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The most probable speed of the molecules in a gas at temperature \(T_{2}\) is equal to the rms speed of the molecules at temperature \(T_{1}\). Find \(T_{2} / T_{1}.\)

Short Answer

Expert verified
\( \frac{T_2}{T_1} = \frac{3}{2} \).

Step by step solution

01

Understand the Terms and Given Conditions

The problem provides two situations involving molecular speeds in a gas: the most probable speed at temperature \( T_2 \) and the root mean square (rms) speed at temperature \( T_1 \). These two speeds are equal.
02

Recall the Formula for the Most Probable Speed

The formula for the most probable speed \( v_p \) of molecules in a gas is given by \( v_p = \sqrt{\frac{2kT}{m}} \), where \( k \) is the Boltzmann constant, \( T \) is the temperature, and \( m \) is the mass of the molecules.
03

Recall the Formula for the Root Mean Square Speed

The formula for the root mean square (rms) speed \( v_{rms} \) is \( v_{rms} = \sqrt{\frac{3kT}{m}} \).
04

Set the Speeds Equal and Solve for Ratio

Since \( v_p = v_{rms} \) at their respective temperatures, we equate their formulas: \[ \sqrt{\frac{2kT_2}{m}} = \sqrt{\frac{3kT_1}{m}} \]. Square both sides to remove the square roots and simplify: \[ \frac{2kT_2}{m} = \frac{3kT_1}{m} \].
05

Simplify and Solve for \( \frac{T_2}{T_1} \)

Cancel out constants and masses from both sides of the equation: \( 2T_2 = 3T_1 \). Solve for \( \frac{T_2}{T_1} \): \[ \frac{T_2}{T_1} = \frac{3}{2} \].

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Key Concepts

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

Most Probable Speed
The concept of most probable speed refers to the speed at which a significant number of molecules are traveling in a gas at a given temperature. It is a key aspect of the kinetic theory of gases that allows us to predict the behavior of gas particles. Unlike average speed or root mean square speed, the most probable speed is specifically the speed that the largest number of molecules are moving at.

Mathematically, the most probable speed (\(v_p\)) can be calculated using the formula:\[v_p = \sqrt{\frac{2kT}{m}}\]where:
  • \(k\) is the Boltzmann constant
  • \(T\) is the absolute temperature of the gas
  • \(m\) is the mass of each molecule
In this formula, \(k\) is a constant that relates the average kinetic energy of particles in a gas with the temperature of the gas. As temperature increases, the most probable speed also rises, indicating faster particle movement. It is important to differentiate this from other speed measures, as it gives insights into how particles are likely distributed in reality.
Root Mean Square Speed
Root Mean Square Speed (\(v_{rms}\)) is another critical concept in the kinetic theory of gases. It provides a measure of the average speed of gas molecules, calculated as if each molecule was traveling at a constant speed at which the kinetic energy of the entire collection of molecules would remain the same.

The formula for root mean square speed is:\[v_{rms} = \sqrt{\frac{3kT}{m}}\]where:
  • \(k\) is the Boltzmann constant
  • \(T\) is the absolute temperature
  • \(m\) is the molecular mass
The rms speed is particularly useful as it aligns closely with the concept of kinetic energy. Since kinetic energy is proportional to the square of speed, averaging the square of speed helps in understanding the dynamics of a gas.

Rms speed increases with temperature, indicating that higher temperatures cause molecules to move faster, increasing their kinetic energy. It is typically larger than the most probable speed, due to being a measure of square-average as opposed to the peak of distribution.
Temperature Ratio
Understanding temperature ratio in the context of molecular speeds bridges the gap between theory and practical observations in physics. When addressing the kinetic theory, the exercise highlights two distinct temperatures: \(T_2\) for most probable speed and \(T_1\) for rms speed. These two are linked through the question of their ratio.

The exercise shows the relationship of these speeds by equalizing their mathematical expressions, leading to:\[\frac{2kT_2}{m} = \frac{3kT_1}{m}\]From this relationship, cancelling out the common factors, we're left with:\[2T_2 = 3T_1\]Solving this, the temperature ratio becomes:\[\frac{T_2}{T_1} = \frac{3}{2}\]This simple ratio illustrates how temperature impacts the speeds of molecules differently. When the most probable speed at one temperature equals the rms speed at another temperature, it highlights how these different measures of molecular speed relate to temperature changes. Through such ratios, we can discern the necessary conditions for various kinetic behaviors of gases.

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Most popular questions from this chapter

Ten particles are moving with the following speeds: four at \(200 \mathrm{~m} / \mathrm{s},\) two at \(500 \mathrm{~m} / \mathrm{s},\) and four at \(600 \mathrm{~m} / \mathrm{s} .\) Calculate their (a) average and (b) rms speeds. (c) Is \(v_{\text {rms }}>v_{\text {avg }} ?\)

(a) An ideal gas initially at pressure \(p_{0}\) undergoes a free expansion until its volume is 3.00 times its initial volume. What then is the ratio of its pressure to \(p_{0} ?\) (b) The gas is next slowly and adiabatically compressed back to its original volume. The pressure after compression is \((3.00)^{1 / 3} p_{0}\). Is the gas monatomic, diatomic, or polyatomic? (c) What is the ratio of the average kinetic energy per molecule in this final state to that in the initial state?

An ideal gas undergoes isothermal compression from an initial volume of \(4.00 \mathrm{~m}^{3}\) to a final volume of \(3.00 \mathrm{~m}^{3} .\) There is \(3.50 \mathrm{~mol}\) of the gas, and its temperature is \(10.0^{\circ} \mathrm{C}\). (a) How much work is done by the gas? (b) How much energy is transferred as heat between the gas and its environment?

We give \(70 \mathrm{~J}\) as heat to a diatomic gas, which then expands at constant pressure. The gas molecules rotate but do not oscillate. By how much does the internal energy of the gas increase?

Adiabatic wind. The normal airflow over the Rocky Mountains is west to east. The air loses much of its moisture content and is chilled as it climbs the western side of the mountains. When it descends on the eastern side, the increase in pressure toward lower altitudes causes the temperature to increase. The flow, then called a chinook wind, can rapidly raise the air temperature at the base of the mountains. Assume that the air pressure \(p\) depends on altitude \(y\) according to \(p=p_{0} \exp (-a y),\) where \(p_{0}=1.00 \mathrm{~atm}\) and \(a=1.16 \times 10^{-4} \mathrm{~m}^{-1} .\) Also assume that the ratio of the molar specific heats is \(\gamma=\frac{4}{3}\). A parcel of air with an initial temperature of \(-5.00^{\circ} \mathrm{C}\) descends adiabatically from \(y_{1}=4267 \mathrm{~m}\) to \(y=1567 \mathrm{~m} .\) What is its temperature at the end of the descent?

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