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Chapter 14: Problem 61

The pressure of sulfur dioxide \(\left(\mathrm{SO}_{2}\right)\) is \(2.12 \times 10^{4} \mathrm{Pa} .\) There are 421 moles of this gas in a volume of \(50.0 \mathrm{m}^{3} .\) Find the translational rms speed of the sulfur dioxide molecules.

Short Answer

Expert verified
The translational rms speed of \(\text{SO}_2\) molecules is approximately 258.4 m/s.

Step by step solution

01

Write Down Given Values

We have the pressure \(P = 2.12 \times 10^4 \, \text{Pa}\), number of moles \(n = 421\), and volume \(V = 50.0 \, \text{m}^3\). We need to find the translational root mean square (rms) speed of \(\text{SO}_2\) molecules.
02

Use the Ideal Gas Equation

The ideal gas law is given by \(PV = nRT\), where \(R = 8.31 \, \text{J/(mol K)}\) is the gas constant. Solving for temperature \(T\), we get \(T = \frac{PV}{nR}\). Substituting the known values:\[T = \frac{2.12 \times 10^4 \, \text{Pa} \times 50.0 \, \text{m}^3}{421 \, \text{mole} \times 8.31 \, \text{J/(mol K)}}\]
03

Calculate Temperature \(T\)

Plug in the values into the equation from Step 2:\[T = \frac{2.12 \times 10^4 \times 50.0}{421 \times 8.31} \approx 307.8 \, \text{K}\].
04

Find Molecular Mass of SO2

The molecular mass of \(\text{SO}_2\) is approximately 64.07 g/mol. Converting to kilograms for SI units, we use \(64.07 \, \text{g/mol} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} = 0.06407 \, \text{kg/mol}\).
05

Apply RMS Speed Formula

The formula for translational rms speed \(v_{rms}\) is given by \[v_{rms} = \sqrt{\frac{3RT}{M}}\], where \(M\) is the molar mass of \(\text{SO}_2\) in kg/mol. Substitute the known values:\[v_{rms} = \sqrt{\frac{3 \times 8.31 \times 307.8}{0.06407}}\].
06

Calculate RMS Speed

Perform the calculation to obtain:\[v_{rms} = \sqrt{\frac{3 \times 8.31 \times 307.8}{0.06407}} \approx 258.4 \, \text{m/s}\]. This is the translational rms speed of the \(\text{SO}_2\) molecules.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in physics and chemistry that describes the behavior of ideal gases. It's expressed as:\[ PV = nRT \]Where:
  • \( P \) is the pressure of the gas in Pascals (Pa).
  • \( V \) is the volume in cubic meters (m\(^3\)).
  • \( n \) is the number of moles of the gas.
  • \( R \) is the universal gas constant, approximately 8.31 J/(mol路K).
  • \( T \) is the temperature in Kelvin (K).
This law applies to ideal gases or, simplifying, to gases in conditions near those of an ideal scenario鈥攙ery low pressure and high temperature. It's widely used to calculate unknowns when the majority of a gas's parameters are known. For instance, if we know the pressure, volume, and moles of gas, we can solve for the temperature, as done in the example exercise with sulfur dioxide.
Root Mean Square Speed
Root Mean Square (RMS) Speed is a concept related to the kinetic energy of gas molecules. It provides an average measure of the speed of molecules in a gas, considering that different molecules can move at different speeds. The formula to calculate RMS speed is:\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]Where:
  • \( R \) is the universal gas constant, 8.31 J/(mol路K).
  • \( T \) is the temperature in Kelvin.
  • \( M \) represents the molar mass of the gas in kg/mol.
RMS speed is essential because it relates to how quickly molecules are moving. The larger the molecules, the slower their RMS speed will be at a given temperature, as mass affects the kinetic energy distribution among particles. In the example exercise, the RMS speed was calculated for sulfur dioxide molecules using their molar mass and the temperature derived from the Ideal Gas Law.
Molecular Mass
Molecular Mass, also known as molar mass, is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It's a critical factor in many calculations, such as determining the amount of a substance in chemical equations or calculating kinetic properties like RMS speed.To convert molecular mass to the units used in most physics equations (such as those involving gases), you must convert grams to kilograms:\[ \text{Molecular Mass in kg/mol} = \frac{\text{Molecular Mass in g/mol}}{1000} \]In our example, sulfur dioxide (\( \text{SO}_2 \)) has a molecular mass of approximately 64.07 g/mol. By dividing by 1000, we find its mass is 0.06407 kg/mol. This conversion is crucial when using it in the equation for RMS speed, where the molecular mass in kg is needed to maintain correct SI units throughout the calculation.

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Most popular questions from this chapter

The average value of the squared speed \(\overline{v^{2}}\) does not equal the square of the average speed \((\bar{v})^{2} .\) To verify this fact, consider three particles with the following speeds: \(v_{1}=3.0 \mathrm{m} / \mathrm{s}, v_{2}=7.0 \mathrm{m} / \mathrm{s},\) and \(\quad v_{3}=9.0 \mathrm{m} / \mathrm{s} . \quad\) Calculate (a) \(\quad \overline{v^{2}}=\frac{1}{3}\left(v_{1}^{2}+v_{2}^{2}+v_{3}^{2}\right) \quad\) and (b) \((\bar{v})^{2}=\left[\frac{1}{3}\left(v_{1}+v_{2}+v_{3}\right)\right]^{2}\)

The drawing shows an ideal gas confined to a cylinder by a massless piston that is attached to an ideal spring. Outside the cylinder is a vacuum. The crosssectional area of the piston is \(A=2.50 \times 10^{-3} \mathrm{m}^{2}\) The initial pressure, volume, and temperature of the gas are, respectively, \(P_{0}, V_{0}=6.00 \times 10^{-4} \mathrm{m}^{3},\) and \(T_{0}=273 \mathrm{K},\) and the spring is initially stretched by an amount \(x_{0}=0.0800 \mathrm{m}\) with respect to its unstrained length. The gas is heated, so that its final pressure, volume, and temperature are \(P_{\mathrm{f}}, V_{\mathrm{f}},\) and \(T_{\mathrm{f}},\) and the spring is stretched by an amount \(x_{\mathrm{f}}=0.1000 \mathrm{m}\) with respect to its unstrained length. What is the final temperature of the gas?

When a gas is diffusing through air in a diffusion channel, the diffusion rate is the number of gas atoms per second diffusing from one end of the channel to the other end. The faster the atoms move, the greater is the diffusion rate, so the diffusion rate is proportional to the rms speed of the atoms. The atomic mass of ideal gas \(\mathrm{A}\) is \(1.0 \mathrm{u}\), and that of ideal gas \(\mathrm{B}\) is \(2.0 \mathrm{u}\). For diffusion through the same channel under the same conditions, find the ratio of the diffusion rate of gas \(A\) to the diffusion rate of gas B.

Compressed air can be pumped underground into huge caverns as a form of energy storage. The volume of a cavern is \(5.6 \times 10^{5} \mathrm{m}^{3}\), and the pressure of the air in it is \(7.7 \times 10^{6} \mathrm{Pa}\). Assume that air is a diatomic ideal gas whose internal energy \(U\) is given by \(U=\frac{5}{2} n R T .\) If one home uses \(30.0 \mathrm{kW} \cdot \mathrm{h}\) of energy per day, how many homes could this internal energy serve for one day?

Insects do not have lungs as we do, nor do they breathe through their mouths. Instead, they have a system of tiny tubes, called tracheae, through which oxygen diffuses into their bodies. The tracheae begin at the surface of an insect's body and penetrate into the interior. Suppose that a trachea is \(1.9 \mathrm{mm}\) long with a cross-sectional area of \(2.1 \times 10^{-9} \mathrm{m}^{2}\) The concentration of oxygen in the air outside the insect is \(0.28 \mathrm{kg} / \mathrm{m}^{3}\), and the diffusion constant is \(1.1 \times 10^{-5} \mathrm{m}^{2} / \mathrm{s} .\) If the mass per second of oxygen diffusing through a trachea is \(1.7 \times 10^{-12} \mathrm{kg} / \mathrm{s},\) find the oxygen concentration at the interior end of the tube.
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