Math Problem Statement

PV= 0.08206nT A 30.0 mole% CO and 70.0 mole% N2 gas mixture is stored in a cylinder with a volume of 3.5 ft3 at a temperature of 85°F. The reading on a Bourdon gauge attached to the cylinder is 500 psi. Calculate the total amount of gas (lb-mole) and the mass of CO(lbm) in the tank.

Solution

We can solve this problem step by step, using the ideal gas law and converting all necessary units to ensure consistency.

Given:

  • The mole fraction of CO (XCOX_{CO}) = 30.0% = 0.30
  • The mole fraction of N2N_2 (XN2X_{N_2}) = 70.0% = 0.70
  • The volume of the gas mixture (VV) = 3.5 ft³
  • The temperature (TT) = 85°F
  • The gauge pressure (PgaugeP_{gauge}) = 500 psi
  • Ideal gas constant (RR) = 0.08206 atm\cdotpLmol\cdotpK\frac{\text{atm·L}}{\text{mol·K}}

Step 1: Convert the pressure and temperature to appropriate units.

  • Convert pressure from psi to atm: Ptotal=Pgauge+Patm=500psi+14.7psi=514.7psiP_{total} = P_{gauge} + P_{atm} = 500 \, \text{psi} + 14.7 \, \text{psi} = 514.7 \, \text{psi} Since 1atm=14.7psi1 \, \text{atm} = 14.7 \, \text{psi}, we can convert the pressure to atm: Ptotal=514.7psi14.7psi/atm35.02atmP_{total} = \frac{514.7 \, \text{psi}}{14.7 \, \text{psi/atm}} \approx 35.02 \, \text{atm}

  • Convert temperature from °F to K: T(K)=59(8532)+273.15=29.44+273.15=302.59KT(K) = \frac{5}{9} (85 - 32) + 273.15 = 29.44 + 273.15 = 302.59 \, \text{K}

Step 2: Convert the volume from cubic feet (ft³) to liters (L).

Since 1ft³=28.3168L1 \, \text{ft³} = 28.3168 \, \text{L}: V=3.5ft³×28.3168L/ft³99.1098LV = 3.5 \, \text{ft³} \times 28.3168 \, \text{L/ft³} \approx 99.1098 \, \text{L}

Step 3: Use the ideal gas law to find the total number of moles of gas (nn).

The ideal gas law is: PV=nRTPV = nRT Solving for nn: n=P×VR×T=35.02×99.10980.08206×302.59140.15molesn = \frac{P \times V}{R \times T} = \frac{35.02 \times 99.1098}{0.08206 \times 302.59} \approx 140.15 \, \text{moles}

Step 4: Convert moles to lb-moles.

Since 1 lb-mole = 453.592 moles: nlbmole=140.15453.5920.309lbmolen_{lb-mole} = \frac{140.15}{453.592} \approx 0.309 lb-mole

Step 5: Calculate the mass of CO in the tank.

The mole fraction of CO is 30%, so the moles of CO (nCOn_{CO}) is: nCO=0.30×140.15=42.045moles of COn_{CO} = 0.30 \times 140.15 = 42.045 \, \text{moles of CO}

The molecular weight of CO is 28.01g/mol28.01 \, \text{g/mol}. To convert to pounds mass (lbm): Mass of CO=42.045moles×28.01g/mol=1177.56g\text{Mass of CO} = 42.045 \, \text{moles} \times 28.01 \, \text{g/mol} = 1177.56 \, \text{g} Convert to lbm: Mass of CO=1177.56453.5922.60lbm\text{Mass of CO} = \frac{1177.56}{453.592} \approx 2.60 \, \text{lbm}

Final Answers:

  • Total amount of gas: 0.309lb-moles0.309 \, \text{lb-moles}
  • Mass of CO: 2.60lbm2.60 \, \text{lbm}

Would you like more details or clarifications?


Relative questions:

  1. How does the ideal gas law change if the gas is not ideal?
  2. What is the importance of the gauge pressure reading in this context?
  3. How would the total mass change if the mole fraction of CO increased?
  4. How do different gas mixtures affect pressure in a cylinder?
  5. How does temperature affect the number of moles of gas in a closed container?

Tip: Always ensure unit consistency when applying gas laws to avoid calculation errors.

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Math Problem Analysis

Mathematical Concepts

Ideal Gas Law
Unit Conversion
Partial Pressure
Mole Fraction

Formulas

Ideal Gas Law: PV = nRT
Pressure conversion: P(atm) = P(psi) / 14.7
Temperature conversion: T(K) = (5/9) × (T(°F) - 32) + 273.15
Volume conversion: 1 ft³ = 28.3168 L
Mass of CO: m(CO) = n(CO) × Molecular weight of CO

Theorems

Ideal Gas Law

Suitable Grade Level

Grades 11-12 (Advanced Chemistry/Physics)