Using the Ideal Gas Law

Minh Nguyen

Hello,

I am not asking for the answer to an example, rather how the book got some numbers. The problem is an example from the book and shows me the solution but does not show the steps.

Given: The compressed air tank has a volume of .84 ft^3. The temperature is 70 F and the atmospheric pressure is 14.7 psi (abs).

Find: When the tank is filled with air at a gage pressure of 50 psi, determine the density of the air and the weight of air in the tank.)

I am told to use the ideal gas law, ρ = p/RT.

Then the book says,

ρ = (50 lb/in2 + 14.7 lb/in2)(144 in2/ft2) / (1716 ft * lb/slug * °R) [(70 + 460) ° R]

So I understand where the (50 lb/in2 + 14.7 lb/in2) is coming from, they are just adding the gage pressure as well as the atmospheric pressure.

BUT, I don't understand where the (144 in2/ft2) came from, as well as the (1716 ft * lb/slug * °R), and [(70+460) °R].

Any help will be appreciated.

Related Mechanical Engineering News on Phys.org

Chestermiller

Mentor
The 1716 is the is the universal gas constant in the Imperial units indicated, and the 70 + 460 is the Rankine temperature. The gas constant value employed calls for the pressure in psf rather than psi. Thats where the 144 comes in.

"Using the Ideal Gas Law"

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving