Solving Pressure Question: Volume, Mass of Air @ 70F & 200psi

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In summary, the conversation is about finding the volume and specific volume of a pressurized tank containing air at 70 F and 200 psi. The ideal gas law is mentioned as a possible solution, but there is confusion about converting units. The correct equation to use is P*v=RT, where P is pressure in lb/ft^2, v is specific volume in ft^3/lbm, R is the universal gas constant for air, and T is temperature in °R. The answer to the question is not in ft^3/lbm and the person asking for help is feeling frustrated.
  • #1
Frustra8ed
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I know this question is easy, but for some reason I am unable to figure it out.

The first part of the problem wants to know what the volume is in a pressurized tank. I figured it and it equals 2.1 ft^3.

The second part asks for the specific volume at 70 F and 200 psi.
So far all I have is v(specific volume) = V/m. Or 1/d = v

The gas in the tank is air, so basically I need to find the mass of air @ 70 F and 200 psi. I have the volume, so that is not the problem.

The answer to the question is .98 ft^3/lbm

Should I be using ideal gas laws? If I do use the ideal, I'm assuming I need to convert units. I'm having a problem converting psi to lbf/ft^2...

I can't come up with this answer, so if anyone can give me a hand I would appreciate it.

Thanks,

Frustr8ed, as the name states. :confused:
 
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  • #2
Frustra8ed said:
I know this question is easy, but for some reason I am unable to figure it out.

The first part of the problem wants to know what the volume is in a pressurized tank. I figured it and it equals 2.1 ft^3.

The second part asks for the specific volume at 70 F and 200 psi.
So far all I have is v(specific volume) = V/m. Or 1/d = v

The gas in the tank is air, so basically I need to find the mass of air @ 70 F and 200 psi. I have the volume, so that is not the problem.

The answer to the question is .98 ft^3/lbm

Should I be using ideal gas laws? If I do use the ideal, I'm assuming I need to convert units. I'm having a problem converting psi to lbf/ft^2...

I can't come up with this answer, so if anyone can give me a hand I would appreciate it.

Thanks,

Frustr8ed, as the name states. :confused:
It would help if you would state the whole question. Use the ideal gas equation of state, PV=nRT to find n (number of moles). The answer won't be in ft^3/lbm.

AM
 
  • #3
You may want to consider using this form of the ideal gas law:

[tex] P \nu = RT [/tex]

where:

[tex]P = Pressure (Lb_{f}/ft^2)[/tex]
[tex]\nu = Specific Volume (ft^3/Lb_{m})[/tex]
[tex]R = Universal Gas Constant (for air: 53.35 Ft*Lb_{f}/°R*LB_{m})[/tex]
[tex]T = Temperature (°R) [/tex]
 
  • #4
Thanks for the responses. I believe that last equation is the one I need to use, but I was having trouble converting units.

Thanks Again,

Frustra8ed
 

1. What is the equation for calculating pressure?

The equation for calculating pressure is P = F/A, where P is pressure, F is force, and A is area.

2. How do I calculate volume using pressure and mass of air?

To calculate volume, you can use the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin.

3. What is the relationship between pressure and temperature in gases?

The relationship between pressure and temperature in gases is known as Charles's Law, which states that at a constant volume, the pressure of a gas is directly proportional to its temperature.

4. How do I convert pressure from psi to other units?

To convert pressure from psi (pounds per square inch) to other units, you can use the following conversion factors: 1 psi = 6.89476 kilopascals (kPa), 1 psi = 0.0689476 bar, 1 psi = 0.068046 atmospheres (atm), and 1 psi = 6894.76 pascals (Pa).

5. How does the mass of air affect pressure?

The mass of air does not directly affect pressure, but it can indirectly affect it by changing the volume of the system. In a closed system, an increase in mass will result in an increase in pressure, and a decrease in mass will result in a decrease in pressure, assuming all other variables are constant.

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