume change of a 2.0-cm³ air bubble released from a depth of 62.4 m, requiring the application of Boyle's Law and hydrostatic pressure calculations. Participants emphasize the importance of understanding hydrostatic pressure to determine the pressure at depth and how it relates to the volume of the bubble at the surface. Overall, the thread provides guidance on applying gas laws to solve these specific physics problems.
tommy1
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1. The density of air is 1.3g/L at standard conditions of 101kPa and 0 degrees celcius. determine the mass of 3.0 L of air at a pressure of 138.2 kPa and a temp of 273 degrees celcius.

2. One atmosphere of air pressure supports a column of water approximately 10.3 m high. A 2.0-cm^3 bubble of air is released by a diver working 62.4 m below the surface of a deep lake. If its temp remains constant, what is the volume of the bubble as it reaches the surface? the pressure on the surface of the lake is one atm.


Homework Equations


i need help with bother of these. alli know is that you use the gas laws.


The Attempt at a Solution

for the first one...i worked at it with a friend at we(mostly him) got i think 21 grams? i left my answer sheet in school but it was around there. He never really got to explain it to me so I am confused. And for the second I am sorry but i have no idea even how to start that one.
 
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tommy1 said:
2. One atmosphere of air pressure supports a column of water approximately 10.3 m high. A 2.0-cm^3 bubble of air is released by a diver working 62.4 m below the surface of a deep lake. If its temp remains constant, what is the volume of the bubble as it reaches the surface? the pressure on the surface of the lake is one atm.

You can use Boyle-Mariotte law here (p_{1}V_{1}=p_{2}V_{2}), since the temperature remains constant. You only need to determine pressures on the surface and at the given depth using the formula for hydrostatic pressure (p=\rho g h).
 
Dr. Jekyll said:
You can use Boyle-Mariotte law here (p_{1}V_{1}=p_{2}V_{2}), since the temperature remains constant. You only need to determine pressures on the surface and at the given depth using the formula for hydrostatic pressure (p=\rho g h).

can you explain that second equation a little more indepth please
 
The second equation is for the hydrostatic pressure. It is caused by the weight of the fluid (density rho) above the certain depth (h).

You say that one atmosphere pressure is equal to hydrostaic pressure 10.3 m below water. For example, the same thing is equal to hydrostatic pressure 735 mm under mercury (735 mmHg).

So, all you need is to calculate pressure under water (since you know the one on the surface, p_{0}): p=p_{0}+\rho g h. You can also do the whole thing without even calculating the pressures (only with heights) if you write the equations in different way.
 
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