Concept explainers
The graph here represents the distribution of molecular speeds of hydrogen and neon at 200 K.
- a Match each curve to the appropriate gas.
- b Calculate the rms speed (in m/s) for each of the gases at 200 K.
- c Which of the gases would you expect to have the greater effusion rate at 200 K? Justify your answer.
- d Calculate the temperature at which the rms speed of the hydrogen gas would equal the rms speed of the neon at 200 K.
(a)
Interpretation:
The curve in the given graph must be matched with the appropriate gas.
Answer to Problem 5.136QP
The taller and narrow curve represents neon atoms
The flatter and wider curve represents hydrogen molecules
Explanation of Solution
Given,
The given graph represents the distribution of molecular speeds of hydrogen and neon at
Matching of given curves with appropriate gases:
In the given graph, the taller and narrow curve whose maximum falls near
The flatter and wider curve whose maximum falls near
The taller and narrow curve matches with neon atoms
The flatter and wider curve matches with hydrogen molecules
(b)
Interpretation:
The rms speed (in
Concept Introduction:
Root-mean-square (rms):
The root-mean-square molecular speed (
Where,
Answer to Problem 5.136QP
The rms speed of neon gas at
The rms speed of hydrogen gas at
Explanation of Solution
Given,
The given graph represents the distribution of molecular speeds of hydrogen and neon at
Calculation of rms speed:
The rms speed of neon is calculated as follows,
The rms speed of hydrogen gas is calculated as follows,
The rms speed of neon gas at
The rms speed of hydrogen gas at
(c)
Interpretation:
The gas that has greater effusion rate at
Concept Introduction:
Graham’s law of effusion:
Answer to Problem 5.136QP
The gas that has greater effusion rate at
Explanation of Solution
Given,
The given graph represents the distribution of molecular speeds of hydrogen and neon at
Gas possessing greater effusion rate:
The rates of effusion are directly related to the rms speed.
Greater the rms speed, greater is the rate of effusion.
Since the rms speed of hydrogen is greater than neon, hydrogen gas will have greater effusion rate.
In the container, the fast moving molecules collide with the holes of the container more often and hence have a higher effusion probability.
The gas that has greater effusion rate at
(d)
Interpretation:
The temperature at which the rms speed of the hydrogen gas equals the rms speed of neon gas at
Concept Introduction:
Root-mean-square (rms):
The root-mean-square molecular speed (
Where,
Answer to Problem 5.136QP
The temperature at which the rms speed of the hydrogen gas equals the rms speed of neon gas at 200 K is
Explanation of Solution
Given,
The given graph represents the distribution of molecular speeds of hydrogen and neon at
Temperature calculation:
The temperature equaling the rms speed of neon is calculated from root mean square equation as follows,
The temperature at which the rms speed of the hydrogen gas equals the rms speed of neon gas at
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