Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) Place the following gases in order of increasing average molecular speed at \(25^{\circ} \mathrm{C}: \mathrm{Ne}, \mathrm{HBr}, \mathrm{SO}_{2}, \mathrm{NF}_{3}, \mathrm{CO}\). (b) Calculate the rms speed of \(\mathrm{NF}_{3}\) molecules at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The order of the gases by increasing average molecular speed at 25°C is Ne > CO > NF3 > SO2 > HBr. The rms speed of NF3 molecules at 25°C is 323.2 m/s.

Step by step solution

01

Calculate molar masses

In this step, we calculate the molar masses for all given gases using the periodic table. Molar mass of Ne: 20.18 g/mol Molar mass of HBr: 80.91 g/mol (1.01 + 79.90) Molar mass of SO2: 64.07 g/mol (32.07 + 2 * 16.00) Molar mass of NF3: 71.00 g/mol (14.01 + 3 * 19.00) Molar mass of CO: 28.01 g/mol (12.01 + 16.00)
02

Order gases by average speed

Now that we have the molar masses, we can compare them to estimate the order of average molecular speed. At a given temperature, gases with lower molar mass have higher average speeds. Order: Ne > CO > NF3 > SO2 > HBr #b. Calculating the rms speed of NF3 molecules#
03

Convert temperature to Kelvin

Before we begin the calculation, we need to convert the given temperature from Celsius to Kelvin. Temperature in Kelvin: 25°C + 273.15 = 298.15 K
04

Use the rms speed formula

The rms speed can be calculated using the formula: \(v_{rms} = \sqrt{\frac{3RT}{M}}\) where: \(v_{rms}\) = rms speed R = gas constant (8.314 J/mol K) T = temperature in Kelvin (298.15 K) M = molar mass in kg/mol (converting the molar mass of NF3 from g/mol to kg/mol) M (NF3) = 71.00 g/mol * 0.001 kg/g = 0.071 kg/mol
05

Calculate the rms speed

Plug in the values into the rms speed formula: \(v_{rms} = \sqrt{\frac{3 (8.314 \,\text{J/mol K})(298.15\, \text{K})}{0.071 \,\text{kg/mol}}}\) \(v_{rms} = \sqrt{\frac{7419.6 \,\text{J/mol}}{0.071 \,\text{kg/mol}}}\) \(v_{rms} = \sqrt{104500 \,\text{m}^{2}\text{/s}^{2}}\) \(v_{rms} = 323.2\, \text{m/s}\) The rms speed of NF3 molecules at 25°C is 323.2 m/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Molar Mass
Molar mass is a fundamental concept in chemistry that represents the mass of one mole of a substance. Measured in grams per mole (g/mol), it is the sum of the atomic masses of all atoms in a molecule. When comparing different gases, the molar mass helps determine their physical properties, like molecular speed at a given temperature. For example, in our exercise, we dealt with the molar masses for several gases such as Ne, HBr, and NF3. To calculate these, you add together the atomic masses of each element in a molecule, as seen in the periodic table. Ul> Atomic mass of hydrogen: ~1.01 g/mol
Atomic mass of bromine: ~79.90 g/mol
Atomic mass of nitrogen: ~14.01 g/mol
Atomic mass of fluorine: ~19.00 g/mol By adding these up, we know that the molar mass of NF3 is 71.00 g/mol.
RMS Speed
Root Mean Square (RMS) speed is a measure of the speed of particles in a gas, which is important for understanding kinetic molecular theory. It gives insight into the average velocity of gas molecules and can be influenced by temperature and the molar mass of the gas. The formula for RMS speed is given by: \[v_{rms} = \sqrt{\frac{3RT}{M}}\] where \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(M\) is the molar mass in kg/mol. Calculating RMS speed allows scientists to predict how gas particles behave under different conditions. For our example with NF3, knowing its RMS speed of 323.2 m/s at 25°C can help determine its behavior in various chemical reactions.
Temperature Conversion
Temperature conversion is crucial when solving problems involving gas laws because the equations typically require temperature to be in Kelvin. Kelvin is the absolute temperature scale often used in scientific calculations to avoid negative temperature values that occur in Celsius. To convert Celsius to Kelvin, add 273.15 to the Celsius temperature: \[\text{Temperature in Kelvin} = \text{Temperature in Celsius} + 273.15\] This conversion is simple but vital for accuracy. For example, in the exercise, the temperature 25°C becomes 298.15 K, ensuring our calculations remain consistent with the principles of thermodynamics.
Gas Constant
The gas constant is a fundamental constant in chemistry represented as \(R\). It appears in a variety of equations relating to the behavior of gases, including the ideal gas law and RMS speed calculations. Its value is approximately 8.314 J/mol K, depending slightly on the units used. \[R = 8.314 \, \text{J/mol K}\] This constant relates energy scale to temperature and is pivotal when calculating kinetic energy or speeds of gas molecules. For instance, when calculating the RMS speed of NF3, the gas constant allows us to link the temperature (in Kelvin) with the molar mass to predict the molecular speed accurately.
Gas Laws
Gas laws are equations that describe the behavior of gases under various conditions, encompassing pressure, volume, temperature, and amount of gas. The simplest is the ideal gas law, given by \(PV = nRT\), which connects these variables using the gas constant \(R\). Understanding these laws helps predict how gases respond to changes in external conditions, crucial for fields such as chemical engineering and meteorology.
  • Boyle’s Law: Relates pressure and volume at constant temperature.
  • Charles’s Law: Involves volume and temperature at constant pressure.
  • Avogadro’s Law: Relates volume and number of moles at a constant temperature and pressure.
These laws set the groundwork for more complex calculations in gas kinetics, such as determining molecular speeds and reaction rates.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A neon sign is made of glass tubing whose inside diameter is \(2.5 \mathrm{~cm}\) and whose length is \(5.5 \mathrm{~m}\). If the sign contains neon at a pressure of \(1.78\) torr at \(35^{\circ} \mathrm{C}\), how many grams of neon are in the sign? (The volume of a cylinder is \(\pi r^{2} h .\)

A \(4.00-\mathrm{g}\) sample of a mixture of \(\mathrm{CaO}\) and \(\mathrm{BaO}\) is placed in a 1.00-L vessel containing \(\mathrm{CO}_{2}\) gas at a pressure of 730 torr and a temperature of \(25^{\circ} \mathrm{C}\). The \(\mathrm{CO}_{2}\) reacts with the \(\mathrm{CaO}\) and \(\mathrm{BaO}\), forming \(\mathrm{CaCO}_{3}\) and \(\mathrm{BaCO}_{3}\). When the reaction is complete, the pressure of the remaining \(\mathrm{CO}_{2}\) is 150 torr. (a) Calculate the number of moles of \(\mathrm{CO}_{2}\) that have reacted. (b) Calculate the mass percentage of \(\mathrm{CaO}\) in the mixture.

Does the effect of intermolecular attraction on the properties of a gas become more significant or less significant if (a) the gas is compressed to a smaller volume at constant temperature; (b) the temperature of the gas is increased at constant volume?

(a) Calculate the density of \(\mathrm{NO}_{2}\) gas at \(0.970 \mathrm{~atm}\) and \(35^{\circ} \mathrm{C}\). (b) Calculate the molar mass of a gas if \(2.50 \mathrm{~g}\) occupies \(0.875 \mathrm{~L}\) at 685 torr and \(35^{\circ} \mathrm{C}\).

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required \(105 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{Lof} \mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of \(1.0 \mathrm{~L} ;\) that is, rate and time are inversely proportional.)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free