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At what temperature is the root-mean-square speed of nitrogen molecules equal to the root-mean-square speed of hydrogen molecules at \(20.0^{\circ} \mathrm{C}\) ? (Hint: Appendix \(\mathrm{D}\) shows the molar mass (in \(\mathrm{g} / \mathrm{mol}\) ) of each element under the chemical symbol for that element. The molar mass of \(\mathrm{H}_{2}\) is twice the molar mass of hydrogen atoms, and similarly for \(\mathrm{N}_{2}\).)

Short Answer

Expert verified
Executing these steps, we have \(T_{N_{2}} = 293.15K * \frac{2.02\times10^{-3} kg/mol}{28.01\times10^{-3} kg/mol} = 20.46 K\). Converting to Celsius gives \(20.46 K - 273.15 = -252.69^{\circ}C\). So, the root-mean-square speed of nitrogen molecules equals the root-mean-square speed of hydrogen molecules at -252.69°C.

Step by step solution

01

Identify Given Information

Given that \(v_{rmsH_{2}} = v_{rmsN_{2}}\) and the temperature \(T_{H_{2}} = 20.0^\circ C = 293.15 K\). We can also find the molar mass of \(H_{2} = 2.02 g/mol = 2.02\times10^{-3} kg/mol\) and \(N_{2} = 28.01 g/mol = 28.01\times10^{-3} kg/mol\). Remember to convert the temperature from Celsius to Kelvin and the molar mass from g/mol to kg/mol, as these are the SI units.
02

Set Up the Equation

We can form an equation by equating the root mean square speed of Nitrogen and Hydrogen gas. \(v_{rmsH_{2}} = v_{rmsN_{2}} \Rightarrow \sqrt{\frac{3kT_{H_{2}}}{m_{H_{2}}}} = \sqrt{\frac{3kT_{N_{2}}}{m_{N_{2}}}}\).
03

Simplify the Equation

Square both sides to get rid of the square root. This gives the equation as: \frac{3kT_{H_{2}}}{m_{H_{2}}} = \frac{3kT_{N_{2}}}{m_{N_{2}}}. Simplify the equation by first canceling out '3k' from both sides.
04

Solve for \(T_{N_{2}}\)

Rearrange the equation to find the temperature of Nitrogen gas: \(T_{N_{2}} = T_{H_{2}} \cdot \frac{m_{H_{2}}}{m_{N_{2}}}\). Substitute the given values into the rearranged equation and compute \(T_{N_{2}}\).
05

Convert \(T_{N_{2}}\) to \(\)°C

The calculated \(T_{N_{2}}\) will be in Kelvin, to convert it to °C, subtract 273.15 from the calculated value.

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Key Concepts

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

Kinetic Theory of Gases
The kinetic theory of gases is a model that helps us understand the behavior of gases on a molecular level. It is based on a few key assumptions:

  • Gases are made up of a large number of tiny particles called molecules which are in constant, random motion.
  • The volume of the individual gas particles is negligible compared to the volume of the container they are in.
  • Gas particles collide with each other and the container walls, but these collisions are perfectly elastic, meaning there is no net loss of kinetic energy.
  • There are no attractive or repulsive forces between the particles; their motion is only influenced by their collisions.
  • The average kinetic energy of gas particles is proportional to the absolute temperature of the gas in Kelvins.

From these principles, we derive the root-mean-square (rms) speed, which is a measure of the speed of particles in a gas. The rms speed is important because it gives us an idea of how fast the particles are moving on average. It is calculated using the formula:
\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]
where \( v_{rms} \) is the root-mean-square speed, \( k \) is the Boltzmann constant, \( T \) is the absolute temperature, and \( m \) is the mass of a gas molecule. By understanding the kinetic theory of gases, we can predict behaviors such as gas pressure, temperature, and volume under different conditions.
Gas Temperature Conversion
For scientific calculations, it's crucial to express temperature in the Kelvin scale, as it is an absolute scale based on theoretical absolute zero, the point where particles theoretically stop moving. In the context of the kinetic theory of gases, the Kelvin scale directly relates to the kinetic energy of the particles.

The Celsius scale, commonly used in daily life, requires conversion to Kelvin for physics-related computations. The following equation is used for the conversion:
\[ K = ^\circ C + 273.15 \]
deconstructed, \( K \) stands for Kelvin and \( ^\circ C \) stands for degrees Celsius. When solving gas-related problems, such as finding the root-mean-square speed, it's imperative to convert the temperature to Kelvins. Hence, in our example, the temperature of hydrogen at \(20.0^\circ C\) is converted into Kelvins as \(293.15 K\) before applying it to equations arising from the kinetic theory of gases.
Molar Mass
The molar mass of a substance is the mass of one mole of that substance. It is a fundamental concept in chemistry and physics as it relates to the stoichiometry of chemical reactions and the behavior of gases. The unit for molar mass is grams per mole (g/mol), but for calculations involving the kinetic theory of gases, it's often necessary to convert this to kilograms per mole (kg/mol) to align with the SI unit system.

In problems involving the root-mean-square speed of gases, the molar mass of the gas molecules needs to be factored in, as seen in the formula for calculating \(v_{rms}\). For example, the molar mass of hydrogen gas \(H_{2}\) is twice that of a single hydrogen atom because a hydrogen molecule consists of two hydrogen atoms. Similarly, the molar mass of nitrogen gas \(N_{2}\) is determined by doubling the atomic mass of a single nitrogen atom.

Knowing the molar masses of hydrogen and nitrogen, as in the case of the exercise comparing their root-mean-square speeds, is essential not only for solving the specific problem at hand but also for a deeper understanding of the behavior of gases in different conditions and the predictions that can be made using the kinetic theory of gases.

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Most popular questions from this chapter

You have several identical balloons. You experimentally determine that a balloon will break if its volume exceeds \(0.900 \mathrm{~L}\). The pressure of the gas inside the balloon equals air pressure \((1.00 \mathrm{~atm}) .\) (a) If the air inside the balloon is at a constant \(22.0^{\circ} \mathrm{C}\) and behaves as an ideal gas, what mass of air can you blow into one of the balloons before it bursts? (b) Repeat part (a) if the gas is helium rather than air.

Modern vacuum pumps make it easy to attain pressures of the order of \(10^{-13}\) atm in the laboratory. Consider a volume of air and treat the air as an ideal gas. (a) At a pressure of \(9.00 \times 10^{-14}\) atm and an ordinary temperature of \(300.0 \mathrm{~K}\), how many molecules are present in a volume of \(1.00 \mathrm{~cm}^{3} ?\) (b) How many molecules would be present at the same temperature but at 1.00 atm instead?

The dark area in Fig. \(\mathbf{P} 18.83\) that appears devoid of stars is a dark nebula, a cold gas cloud in interstellar space that contains enough material to block out light from the stars behind it. A typical dark nebula is about 20 light-years in diameter and contains about 50 hydrogen atoms per cubic centimeter (monatomic hydrogen, not \(\mathrm{H}_{2}\) ) at about \(20 \mathrm{~K}\). (A lightyear is the distance light travels in vacuum in one year and is equal to \(\left.9.46 \times 10^{15} \mathrm{~m} .\right)\) (a) Estimate the mean free path for a hydrogen atom in a dark nebula. The radius of a hydrogen atom is \(5.0 \times 10^{-11} \mathrm{~m}\). (b) Estimate the rms speed of a hydrogen atom and the mean free time (the average time between collisions for a given atom). Based on this result, do you think that atomic collisions, such as those leading to \(\mathrm{H}_{2} \mathrm{~mol}-\) ecule formation, are very important in determining the composition of the nebula? (c) Estimate the pressure inside a dark nebula. (d) Compare the rms speed of a hydrogen atom to the escape speed at the surface of the nebula (assumed spherical). If the space around the nebula were a vacuum, would such a cloud be stable or would it tend to evaporate? (e) The stability of dark nebulae is explained by the presence of the interstellar medium (ISM), an even thinner gas that permeates space and in which the dark nebulae are embedded. Show that for dark nebulae to be in equilibrium with the ISM, the numbers of atoms per volume \((N / V)\) and the temperatures \((T)\) of dark nebulae and the ISM must be related by $$ \frac{(N / V)_{\text {nebula }}}{(N / V)_{\text {ISM }}}=\frac{T_{\text {ISM }}}{T_{\text {nebula }}} $$ (f) In the vicinity of the sun, the ISM contains about 1 hydrogen atom per \(200 \mathrm{~cm}^{3} .\) Estimate the temperature of the ISM in the vicinity of the sun. Compare to the temperature of the sun's surface, about \(5800 \mathrm{~K}\). Would a spacecraft coasting through interstellar space burn up? Why or why not?

Three moles of an ideal gas are in a rigid cubical box with sides of length \(0.300 \mathrm{~m}\). (a) What is the force that the gas exerts on each of the six sides of the box when the gas temperature is \(20.0^{\circ} \mathrm{C} ?\) (b) What is the force when the temperature of the gas is increased to \(100.0^{\circ} \mathrm{C} ?\)

A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains \(499 \mathrm{~cm}^{3}\) of air at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{~Pa}\right)\) and a temperature of \(27.0^{\circ} \mathrm{C}\). At the end of the stroke, the air has been compressed to a volume of \(46.2 \mathrm{~cm}^{3}\) and the gauge pressure has increased to \(2.72 \times 10^{6} \mathrm{~Pa}\). Compute the final temperature.

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