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(II) (a) For an ideal gas at temperature \(T\) show that $$ \frac{d v_{\mathrm{rms}}}{d T}=\frac{1}{2} \frac{v_{\mathrm{rms}}}{T} $$ and using the approximation \(\Delta v_{\mathrm{rms}} \approx \frac{d v_{\mathrm{rms}}}{d T} \Delta T,\) show that $$ \frac{\Delta v_{\mathrm{rms}}}{v_{\mathrm{rms}}} \approx \frac{1}{2} \frac{\Delta T}{T} $$ (b) If the average air temperature changes from \(-5^{\circ} \mathrm{C}\) in winter to \(25^{\circ} \mathrm{C}\) in summer, estimate the percent change in the rms speed of air molecules between these seasons.

Short Answer

Expert verified
The percent change in rms speed of air molecules is approximately 5.3%.

Step by step solution

01

Define the rms Speed Formula

The root-mean-square (rms) speed of an ideal gas is given by the formula \( v_{\mathrm{rms}} = \sqrt{\frac{3kT}{m}} \), where \( k \) is the Boltzmann constant, \( T \) is the temperature in Kelvin, and \( m \) is the molecular mass of the gas.
02

Differentiate v_rms with Respect to T

The derivative of \( v_{\mathrm{rms}} = \sqrt{\frac{3kT}{m}} \) with respect to \( T \) can be found using the chain rule. This results in:\[ \frac{d v_{\mathrm{rms}}}{d T} = \frac{1}{2} \left( \frac{3k}{m} \right)^{1/2} T^{-1/2} \] which simplifies to \[ \frac{1}{2} \frac{v_{\mathrm{rms}}}{T} \] confirming the given equation.
03

Approximate Delta v_rms

Using the linear approximation \( \Delta v_{\mathrm{rms}} \approx \frac{d v_{\mathrm{rms}}}{d T} \Delta T \), substitute the expression for \( \frac{d v_{\mathrm{rms}}}{d T} \) to get:\[ \Delta v_{\mathrm{rms}} \approx \frac{1}{2} \frac{v_{\mathrm{rms}}}{T} \Delta T \] which simplifies to \[ \frac{\Delta v_{\mathrm{rms}}}{v_{\mathrm{rms}}} \approx \frac{1}{2} \frac{\Delta T}{T} \].
04

Convert Temperatures to Kelvin

Convert the winter and summer temperatures from Celsius to Kelvin: 1. Winter: \(-5^{\circ} C = 268\, K\)2. Summer: \(25^{\circ} C = 298\, K \).
05

Calculate Temperature Change

Subtract the winter temperature from the summer temperature to find the change in temperature: \( \Delta T = 298\, K - 268\, K = 30\, K \).
06

Estimate Percent Change in rms Speed

Use the formula \( \frac{\Delta v_{\mathrm{rms}}}{v_{\mathrm{rms}}} \approx \frac{1}{2} \frac{\Delta T}{T} \) and substitute \( \Delta T = 30\, K \) and an average \( T = 283\, K \) (midpoint of 268K and 298K): \[ \frac{\Delta v_{\mathrm{rms}}}{v_{\mathrm{rms}}} \approx \frac{1}{2} \frac{30}{283} \approx 0.053 \] giving approximately a 5.3% increase in rms speed.

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Key Concepts

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

Root-Mean-Square Speed
The root-mean-square speed, commonly abbreviated to \( v_{\text{rms}} \), is an important concept when discussing the behavior of gases. In the world of physics, the rms speed refers to the square root of the average of the squares of the speeds of all the molecules in a gas sample.
This measure gives us a meaningful average speed since it's directly related to kinetic energy. Unlike average speed, which is simply the mean of velocities, the rms speed considers the velocities squared, which is crucial for deriving accurate results.
For an ideal gas, the formula for rms speed is:
  • \[ v_{\mathrm{rms}} = \sqrt{\frac{3kT}{m}} \]
where \( k \) is the Boltzmann constant, \( T \) is the absolute temperature in Kelvin, and \( m \) is the molecular mass of the gas. This equation links microscopic particle behavior to macroscopic properties like temperature.
Temperature Change
Temperature plays a decisive role in the behavior of gases. As per the ideal gas law, when the temperature of a gas changes, so does the speed of its molecules. A rise in temperature causes the gas molecules to move more energetically, increasing their speed.
In the context of this problem, the temperature change from winter to summer affects the root-mean-square speed of air molecules. Converting temperatures from \(^\circ\)C to Kelvin is crucial because the Kelvin scale directly relates temperature to energy. This conversion accounts for the absolute measure of thermal energy:
  • Winter: \(-5^{\circ}C = 268 K\)
  • Summer: \(25^{\circ}C = 298 K\)
In this situation, \( \Delta T \), the change in temperature, is 30 K (i.e., 298 K - 268 K). This difference is essential for understanding how temperature variation can impact molecular speeds in gases.
Calculus Differentiation
Understanding how changes in temperature affect \( v_{\text{rms}} \) involves calculus differentiation. By differentiating the rms speed formula with respect to temperature, we can predict how small changes in temperature could influence molecular speed.
The derivation begins with the formula \( v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \). Applying the chain rule in differentiation allows us to find:
  • \[ \frac{d v_{\mathrm{rms}}}{d T} = \frac{1}{2} \frac{v_{\mathrm{rms}}}{T} \]
This expression tells us how the rms speed changes concerning the temperature of the gas. It shows that rms speed is proportional to the square root of temperature, a crucial point for understanding any dynamic process involving gases.
RMS Speed Approximation
To estimate practical changes in rms speed, we use an approximation method. The small changes in rms speed \( \Delta v_{\mathrm{rms}} \) in response to temperature changes can be related using a linear approximation. The expression is:
  • \[ \Delta v_{\mathrm{rms}} \approx \frac{1}{2} \frac{v_{\mathrm{rms}}}{T} \Delta T \]
This holds under the assumption that \( \Delta T \) is small relative to \( T \). From this, we find the relative change in rms speed:
  • \[ \frac{\Delta v_{\mathrm{rms}}}{v_{\mathrm{rms}}} \approx \frac{1}{2} \frac{\Delta T}{T} \]
This approximation simplifies calculations by relating changes in speed directly to proportional changes in temperature, making it easy to estimate percentages, such as a 5.3% increase in rms speed due to the temperature rise from winter to summer.

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Most popular questions from this chapter

A sample of liquid cesium is heated in an oven to \(400^{\circ} \mathrm{C}\) and the resulting vapor is used to produce an atomic beam. The volume of the oven is \(55 \mathrm{~cm}^{3},\) the vapor pressure of \(\mathrm{Cs}\) at \(400^{\circ} \mathrm{C}\) is \(17 \mathrm{~mm}-\mathrm{Hg}\), and the diameter of cesium atoms in the vapor is \(0.33 \mathrm{nm} .(a)\) Calculate the mean speed of cesium atoms in the vapor. ( \(b\) ) Determine the number of collisions a single Cs atom undergoes with other cesium atoms per second. (c) Determine the total number of collisions per second between all of the cesium atoms in the vapor. Note that a collision involves two Cs atoms and assume the ideal gas law holds.

A scuba tank has a volume of \(3100 \mathrm{~cm}^{3}\). For very deep dives, the tank is filled with \(50 \%\) (by volume) pure oxygen and \(50 \%\) pure helium. (a) How many molecules are there of each type in the tank if it is filled at \(20^{\circ} \mathrm{C}\) to a gauge pressure of \(12 \mathrm{~atm} ?(b)\) What is the ratio of the average kinetic energies of the two types of molecule? (c) What is the ratio of the rms speeds of the two types of molecule?

(a) Estimate the rms speed of an amino acid, whose molecular mass is \(89 \mathrm{u}\), in a living cell at \(37^{\circ} \mathrm{C}\). \((b)\) What would be the \(\mathrm{rms}\) speed of a protein of molecular mass \(85,000 \mathrm{u}\) at \(37^{\circ} \mathrm{C} ?\)

(I) Twelve molecules have the following speeds, given in arbitrary units: \(6.0,2.0,4.0,6.0,0.0,4.0,1.0,8.0,5.0,3.0,7.0,\) and \(8.0 .\) Calculate \((a)\) the mean speed, and \((b)\) the rms speed.

(I) Calculate the rms speed of helium atoms near the surface of the Sun at a temperature of about \(6000 \mathrm{~K}\).

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