Problem 36
Question
A neon sign is made of glass tubing whose inside diameter is \(3.0 \mathrm{~cm}\) and length is \(10.0 \mathrm{~m}\). If the sign contains neon at a pressure of \(265 \mathrm{~Pa}\) at \(30^{\circ} \mathrm{C}\), how many grams of neon are in the sign?
Step-by-Step Solution
Verified Answer
The neon sign contains approximately 0.0155 g of neon.
1Step 1: Understand the Given Data
We have a glass tube with an inside diameter of \(3.0 \, \mathrm{cm}\), a length of \(10.0 \, \mathrm{m}\), containing neon gas. The pressure is \(265 \, \mathrm{Pa}\) and the temperature is \(30^{\circ}\mathrm{C}\). We need to find the mass of neon in the tube.
2Step 2: Calculate the Volume of the Tube
The tube is a cylinder, and its volume can be calculated using the formula \(V = \pi r^2 h\), where \(r\) is the radius and \(h\) is the length. The radius is half of the diameter, so \(r = 1.5 \mathrm{~cm} = 0.015 \mathrm{~m}\). The volume is \(V = \pi (0.015)^2 (10) \, \mathrm{m}^3\).
3Step 3: Use Ideal Gas Law to Find Moles of Neon
The ideal gas law is \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant (\(8.314 \, \mathrm{m^3 \, Pa \, K^{-1} \, mol^{-1}}\)), and \(T\) is the temperature in Kelvin. \(T = 30^{\circ} \mathrm{C} + 273.15 = 303.15 \mathrm{~K}\). Rearrange to find \(n = \frac{PV}{RT}\).
4Step 4: Convert Moles to Grams
The molar mass of neon is approximately \(20.18 \, \mathrm{g/mol}\). Convert moles to grams using \(\text{mass} = n \times \text{molar mass}\).
5Step 5: Conduct Calculations
Calculate the volume: \(V = \pi (0.015)^2 (10) = 0.00707 \, \mathrm{m^3}\). Calculate moles: \(n = \frac{265 \, \mathrm{Pa} \times 0.00707 \, \mathrm{m^3}}{8.314 \, \mathrm{m^3 \, Pa \, K^{-1} \, mol^{-1}} \times 303.15 \, \mathrm{K}}\). This gives \(n \approx 7.68 \times 10^{-4} \, \mathrm{mol}\). Calculate mass: mass = \(7.68 \times 10^{-4} \, \mathrm{mol} \times 20.18 \, \mathrm{g/mol} \approx 0.0155 \, \mathrm{g}\).
6Step 6: Conclusion
The mass of neon in the sign is approximately \(0.0155 \, \mathrm{g}\).
Key Concepts
Gas Volume CalculationMolar Mass ConversionPressure and Temperature Relations
Gas Volume Calculation
Calculating the volume of gas inside a container is a key skill when dealing with the Ideal Gas Law. In this exercise, we're working with a cylindrical glass tube. The volume of a cylinder can be determined using the formula:
Plug this into the formula with the length of the tube, 10.0 m, to find the volume:
- \[ V = \pi r^2 h \]
Plug this into the formula with the length of the tube, 10.0 m, to find the volume:
- \[ V = \pi (0.015)^2 (10) = 0.00707 \, \text{m}^3 \]
Molar Mass Conversion
Once you've determined the amount of substance in moles, converting moles to grams is straightforward if the molar mass is known. Molar mass is the mass of one mole of a substance, and for neon, it's about 20.18 g/mol. This means every mole of neon weighs 20.18 grams.
In this problem, once we solve for the number of moles of neon using the Ideal Gas Law, we use the equation:
In this problem, once we solve for the number of moles of neon using the Ideal Gas Law, we use the equation:
- \[ \text{mass} = n \times \text{molar mass} \]
- \[ \text{mass} = 7.68 \times 10^{-4} \, \text{mol} \times 20.18 \, \text{g/mol} = 0.0155 \, \text{g} \]
Pressure and Temperature Relations
In the Ideal Gas Law, the relations between pressure and temperature are essential for calculations. The law is expressed as:
The pressure of the neon in the tube is given as 265 Pa. Plugging these into the Ideal Gas Law allows us to solve for moles of gas. This step connects the physical state of the gas (pressure and temperature) to its quantity (moles), allowing for further conversion to mass.
- \[ PV = nRT \]
- \( P \) is pressure,
- \( V \) is volume,
- \( n \) is the number of moles,
- \( R \) is the ideal gas constant, and
- \( T \) is temperature in Kelvin.
The pressure of the neon in the tube is given as 265 Pa. Plugging these into the Ideal Gas Law allows us to solve for moles of gas. This step connects the physical state of the gas (pressure and temperature) to its quantity (moles), allowing for further conversion to mass.
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