/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 A neon sign is made of glass tub... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 36

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?

Short Answer

Expert verified
The neon sign contains approximately 0.0155 g of neon.

Step by step solution

01

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.
02

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 \(V = \pi (0.015)^2 (10) \, \mathrm{m}^3\).
03

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 \(n = \frac{PV}{RT}\).
04

Convert Moles to Grams

The molar mass of neon is approximately \(20.18 \, \mathrm{g/mol}\). Convert moles to grams using \(\text{mass} = n \times \text{molar mass}\).
05

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 = \(7.68 \times 10^{-4} \, \mathrm{mol} \times 20.18 \, \mathrm{g/mol} \approx 0.0155 \, \mathrm{g}\).
06

Conclusion

The mass of neon in the sign is approximately \(0.0155 \, \mathrm{g}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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:
  • \[ V = \pi r^2 h \]
Where \( r \) is the radius of the cylinder and \( h \) is its height (or length in this context). Here, the inside diameter of the tube is given as 3.0 cm, which means the radius is half of the diameter. So, it's 1.5 cm or 0.015 m when converted to meters.
Plug this into the formula with the length of the tube, 10.0 m, to find the volume:
  • \[ V = \pi (0.015)^2 (10) = 0.00707 \, \text{m}^3 \]
Using the correct units is crucial here, as we'll see when applying the Ideal Gas Law.
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:
  • \[ \text{mass} = n \times \text{molar mass} \]
Where \( n \) is the number of moles. For our calculation, it was approximately 7.68 x 10-4 moles. Through multiplication, we find the mass of neon in grams:
  • \[ \text{mass} = 7.68 \times 10^{-4} \, \text{mol} \times 20.18 \, \text{g/mol} = 0.0155 \, \text{g} \]
This value tells us how heavy the neon gas inside the sign is.
Pressure and Temperature Relations
In the Ideal Gas Law, the relations between pressure and temperature are essential for calculations. The law is expressed as:
  • \[ PV = nRT \]
Where:
  • \( P \) is pressure,
  • \( V \) is volume,
  • \( n \) is the number of moles,
  • \( R \) is the ideal gas constant, and
  • \( T \) is temperature in Kelvin.
It's critical to remember that temperature must be in Kelvin for these calculations. Convert it from Celsius by adding 273.15. So, 30掳C is 303.15 K.
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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Most popular questions from this chapter

In the contact process, sulfur dioxide and oxygen gas react to form sulfur trioxide as follows: $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) $$ At a certain temperature and pressure, \(50 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) reacts with \(25 \mathrm{~L}\) of \(\mathrm{O}_{2}\). If all the \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are consumed, what volume of \(\mathrm{SO}_{3}\), at the same temperature and pressure, will be produced?

When a large evacuated flask is filled with argon gas, its mass increases by \(3.224 \mathrm{~g}\). When the same flask is again evacuated and then filled with a gas of unknown molar mass, the mass increase is \(8.102 \mathrm{~g}\). (a) Based on the molar mass of argon, estimate the molar mass of the unknown gas. (b) What assumptions did you make in arriving at your answer?

The atmospheric concentration of \(\mathrm{CO}_{2}\) gas is presently 407 ppm (parts per million, by volume; that is, 407 L of every \(10^{6} \mathrm{~L}\) of the atmosphere are \(\mathrm{CO}_{2}\) ). What is the mole fraction of \(\mathrm{CO}_{2}\) in the atmosphere?

Suppose you are given two flasks at the same temperature, one of volume \(2 \mathrm{~L}\) and the other of volume \(3 \mathrm{~L}\). The 2 -L flask contains \(4.8 \mathrm{~g}\) of gas, and the gas pressure is \(x \mathrm{kPa}\). The 3 -L flask contains \(0.36 \mathrm{~g}\) of gas, and the gas pressure is \(0.1 x\). Do the two gases have the same molar mass? If not, which contains the gas of higher molar mass?

An ideal gas at a pressure of \(152 \mathrm{kPa}\) is contained in a bulb of unknown volume. A stopcock is used to connect this bulb with a previously evacuated bulb that has a volume of \(0.800 \mathrm{~L}\) as shown here. When the stopcock is opened, the gas expands into the empty bulb. If the temperature is held constant during this process and the final pressure is \(92.66 \mathrm{kPa}\), what is the volume of the bulb that was originally filled with gas?
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