/*! 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 19 Suppose you have a gas containin... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose you have a gas containing hydrogen molecules and oxygen molecules, in thermal equilibrium. Which molecules are moving faster, on average? By what factor?

Short Answer

Expert verified
Hydrogen molecules move faster; they are 4 times faster than oxygen molecules.

Step by step solution

01

Understanding Molecular Speed

The average speed of a gas molecule is related to its mass and the temperature of the gas. According to the kinetic theory of gases, lighter molecules move faster than heavier molecules at the same temperature because they have more kinetic energy.
02

Using the Kinetic Energy Formula

The average kinetic energy of a gas molecule is given by the formula: \[ \frac{1}{2} m v^2 = \frac{3}{2} k T \] where \( m \) is the mass of the molecule, \( v \) is the speed, \( k \) is the Boltzmann constant, and \( T \) is the absolute temperature. For molecules at the same temperature, the kinetic energy is equal.
03

Expressing Speed in Terms of Mass

Rearranging the kinetic energy formula to solve for speed gives:\[ v = \sqrt{\frac{3kT}{m}} \]Since the temperature \( T \) and the Boltzmann constant \( k \) are constants for both gases, \( v \) varies inversely with the square root of the mass \( m \).
04

Calculating the Speed Ratio

The molecular mass of hydrogen (\(H_2\)) is approximately 2 amu, and for oxygen (\(O_2\)), it is approximately 32 amu. Thus, the ratio of speeds is determined by the inverse square root of their masses:\[ \frac{v_{H_2}}{v_{O_2}} = \sqrt{\frac{m_{O_2}}{m_{H_2}}} = \sqrt{\frac{32}{2}} = 4 \]
05

Conclusion

Hydrogen molecules move faster on average than oxygen molecules in thermal equilibrium. The speed of hydrogen molecules is \(4\) times the speed of oxygen molecules.

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Key Concepts

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

Molecular Speed
In the realm of gases, molecular speed refers to how quickly individual molecules zip around their container. The kinetic theory of gases gives us a deeper understanding of this concept. It tells us that molecules in a gas are in constant, random motion. The speed at which they move is influenced by two key factors:
  • The mass of the molecules
  • The temperature of the gas
Lighter molecules tend to move faster than heavier ones when they are at the same temperature. This is because lighter molecules need to travel faster to have the same kinetic energy as their heavier counterparts.
In our given example, hydrogen molecules, which are lighter than oxygen molecules, move faster in thermal equilibrium. By analyzing the mass of hydrogen ( 2 atomic mass units, or amu) and oxygen (32 amu), you see why hydrogen zips by more quickly.
Kinetic Energy
Kinetic energy is the energy of motion, and for gas molecules, it measures how much energy is in their movements. According to the formula:\[\frac{1}{2} m v^2 = \frac{3}{2} k T\]This states that the average kinetic energy of a gas molecule depends on its mass (\(m\)) and speed (\(v\)), alongside the temperature (\(T\)) and the Boltzmann constant (\(k\)).

When you're exploring gases, the fascinating aspect is that all molecules at the same temperature will have the same average kinetic energy. It's about balance—faster speeds for lighter molecules and slower speeds for heavier ones.

In our case, even though hydrogen molecules move faster, oxygen molecules have more mass, and thus, both have the same average kinetic energy. This balance is achieved because each molecule's \(v^2\) compensates for its mass to maintain equilibrium.
Thermal Equilibrium
Thermal equilibrium is a state where two or more systems interacting with each other don't exchange heat—they've reached a balanced state at the same temperature. In this state, energy is still in motion, but there is no net change in energy between systems.
This equilibrium underlies our understanding of molecular speeds and kinetic energy.

When gases like hydrogen and oxygen reach thermal equilibrium, their molecules settle into a state where their average kinetic energies are equal. However, because these gases differ in mass, their molecular speeds differ even though they're in equilibrium. This concept is essential to understanding gas behaviors because the energy distribution becomes stable across different gases.
  • Each molecule's speed is influenced by its mass and the prevailing temperature.
  • Energy is uniformly distributed across the molecules at equilibrium, though their speeds may differ.
These principles allow us to predict behaviors in various mixtures of gases—highlighting why lighter gases move faster even though they share the same kinetic energy as heavier gases.

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Most popular questions from this chapter

Even at low density, real gases don't quite obey the ideal gas law. A systematic way to account for deviations from ideal behavior is the virial expansion, $$P V=n R T\left(1+\frac{B(T)}{(V / n)}+\frac{C(T)}{(V / n)^{2}}+\cdots\right)$$ where the functions \(B(T), C(T),\) and so on are called the virial coefficients. When the density of the gas is fairly low, so that the volume per mole is large, each term in the series is much smaller than the one before. In many situations it's sufficient to omit the third term and concentrate on the second, whose coefficient \(B(T)\) is called the second virial coefficient (the first coefficient being 1 ). Here are some measured values of the second virial coefficient for nitrogen \(\left(\mathrm{N}_{2}\right):\) $$\begin{array}{cc} T(\mathrm{K}) & B\left(\mathrm{cm}^{3} / \mathrm{mol}\right) \\ \hline 100 & -160 \\ 200 & -35 \\ 300 & -4.2 \\ 400 & 9.0 \\ 500 & 16.9 \\ 600 & 21.3 \end{array}$$ (a) For each temperature in the table, compute the second term in the virial equation, \(B(T) /(V / n),\) for nitrogen at atmospheric pressure. Discuss the validity of the ideal gas law under these conditions. (b) Think about the forces between molecules, and explain why we might expect \(B(T)\) to be negative at low temperatures but positive at high temperatures. (c) Any proposed relation between \(P, V,\) and \(T,\) like the ideal gas law or the virial equation, is called an equation of state. Another famous equation of state, which is qualitatively accurate even for dense fluids, is the van der Waals equation, $$\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)=n R T$$ where \(a\) and \(b\) are constants that depend on the type of gas. Calculate the second and third virial coefficients \((B \text { and } C)\) for a gas obeying the van der Waals equation, in terms of \(a\) and \(b\). (Hint: The binomial expansion says that \((1+x)^{p} \approx 1+p x+\frac{1}{2} p(p-1) x^{2},\) provided that \(|p x| \ll 1 .\) Apply this approximation to the quantity \([1-(n b / V)]^{-1}\).) (d) Plot a graph of the van der Waals prediction for \(B(T),\) choosing \(a\) and \(b\) so as to approximately match the data given above for nitrogen. Discuss the accuracy of the van der Waals equation over this range of conditions. (The van der Waals equation is discussed much further in Section 5.3.)

A battery is connected in series to a resistor, which is immersed in water (to prepare a nice hot cup of tea). Would you classify the flow of energy from the battery to the resistor as "heat" or "work"? What about the flow of energy from the resistor to the water?

Uranium has two common isotopes, with atomic masses of 238 and \(235 .\) One way to separate these isotopes is to combine the uranium with fluorine to make uranium hexafluoride gas, UF \(_{6}\), then exploit the difference in the average thermal speeds of molecules containing the different isotopes. Calculate the rms speed of each type of molecule at room temperature, and compare them.

Calculate the mass of a mole of dry air, which is a mixture of \(\mathrm{N}_{2}\) \((78 \% \text { by volume }), \mathrm{O}_{2}(21 \%),\) and argon \((1 \%)\).

Two identical bubbles of gas form at the bottom of a lake, then rise to the surface. Because the pressure is much lower at the surface than at the bottom, both bubbles expand as they rise. However, bubble \(A\) rises very quickly, so that no heat is exchanged between it and the water. Meanwhile, bubble \(B\) rises slowly (impeded by a tangle of seaweed), so that it always remains in thermal equilibrium with the water (which has the same temperature everywhere). Which of the two bubbles is larger by the time they reach the surface? Explain your reasoning fully.

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