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.

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 (\(X_{CO}\)) = 30.0% = 0.30
- The mole fraction of \(N_2\) (\(X_{N_2}\)) = 70.0% = 0.70
- The volume of the gas mixture (\(V\)) = 3.5 ft³
- The temperature (\(T\)) = 85°F
- The gauge pressure (\(P_{gauge}\)) = 500 psi
- Ideal gas constant (\(R\)) = 0.08206 \(\frac{\text{atm·L}}{\text{mol·K}}\)

#### Step 1: Convert the pressure and temperature to appropriate units.
- **Convert pressure** from psi to atm:
  \[
  P_{total} = P_{gauge} + P_{atm} = 500 \, \text{psi} + 14.7 \, \text{psi} = 514.7 \, \text{psi}
  \]
  Since \(1 \, \text{atm} = 14.7 \, \text{psi}\), we can convert the pressure to atm:
  \[
  P_{total} = \frac{514.7 \, \text{psi}}{14.7 \, \text{psi/atm}} \approx 35.02 \, \text{atm}
  \]

- **Convert temperature** from °F to K:
  \[
  T(K) = \frac{5}{9} (85 - 32) + 273.15 = 29.44 + 273.15 = 302.59 \, \text{K}
  \]

#### Step 2: Convert the volume from cubic feet (ft³) to liters (L).
Since \(1 \, \text{ft³} = 28.3168 \, \text{L}\):
\[
V = 3.5 \, \text{ft³} \times 28.3168 \, \text{L/ft³} \approx 99.1098 \, \text{L}
\]

#### Step 3: Use the ideal gas law to find the total number of moles of gas (\(n\)).
The ideal gas law is:
\[
PV = nRT
\]
Solving for \(n\):
\[
n = \frac{P \times V}{R \times T} = \frac{35.02 \times 99.1098}{0.08206 \times 302.59} \approx 140.15 \, \text{moles}
\]

#### Step 4: Convert moles to lb-moles.
Since 1 lb-mole = 453.592 moles:
\[
n_{lb-mole} = \frac{140.15}{453.592} \approx 0.309 lb-mole
\]

### Step 5: Calculate the mass of CO in the tank.
The mole fraction of CO is 30%, so the moles of CO (\(n_{CO}\)) is:
\[
n_{CO} = 0.30 \times 140.15 = 42.045 \, \text{moles of CO}
\]

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

### Final Answers:
- Total amount of gas: \(0.309 \, \text{lb-moles}\)
- Mass of CO: \(2.60 \, \text{lbm}\)

Would you like more details or clarifications?

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### 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.

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.

This problem involves calculating the total gas amount in lb-moles and the mass of CO in a gas mixture of CO and N2, using the ideal gas law. With given pressure, volume, and temperature data, this solution provides a step-by-step guide, including unit conversions. Suitable for advanced high school chemistry or physics students.

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