What are the units for the Universal Gas Constant?

AI Thread Summary
The discussion focuses on understanding the units and values used in the ideal gas law for a specific problem involving a compressed air tank. Key points include the calculation of air density using the formula ρ = p/RT, where the total pressure is the sum of gauge and atmospheric pressure. The conversion factor of 144 in²/ft² is explained as necessary for unit consistency when calculating pressure in psi to lb/ft². Additionally, the conversion from Fahrenheit to Rankine is clarified, with the correct formula being °R = °F + 460. The Universal Gas Constant (R) is highlighted, with various alternative units provided for clarity.
Minh Nguyen
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Homework Statement


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.)

Homework Equations


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

The Attempt at a Solution


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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.
 
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I don't understand where the (144 in2/ft2) came from, as well as the (1716 ft * lb/slug * °R), and [(70+460) °R].
The units of that 144 term are a hint.

What value would you have used for R in the formula ##\rho=p/RT## ? You would have probably used a google search to determine R. What name is R known as?

How would you convert 70° F to degrees Rankine to use in your Gas Law formula?
 
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I was going to use 8.314 K-1mol-1 since that is the gas constant for air.

I looked through my textbook and found that °R = °F + 459.67, so I would do 70°F + 460 = °R.

Would I do something with the volume to get the 144? I am not even sure what that number is doing there since the equation just says absolute pressure should be the numerator.
 
What is the result if you take the square root of everything in the brackets here: ##\left( 144\ \dfrac {in^2}{ft^2} \right)##
 
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You would get 12 in/ft.
 
Minh Nguyen said:
I was going to use 8.314 K-1mol-1 since that is the gas constant for air.
Almost.

Choose whichever of the alternative forms best suits you and your data, e.g.,
The Universal Gas Constant - Ru - in alternative Units
  • atm.cm3/(mol.K) : 82.0575
  • atm.ft3/(lbmol.K) : 1.31443
  • atm.ft3/(lbmol.°R) : 0.73024
  • atm.l/(mol.K) : 0.08206
  • bar.cm3/(mol.K) : 83.14472
  • bar.l/(mol.K) : 0.08314472
  • Btu/(lbmol.°R) : 1.9859
  • cal/(mol.K) : 1.9859
  • erg/(mol.K) : 83144720
  • hp.h/(lbmol.°R) : 0.0007805
  • inHg.ft3/(lbmol.°R) : 21.85
  • J/(mol.K) : 8.3144598
  • kJ/(kmol.K) : 8.3144598
  • J/(kmol.K) : 8314.472
  • (kgf/cm2).l/(mol.K) : 0.084784
  • kPa.cm3/(mol.K) : 8314.472
  • kWh/(lbmol.°R) : 0.000582
  • lbf.ft/(lbmol.°R) : 1545.349
  • mmHg.ft3/(lbmol.K) : 999
  • mmHg.ft3/(lbmol.°R) : 555
  • mmHg.l/(mol.K) : 62.364
  • Pa.m3/(mol.K) : 8.314472
  • psf.ft3/(lbmol.°R) : 1545.349
  • psi.ft3/(lbmol.°R) : 10.73
  • Torr.cm3/(mol.K) : 62364
from http://www.engineeringtoolbox.com/individual-universal-gas-constant-d_588.html
 
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