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Chapter 12: Q53P (page 472)

Figure 12.59 shows the distribution of speeds of atoms in a particular gas at a particular temperature. Approximately what is the average speed? Is the RMS (root-mean-square) speed bigger or smaller than this? Approximately what fraction of the molecules have speeds greater than 1000 m/s?

Short Answer

Expert verified
  • The average speed of an atom isν-=600m/s.
  • The RMS speed is bigger than the average speed.
  • The fraction of the molecules has a speed greater than 1000 m/s is roughly 110.

Step by step solution

01

Identification of given data

  • The graph between v(m/s) and f(v) .
02

Concept of root mean square speed 

Molecules in gas travel at an average velocity-squared, which is defined as the square root of the velocity-squared squared for each molecule.

03

Determination of root mean square speed of gas

The average speed of gas from the graph is .

The RMS speed is always larger than the average speed.

The root mean square speed is given by,

Vrms=3RTM (i)

The average speed of a gas is given by,

Vav=8RTπM (ii)

Comparing equation (ii) with equation (i), we get,

2=8π144:1.595

Thus, we can conclude that the RMS value is always bigger than the average speed.

From looking at the graph, we can conclude that the fraction of the molecules has a speed greater than 1000m/s is approximate 110.

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Most popular questions from this chapter

In Chapter 4 you determined the stiffness of the interatomic “spring” (chemical bond) between atoms in a block of lead to be 5 N/m, based on the value of Young’s modulus for lead. Since in our model each atom is connected to two springs, each half the length of the interatomic bond, the effective “interatomic spring stiffness” for an oscillator is

4 × 5 N/m = 20 N/m. The mass of one mole of lead is 207 g (0.207 kg). What is the energy, in joules, of one quantum of energy for an atomic oscillator in a block of lead?

List explicitly all the ways to arrange 2 quanta among 4 one-dimensional oscillators.

This question follows the entire chain of reasoning involved in determining the specific heat of an Einstein solid. Start with two metal blocks, one consisting of one mole of aluminum (27 g) and the other of one mole of lead (207 g), both initially at a temperature very near absolute zero (0 K). From measurements of Young’s modulus one finds that the effective stiffness of the interatomic bond modeled as a spring is 16N/mfor aluminum and 5 N/m for lead. (a) Is the number of quantized oscillators in the aluminum block greater, smaller, or the same as the number in the lead block? (b) What is the initial entropy of each block? (c) In which metal is the energy spacing of the quantized harmonic oscillators larger? (d) If we add 1 J of energy to each block, which metal now has the larger number of energy quanta? (e) In which block is the number of possible ways of arranging this of energy greater? (f) Which block now has the larger entropy? (g) Which block experienced a greater entropy change? (h) Which block experienced the larger temperature change? (i) Which metal has the larger specific heat at low temperatures? (j) Does your conclusion agree with the actual data given in Figure 12.33? (The numerical data are given in a table accompanying Problem P64.)

The entropy S of a certain object (not an Einstein solid) is the following function of the internal energy E:S=bE1/2, where b is a constant. (a) Determine the internal energy of this object as a function of the temperature.

(b) What is the specific heat of this object as a function of the temperature?

The reasoning developed for counting microstates applies to many other situations involving probability. For example, if you flip a coin 5 times, how many different sequences of 3 heads and 2 tails are possible? Answer: 10 different sequences, such as HTHHT or TTHHH. In contrast, how many different sequences of 5 heads and 0 tails are possible? Obviously only one, HHHHH, and our equation gives , using the standard definition thatis defined to equal 1.

If the coin is equally likely on a single throw to come up heads or tails, any specific sequence like HTHHT or HHHHH is equally likely. However, there is only one way to get HHHHH, while there are 10 ways to get 3 heads and 2 tails, so this is 10 times more probable than getting all heads.

Use the expression to calculate the number of ways to get 0 heads, 1 head, 2 heads, 3 heads, 4 heads, or 5 heads in a sequence of 5 coin tosses. Make a graph of the number of ways vs. the number of heads.
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